Electrical Technology

                                     CONTENTS

 1. Electric Current and Ohm's Law 

 Electron Drift Velocity-Charge Velocity and Velocity of Field Propagation-The Idea of Electric Potential-Resistance-Unit of Resistance-Law of Resistance-Units of Resistivity-Conductance and Conductivity-Effect of Temperature on Resistance-Temperature Coefficient of Resistance-Value of a at Different Temperatures-Variation of Resistivity with Temperature-Ohm's Law-Resistance in Series-Voltage Divider Rule-Resistance in ParalleJ-Types of Resistors-Nonlinear Resistors-Varistor-Short and Open Circuits- 'Shorts' in a Series Circuit-'Opens' in Series Circuit-'Open's in a Parallel Circuit- 'Shorts' in Parallel Circuits-Division of Current in Parallel Circuits-Equivalent Resistance-Duality Between Series and Parallel Circuits-Relative Potential-Voltage Divider Circuits-Objective Tests. 

2. DC Network Theorems 

 Electric Circuits and Network Theorems-Kirchhoff's Laws-Determination of Voltage Sign-Assumed Direction f Current-Solving Simultaneous Equations-DeterminantsSolving Equations with Two Unknowns-Solving Equations With Three Unknowns-Independent and Dependent Sources -Maxwell's Loop Current Method-Mesh Analysis Using Matrix Form-Nodal Analysis with Voltage Sources-Nodal Analysis with Current SourcesSource Conversion-Ideal Constant-Voltage Source-Ideal Constant-Current Source-Superposition Theorem- Thevenin Theorem-How to Thevenize a Given Circuit ?-General Instructions for Finding Thevenin Equivalent Circuit-Reciprocity Theorem-Delta/Star Transformation-StarlDelta Transformation-Compensation Theorem-Norton' s TheoremHow to Nortanize a Given Circuit ?-General Instructions for Finding Norton Equivalent Circuit-Millman's Theorem-Generalised Form of Millman's Theorem-Maximum Power Transfer Theorem-Power Transfer Efficiency-Objective Tests. 

3. Work, Power and Energy n. 168-179 Effect of Electric Current-Joule's Law of Electric Heating-Thermal Efficiency-S-I. Units-Calculation of Kilo-watt Power of a Hydroelectric Station-Objective Tests. 

4. Electrostatics 180-20 Static Electricity-Absolute and Relative Permittivity of a Medium-Laws of Electrostatics-Electric Field-Electrostatic Induction-Electric Flux and Faraday Tubes-Field Strength or Field Intensity or Electric Intensity (E)-Electric Flux Density or Electric Displacement D-Gauss Law-The Equations of Poisson and Laplace-Electric Potential and Energy-Potential and Potential Difference-Potential at a Point-Potential of a Charged Sphere-Equipotential Surfaces-Potential and Electric Intensity Inside a Conducting SpherePotential Gradient-Breakdown Voltage and Dielectric Strength-Safety Factor of Dielectric-Boundary Conditions-Objective Tests. 

5. Capacitance ... 201-240 Capacitor-Capacitance-Capacitance of an Isolated Sphere-Spherical Capacitor - Parallel-plate Capacitor-Special Cases of Parallel-plate Capacitor-Multiple and Variable (ix) Capacitors-Cylindrical Capacitor-Potential Gradient in Cylindrical Capacitor-Capacitance Between two Parallel Wires-Capacitors in Series-Capacitors in Parallel-Cylindrical Capacitor with Compound Dielectric-Insulation Resistance of a Cable CapacitorEnergy Stored in a Capacitor-Force of Attraction Between Oppositely-charged PlatesCurrent-Voltage Relationships in a Capacitor-Charging of a Capacitor-Time ConstantDischarging of a Capacitor-Transient Relations during Capacitor Charging Cycle-Transient Relations during Capacitor Discharging Cycle-Charging and Discharging of a Capacitor with Initial Charge-Objective Tests. 6. Magnetism and Electromagnetism ... 241-278 Absolute and Relative Permeabilities of a Medium-Laws of Magnetic Force-Magnetic Field Strength (H)-Magnetic Potential-Flux per Unit Pole-Flux Density (B)-Absolute Parmeability 0.1)and Relative Permeability (J.l)-Intensity of Magnetisation (I)-Susceptibility (K)-Relation Between B, H, I and K-Boundary Conditions-Weber and Ewing's Molecular Theory-Curie Point. Force on a Current-carrying Conductor Lying in a Magnetic Field-Ampere's Work Law or Ampere's Circuital Law-Biot-Savart Law-Application of Biot-Savart Law-Force Between two Parallel Conductors-Magnitude of Mutual ForceDefinition of Ampere-Magnetic Circuit-Definitions-Composite Series Magnetic Circuit-How to Find Ampere-turns ?-Comparison Between Magnetic and Electric CircuitsParallel Magnetic Circuits-Series-Parallel Magnetic Circuits-Leakage Flux and Hopkinson's Leakage Coefficient-Magnetisation Curves-Magnetisation curves by Ballistic Galvanometer-Magnetisation Curves by FIuxmete-Objective Tests. 7. Electromagnetic Induction ... 279-297 Relation Between Magnetism and Electricity-Production of Induced E.M.F. and CurrentFaraday's Laws of Electromagnetic Induction-Direction of Induced E.M.F. and CurrentLenz's Law-Induced E.M.F.-Dynamically-induced E.M.F.-Statically-induced E.M.F.- Self-Inductance-Coefficient of Self-Inductance (L)-Mutual Inductance-Coefficient of Mutual Inductance (M)-Coefficient of Coupling-Inductances in Series-Inductances in Parallel-Objective Tests. 8. Magnetic Hysteresis ... 298-318 Magnetic Hysteresis_Area of Hysteresis Loop-Properties and Application of Ferromagnetic Materials-Permanent Magnet Materials-Steinmetz Hysteresis Law-Energy Stored in Magnetic Field-Rate of Change of Stored Energy-Energy Stored per Unit VolumeLifting Power of Magnet-Rise of Current in Inductive Circuit-Decay of Current in Inductive Circuit-Details of Transient Current Rise in R-L Circuit-Details of Transient Current Decay in R-L Circuit-Automobile Ignition System-Objective Tests. 9. Electrochemical Power Sources ... 319-351 Faraday's Laws of electrolysis-Polarisation or Back e.m.f.-Value of Back e.m.f.-:- Primary and Secondary Batteries-Classification of Secondary Batteries base on their UseClassification of Lead Storage Batteries-Parts of a Lead-acid Battery-Active Materials of Lead-acid Cells-Chemical. Changes-Formation of Plates of Lead-acid Cells-Plante Process-Structure of Plante Plates-Faure Process-Positive Pasted Plates-Negative (x) Pasted Plates-Structure of Faure Plates-Comparison : Plante and Faure Plates-Internal Resistance and Capacity of a Cell-Two Efficiencies of the Cell-Electrical Characteristics of the Lead-acid Cell-Battery Ratings-Indications of a Fully-Charged Cell-Application of Lead-acid Batteries-Voltage Regulators-End-cell Control System-Number of Endcells-Charging Systems-Constant-current System-Constant-voltage System-Trickle Charging-Sulphation-Causes and Cure-Maintenance of Lead-acid Cells-Mains operated Battery Chargers-Car Battery Charger-Automobile Battery Charger-Static Uninterruptable Power Systems-Alkaline Batteries-Nickel-iron or Edison Batteries-Chemical ChangesElectrical Characteristics-Nickel-Cadmium Batteries-Chemical Changes--C;omparison : Lead-a~id and Edison Cells-Silver-zinc Batteries-High Temperature Batteries-Secondary Hybrid Cells-Fuel Cells-Hydrogen-Oxygen Fuel Cells-Batteries for AircraftBatteries for Submarines-Objective Tests. 10. Electrical Instruments and Measurements .., 352-425 Absolute and Secondary Instruments-Electrical Principles of Operation-Essentials of Indicating Instruments-Deflecting Torque-Controlling Torque-Damping Torque-Moving-iron Ammeters and Voltmeters-Attraction Type-Repulsion Type-Sources of ErrorAdvantages and Disadvantages-Deflecting Torque in terms of Change in Self-inductionExtension of Range by Shunts and Multipliers-Moving-coil Instruments-Permanent Magnet Type Instruments-Advantages and Disadvantages-Extension of Range-Voltmeter Sensitivity-Multi-range Voltmeter-Electrodynamic or Dynamometer Type InstrumentsHot-wire Instruments- Magnification of the Expansion-Thermocouple Ammeter-MeggerInduction type Voltmeters and Ammeters-Induction Ammeters-Induction VoltmeterErrors in Induction Instruments-Advantages and Disadvantages--Electrostatic Voltmeters-Attracted-disc Type Voltmeter-Quardant Type Voltmeter-Kelvin's Multicellular Voltmeter-Advantages and Limitation of Electrostatic Voltmeters-Range Extension of Electrostatic Voltmeters-W attmeters-Dynamometer Wattmeter-Wattmeter Errors-Induction. Wattmeters-Advantages and Limitations ofInduction Wattmeters-Energy MetersElectrolytic Meter-Motor Meters-Errors in Motor Meters-Quantity or Ampere-hour Meters-Ampere-hour Me~ury Motor Meter-Friction Compensation-Mercury Meter Modified as Watt-hour Meter-Commutator Motor Meters-Induction Type Single-phase Watthour Meter-Errors in Induction Watthour Meters-Ballistic Galvanometer-Vibration Galvanometer-Vibrating-reed Frequency Meter-Electrodynamic Frequency Meter-Movingiron Frequency Meter-Electrodynamic Power Factor Meter-Moving-;iron Power Factor Meter-Nalder-Lipman Moving-iron Power Factor Meter-D.C. Potentiometer-Direct Reading Potentiometer-Standardising the Potentiometer-Calibration of Ammeters-Calibration of Voltmeters-A.C. Potentiometers-Drysdale Potentiometer-Gall Co-ordinate Potentiometer-Instrument Transformers-Ratio and Phase-angle Errors-Current Transformers-Theory of Current Transformer-Clip-on Type Current Transformer-Potential Transformers. Objective Tests. 11. A.C. Fundamentals ... 426-465 Generation of Alternating Voltages and Currents-Equations of the Alternating VoJtages and Currents-Alternate Method for the Equations of Alternating Voltages and currentsSimple Waveforms-Complex Waveforms-Cycle- Time-Period-Frequency-AmplitudeDifferent Forms of E.M.F. Equation-Phase-Phase Difference-Root Mean Square (R.M.S.) (xi) Value-Mid-ordinate Method-Analytical Method-R.M.S. Value of a Complex WaveAverage Value-Form Factor-Crest or Peak Factor-R.M.S. Value of H.W. Rectified A.C.-Average Value-Form Factor of H.W. Rectified -Representation of Alternating Quantities-Vector Diagrams Using R.M.S. Values-Vector Diagrams of Sine Waves of Same Frequency-Addition of Two Alternating Quantities-Addition and Subtraction of Vectors-A.C. Through Resistance, Inductance and Capacitance-A.C. through Pure Ohmic Resistance alone-A.C. through Pure Inductance alone-Complex Voltage Applied to Pure Inductance-A.C. through Capacitance alone Objective Tests. 12. Complex Numbers ... 466-474 Mathematical Representation of Vectors-Symbolic Notation-Significance of Operator jConjugate Complex Numbers-Trigonometrical Form. of Vector-Exponential Form of Vector-Polar Form of Vector Representation-Addition and Subtraction of Vector Quantities-Multiplication and Division of Vector Quantities-Power and Root of Vectors-The 120° Operator-Objective Tests. . 13. Series A.C. Circuits ... 475-522 A.C. through Resistance and Inductance-Power Factor-Active and Reactive Components of Circuit Current-I-Active, Reactive and Apparent Power-Q-factor of a Coil-Power in an Iron-cored Chocking Coil-A.C. Through Resistance and Capacitance-Dielectric Loss and Power Factor of a Capacitor-Resistance, Inductance and Capacitance in Series-Resonance in R-L-C Circuits-Graphical Representation of Resonance-Resonance Curve-Half-power Bandwidth of a Resonant Circuit-Bandwidth B at any Off-resonance Frequency-Determination of Upper and Lower Half-Power Frequencies-Values of Edge Frequencies-Q-Factor of a Resonant Series Circuit-Circuit Current at Frequencies Other than Resonant Frequencies-Relation Between Resonant Power Po and Off-resonant Power P-Objective Test. 14. Parallel A.C. Circuits ... 523-562 Solving Parallel Circuits-Vector or Phasor Method-Admittance Method-Application of Admittance Method-Complex or Phasor Algebra-Series-Parallel Circuits-Series Equivalent of a Parallel Circuit-Parallel Equaivalent of a Series Circuit-Resonance in Parallel Circuits-Graphic Representation of Parallel Resonance-Points to Remember;-Bandwidth of a Parallel Resonant Circuit-Q-factor of a Parallel Circuit-Objective Tests. 15. A.C. Network Analysis 563-586 Introduction-Kirchhoffs Laws-Mesh Analysis-Nodal Analysis-Superposition Theorem-Thevenin's Theorem-Reciprocity Theorem-Norton's Theorem-Maximum Power Transfer Theorem-MiJlman's Theorem. l6. A.C. Bridges ... 587-598 A.C. Bridges-Maxwell's Inductance Bridge-Maxwell-Wien Bridge-Anderson BridgeHay's Bridge-The Owen Bridge-Heaviside Compbell Equal Ratio Bridge-Capacitance Bridge-De Sauty Bridge-Schering Bridge-Wien Series Bridge-Wien Parallel BridgeObjective Tests. (xii) 17. A.C. Filter Networks ... 599-610 Introduction-Applications-Different Types of Filters-Octaves and Decades of frequency-Decible System-Value of 1 dB-Low-Pass RC Filter-Other Types of Low-Pass Filters-Low-Pass RL Filter-High-Pass R C Filter-High Pass R L Filter-R-C Bandpass Filter-R-C Bandstop Filter-The-3 dB Frequencies-Roll-off of the Response CurveBandstop and Bandpass Resonant Filter Circuits-Series-and Parallel-Resonant Bandstop Filters-Parallel-Resonant Bandstop Filter-Series-Resonant Bandpass Filter-Parallel-Resonant Bandpass Filter-Objective Test. 18. Circle Diagrams ... 611-618 Circle Diagram of a Series Circuit-Rigorous Mathematical Treatment-Constant Resistance but Variable Reactance-Properties of Constant Reactance But Variable Resistance Circuit-Simple Transmission Line Circuit. 19. Polyphase Circuits ... 619-702 Generation of Polyphase Voltages-Phase Sequence-Phases Sequence At Load-Numbering of Phases-Interconnection of Three Phases-Star or Wye (Y) Connection-Values of Phase Currents-Voltages and Currents in Y-Connection-Delta (A) or Mesh ConnectionBalanced Y/A and AIY Conversions-Star and Delta Connected Lighting Loads-Power Factor Improvement-Power Correction Equipment-Parallel" Loads-Power Measurement in 3-phase Circuits-Three Wattmeter Method-Two Wattmeter Method-Balanced or Unbalanced load-Two Wattmeter Method-Balanced Load-Variations in Wattmeter Readings-Leading Power Factor-Power Factor-Balanced Load-Balanced Load-LPF-Reactive Voltamperes with One Wattmeter-One Wattmeter Method-Copper Required for Transmitting Power Under Fixed Conditions-Double Subscript Notation-Unbalance~ Loads-Unbalanced A-connected Load-Four-wire Star-connected Unbalanced Load-Unbalanced Y-connected Load Without Neutral-Millman' s Thereom-Application of Kirchhoff's Laws-Delta/Star and StarlDelta Conversions-Unbalanced Star-connected Non-inductive Load-Phase Sequence Indicators-Objective Tests. 20. Harmonics ... 703-724 Fundamental Wave and Harmonics-Different Complex Waveforms-General Equation of a Complex Wave-R.M.S. Value of a Complex Wave-Form Factor of a Copmplex Wave- . Power Supplied by a Complex Wave-Harmonics in Single-phase A.C Circuits-Selective Resonance Due to Harmonics-Effect of Harmonics on Measurement of Inductance and Capacitance-Harmonics in Different Three-phase Systems-Harmonics in Single and 3- Phase Transformers-Objective Tests. 21. Fourier Series ... 725-757 Harmonic Analysis-Periodic Functions-Trigonometric Fourier Series-Alternate Forms of Trigonometric Fourier Series-Certain Useful Integral Calculus Theorems-Evalulation of Fourier Constants-Different Types of Functional Symmetries-Line or Frequency Spectrum-Procedure for Finding the Fourier Series of a Given Function-Wave AnalyzerSpectrum Analyzer-Fourier Analyzer-Harmonic Synthesis-Objective Tests. (xiii) 22. Transients ... 758-774 Introduction-Types of Transients-Important Differential Equations-Transients in R-L Circuits (D.C.),-Short Circuit Current-Time Constant-Transients in R-L Circuits (A.C.)-Transients in R-C Series Circuits (D.C.)-Transients in R-C Series Circuits (A.C)- Double Energy Transients-Objective Tests. 23. Symmetrical Components ... 775-791 Introduction-The Positive-sequence Components-The Negative-sequence ComponentsThe Zero-sequence Components-Graphical Composition of Sequence Vectors-Evaluation of VAlor VI-Evaluation of VA2or V2-Evaluation VAOor vo-Zero Sequence Components of Current and Voltage-Unbalanced Star Load form Unbalanced Three-phase Three-Wire System-Unbalanced Star Load Supplied from Balanced Three-phase Three-wire SystemMeasurement of Symmetrical Components of Circuits-Measurement of Positive and Negativesequence Voltages-Measurement of Zero-sequence Component of Voltage---Objective Tests. 24. Introduction to Electrical Energy Generation ... 792-800 Preference for Electricity-Comparison of Sources of Power-Sources for Generation of Electricity-Brief Aspects of Electrical Energy Systems-Utility and Consumers-Why is the Three-phase a.c. system Most Popular?-Cost of Generation-Staggering of Loads during peak-demand Hours-Classifications of Power Transmission-Selecting A.C. Transmission Voltage for a Particular Case-Conventional Sources of Electrical Energy-Steam Power Stations (Coal-fired)-Nuclear Power Stations-Advantages of Nuclear Generation-Disadvantages-Hydroelectric Generation-Non-Conventional Energy Sources-Photo Voltaic Cells (P.V. Cells or SOLAR Cells)-Fuel Cells-Principle of Operation-Chemical Process (with Acidic Electrolyte)-Schematic Diagram-Array for Large outputs-High LightsWind Power-Background-Basic Scheme-Indian Scenario. 1 . ELECTRIC CURRENT AND OHM'S LAW 1.1. Electron Drift Velocity Suppose that in a conductor, the number of free electrons available per m3 of the conductor material is n and let their axial drift velocity be v metres/second. In time dt, distance travelled would be v x dt. If A.is area of cross-section of the conductor, then the volume is vAdt and the number of electrons contained in this volume is vA dt. Obviously, all these electrons will cross the conductor cross-section in time dt. If e is the charge of each electron, then total charge which crosses the section in time dt is dq =nAev dt. Since current is the rate of flow of charge, it is given as = - = :. dq nAev dt . I =nAev dt dt Current density J = ilA =ne v amperelmetre2 Assuming a normal current density J =1.55 X 106 Alm2, n = 1029for a copper conductor 6 -19 and e =1. x 10 coulomb, we get 1.55 x 106 = 1029x 1.6 X 10-19x v :. v = 9.7 X 10-5m/s = 0.58 cm/min It is seen that contrary to the common but mistaken view, the electron drift velocity is rather very slow and is independent of the current flowing and the area of the conductor. N.H.Current density i.e., the current per unit area. is a vector quantity. It is denoted by the symbol J . --> Therefore. in vector notation, the relationship between current I and J is: -->--> --> I = J. a [where a is the vector notation for area 'a'] For extending the scope of the above relationship. so that it becomes applicable for area of any shape, we write: f --> --> J = J .d a The magnitudeof the currentdensitycan. therefore.be writtenas J.a. Example 1.1. A conductor material has a free-electron density of J{j4 electrons per metrl. When a voltage is applied, a constant drift velocity of 1.5 X ]0-2 metre/second is attained by the electrons. If the cross-sectional area of the material is 1 cm2,calculate the magnitude of the current. Electronic charge is 1.6 x ]0-19coulomb. (Electrical Engg. Aligarh Muslim University 1981) Solution. The magnitude of the current is = nAev amperes n = 1024;A = 1cm2 = 10-4m2 e = 1.6 x 10-19C ; v = 1.5 X 10-2m/s = 1024X 10-4x 1.6 X 10-19x 1.5 X 10-2=0.24 A Here, .. 1.2. Charge Velocity and Velocity of Field Propagation The speed with which charge drifts in a conductor is called the velocity of charge. As seen from 1 2 Electrical Technology above. its value is quite low, typically fraction of a metre per second. However. the speed with which the effect of e.m.f. is experienced at all parts of the conductor resulting in the flow of current is called the velocitv of propagation of electricalfield. It is indepen- . 8 dent of current and voltage and has high but constant value of nearly 3 x 10 mls. Example 1.2. Find the velocity of charge leading to 1 A current which flows in a copper conductor of cross-section 1 em2and length 10 km. Free electron dellSityof copper =8.5.x 1028per m3. How long will it take the electric charge to travelfrom one end of the conductor to the other. Solution. i =neAv or v =ih,('A .. v = 1/(85 y I02X) x 1.6 x 10 1'1X (I X 10-4) = 7.35 X 10-7 mls =0.735 Jlmls Time taken by the charge 10lravd conductor length of 10 kIn is t = distance _ 10x 103 =1.36 x 1010S velocity 7.35 x 10-7 Now. I year::: 365 x 24 x 3600 = 31,536.000 s 10 t = 1.36 x 10 /31.536.000 = 431 years 1.3. The Idea of Electric Potential In Fig. 1.1is shown a simple voltaic cell. It consists of copper plate (known as anode) and a zinc rod (I.e. cathode) immersed in dilute sulphuric acid (H2S04) contained in a suitable vessel. The chemical action taking place within the cell causes the electrons to be removed from Cu plate and to be deposited on the zinc rod at the same time. This transfer of electrons is accomplished through the agency of the diluted H2S04 which is known as the electrolyte. The result is that zinc rod becomes negative due to the deposition of electrons on it and the Cu plate becomes positive due to the removal of electrons from it. The large number of electrons collected on the zinc rod is being attracted by anode but is prevented from returning to it by the force set up by the chemical action within the cell. Conventional Direction of Current --0 --0 --0. Directionof '" ElectronFlow . --- 0- '" 1 1 Water ,'-:' -- - --- --- -- - -- - - -- - - - - :-, r Direction of Flow 1 Cu Zn Fig. 1.1. Fig. 1.2 But if the two electrodes are joined by a wire externally. then electrons rush to the anode thereby equalizing the charges of the two electrodes. However. due to the continuity of chemical action. a continuous difference in the number of electrons on the two electrodes is maintained which keeps up a continuous flow of current through the external circuit. The action of an electric cell is similar to that of a water pump which. while working. maintains a continuous flow of water i.e. water current through the pipe (Fig. 1.2). Electric Current and Ohm's Law 3 It should be particularly noted that the direction of electronic current is from zinc to copper in the external circuit. However, the direction of conventional current (which is given by the direction of flow of positive charge) is from Cu to zinc. In the present case, there is no flow of positive charge as such from one electrode to another. But we can look upon the arrival of electrons on copper plate (with subsequent decrease in its positive charge) as equivalent to an actual departure of positive charge from it. When zinc is negatively charged, it is said to be at negativepotential with respect to theelectrolyte, whereas anode is said to be at positive potential relative to the electrolyte. Between themselves, Cu plateis assumed to be at a higher potential than the zinc rod. The differencein potential is continuously maintainedby the chemical action going on in the cell which supplies energy to establish this potential difference. 1.4. Resistance It may be defmed as the property of a substance due to which it opposes (or restricts) the flow of electricity (i.e., electrons) through it. Metals (as a class), acids and salts solutions are good conductors of electricity. Amongst pure metals,silver, copper and aluminium are very good conductors in the given order.* This, as discussed earlier, is due to the presence of a large number of free or loosely-attached electrons in their atoms. Thesevagrantelectrons assume a directed motion on the application of an electric potential difference. These electrons while flowing pass through the molecules or the atoms of the conductor, collide and other atoms and electrons, thereby producing heat. Those substances which offer relatively greater difficulty or hindrance to the passage of these electrons are said to be relatively poor conductors of electricity like bakelite, mica, glass, rubber, p.v.c. (polyvinyl chloride) and dry wood etc. Amongst good insulators can be included fibrous substances such as paper and cotton when dry, mineral oils free from acids and water, ceramics like hard porcelain and asbestos and many other plastics besides p.v.c. It is helpful to remember that electric friction is similar to friction in Mechanics. 1.5. The Unit of Resistance The practical unit of resistance is ohm.** A conductor is said to have a resistance of one ohm if it permits one ampere current to flow through it when one volt is impressed across its terminals. For insulators whose resistances are very high, a much bigger unit is used i.e. megaohm =106 ohm (the prefix 'mega' or mego meaning a million) or kilohm = 103ohm (kilo means thousand). In the case of very small resistances, smaller units like milli-ohm = 10-3 ohm or microhm = 10-6 ohm are used. The symbol for ohm is Q. Table 1.1. Multiples and Sub-multiples of Ohm However. for the same resistance per unit length, cross-sectional area of aluminium conductor has to be 1.6 times that of the copper conductor but it weighs only half as much. Hence, it is used where economy of weight is more important than economy of space. ** After George Simon Ohm (1787-1854), a German mathematician who in about 1827 formulated the law of known after his name as Ohm's Law. Prefix Its meaning Abbreviation Equal to Mega- One million MO 1060 Kilo- One thousand kO 1030 Centi- One hundredth - - MiIIi- One thousandth mO 10-30 Micro- One millionth flO 10-<>0 4 Electrical Technology 1.6. Laws of Resistance The resistance R offered by a conductor depends on the following factors : (i) It varies directly as its length, I. (ii) It varies inversely as the cross-section A of the conductor. (iii) It depends on the nature of the material. (iv) It also depends on the temperature of the conductor. Smaller I Larger A Low R Larger I Smaller A Greater R Fig. 1.3. Fig. 1.4 Neglecting the last factor for the time being, we can say that R oc ~ or R = P ~ ...(i) where p is a constant depending on the nature of the material of the conductor and is known as its specific resistance or resistivity. If in Eq. (i), we put I = I metre and A =I metre2,then R =P (Fig. 1.4) Hence, specific resistance of a material may be defined as the resistance between the oppositefaces of a metre cube of that material. 1.7. Units of Resistivity From Eq. (i), we have p = ARI In the S.I. system of units, A metre2 x R ohm = AR ohm-metre p = I metre 1 Hence, the unit of resistivity is ohm-metre (Q-m). It may, however, be noted that resistivity is sometimes expressed as so many ohm per m3. Although, it is incorrect to say so but it means the same thing as ohm-metre. If I is in centimetres and A in cm2,then p is in ohm-centimetre (Q-cm). Values of resistivity and temperature coefficients for various materials are given in Table 1.2. The resistivities of commercial materials may differ by several per cent due to impurities etc. Electric Current and Ohm's Law 5 Table 1.2. Resistivities and Temperature Coefficients Example 1.3. A coil consists of 2000 turns of copper wire having a cross-sectional area of 0.8 J mm-. The mean length per turn is 80 em and the resistivity of copper is 0.02 J1Q~m. Find the resistance of the coil and power absorbed by the coil when connected across 110 V d.c. supply. (F.Y. Engg. Pone Univ. May 1990) . Solution. Length of the coil, I =0.8 x 2000 =1600 m ; A =0.8 mm2=0.8 x 10-6m2. R = P~ =0.02X 10-6x 1600/0.8X 10-6=40 Q Power absorbed = V/ R =1102/40=302.5 W Example 1.4. An aluminium wire 7.5 m long is connected in a parallel with a copper wire 6 m long. When a current of 5 A is passed through the combination, it isfound that the current in the aluminium wire is 3 A. The diameter of the aluminium wire is 1 mm. Determine the diameter of the copper wire. Resistivity of copper is 0.017 J1Q-m..that of the aluminium is 0.028 J1Q-m. (F.Y. Engg. Pone Univ. May 1991) Solution. Let the subscript I represent aluminium and subscript 2 represent copper. I I R P I a R I = pi and R2=P2 --1.. :. ...1.=...1.. .1.. -L al a2 RI PI II a2 RI P2. 12 . .. a2 = al.R. -'T ...(1) 2 PI 1 Now /1 = 3 A ; /2 =5 - 3=2 A. If Vis thecommonvoltageacrosstheparallelcombinationof aluminiumandcopperwires,then Material Resistivity in ohm-metre Temperature coefficient at at 20.C (x 10-JI) 20.C (x 10-4) Aluminium, commercial 2.8 40.3 Brass 6-8 20 Carbon 3000 - 7000 -5 Constantan or Eureka 49 +0.1 to-o.4 Copper (annealed) 1.72 39.3 German Silver 20.2 2.7 (84% Cu; 12% Ni; 4% Zn) Gold 2.44 36.5 Iron 9.8 65 Manganin 44 - 48 0.15 (84% Cu; 12% Mn; 4% Ni) Mercury 95.8 8.9 Nichrome 108.5 1.5 (60% Cu ; 25% Fe ; 15% Cr) Nickel 7.8 54 Platinum 9 - 15.5 36.7 Silver 1.64 38 Tungsten 5.5 47 Amber 5 x 1014 Bakelite 1010 Glass 1010 _ 1012 Mica 1015 Rubber 1016 Shellac 1014 Sulphur 1015 6 Electrical Technology V = II Rt =12R2 :. R/R2 = I.P. =2/3 rrd2 7tX12 7t 2 at = -=-=-mm 444 Substituting the given values in Eq. (i), we get _ ~x2x 0.017 x..i...= 02544 2 a2 - 4 3 0.028 7.5 . m 2 . . 7tXd2 14 = 0.2544 or d2 = 0.569 mm Example 1.5. (a) A rectangular carbon block has dimensions 1.0 cm x 1.0 cm x 50 cm. (i) What is the resistance measured between the two square ends? (ii) between two opposing rectangularfaces / Resistivity of carbon at 20°C is 3.5 X /0-5 Q-m. (b) A current of 5 A exists in a /O-Q resistance for 4 minutes (i) how many coulombs and (ii) how many electrons pass through any section oftlze resistor in this time? Charge of the electron =1.6 x /0-19 C. (M.S. Univ. Baroda 1989) Example 1.6. Calculate the resistance of 1 km long cable composed of 19 stands of similar copper conductors. each strand being 1.32mm in diameter. Allow 5% increase in lengthfor the 'lay' (twist) of each strand in completea cable. Resistivity of copper may be taken as 1.72 x 10-8Q-m. Solution. Allowing for twist, the length of the stands. = 1000 m + 5% of 1000m =1050 m Area of cross.section of 19 strands of copper conductors is 19 x 7tx tfl4 = 19 7tx (1.32 x 10-3)2/4m2 I 1.72x 10-8x 1050x 4 Now, R = P_ A = 2 -6 = 0.694a 197tX1.32 x 10 Example 1.7. A lead wire and an iron wire are connected in parallel. Their respective ~pecific resistances are in the ratio 49 : 24. Theformer carries 80 percent more current than the latter and the latter is 47 percent longer than theformer. Determine the ratio of their cross sectional areas. (Elect. Engg. Nagpur Univ. 1993) Solution. Let suffix I represent lead and suffix 2 represent iron. We are given that Pl/pz = 49/24; if i2 =1, i. =1.8; if I. =1,12 = 1.47 I. 12 Now, RI = Pt A: and R2 =P2 ~ Since the two wires are in parallel, il = VlRt and i2=VlR2 i2 Rl _ Pili Az - = ---xi. R2 A' Pi2 Az = i2 Xpzl2 =...L x 24 x 1.47= 0.4 A i, Pili 1.8 49 .. .. Solution. (a) (i) R = P VA Here, A = I x I = I cm2= 10-4m2; I = 0.5 m .. R = 3.5 X 10-5 x 0.5110-4 = 0.175 a (ii) Here, I = I cm; A = I x 50 = 50 cm2 = 5 x 10-3m2 R = 3.5 X 10-5 x 10-2/5 X 10-3 =7 x 10-5a (b) (i) Q = It =5 x (4 x 60)=1200 C (ii) Q _ 1200 _ 20 n = e - 1.6x10-t9- 75x 10 Example 1.8. A piece of silver }'Virehas a resistance of I Q. What will be the resistance of manganin wireof one-third the length and one-third thediameter. if thespecific resistanceofmanganin is 30 times that of silver. (Electrical Engineering-I, Delhi Univ. 1978) I Solution. For silver wire RJ = .i; For manganin wire, R =P21 ~ 2 - R2 = -x-x- P2 12 ~ RI PI I] ~ 2 2 A] = 1td] /4 and A2 =7t d2 /4 R2 = P2x/2x ( !!l ) 2 R, p, I. d2 2 2 R, = I Q; li/, =113,(d,/d2) =(3/1) =9; pip, =30 .. R2 = I x 30 x (113)x 9 =900 Example 1.9. Theresistivityof aferric-chromium-aluminiumalloyis 51 X /0-8 Q-m. A sheet of the material is 15 em long. 6 em wide and 0.014 em thick. Determine resistance between (a) opposite ends and (b) opposite sides. (Electric Circuits, Allahabad Univ. 1983) Solution. (a) As seen from Fig. 1.5 (a) in this case, I = 15cm =0.15 cm A = 6 x 0.014 =0.084 cm2 = 0.084 x 10-4m2 Electric Current and Ohm's Law .. Now .. .. I 51x 10-8x 0.15 R P - = A - 0.084X 10-4 =9.1 X 10-30 As seen from Fig. 1.5 (b) here =0.014 cm = 14 x 10-5m = 15x 6 =90 cm2=9 x 10-3m2 =51 x 10-8x 14 x 10-5/9x 10-3=79.3 X 10-100 -6cm- (b) I A R (a) 7 .. (b) Fig. 1.5 Example 1.10. Theresistanceof the wire usedfor telephone is 35 Q per kilometre when the weightof the wire is 5 kg per kilometre. If the specific resistance of the material is 1.95 x /0-8 Q-m. what is the cross-sectional area of the wire? What will be the resistance of a loop to a subscriber 8 km from the exchange if wire of the same material but weighing 20 kg per kilometre is used? Solution. Here R = 35 Q; I=I km =1000 m; P =1.95 x 10-8Q-m N R - l. A - pi . A _ 1.95X10-8x 1000 - 5 ' 5 7 10-8 2 ow. - P or - .. - A R - - . x m If the second case, if the wire is of the material but weighs 20 kglkm, then its cross-section must be greater than that in the first case. 20 0-8 2 8 0-8 2 Cross-section in the second case = 5 x 55.7 x 1 =22 . x I m I 1.95x 10-8x 16000 Lengthofwire=2x8=16km=16000m:. R=P- A = 8 -=140.10 222.8x 10- Tutorial Problems No. 1.1 1. Calculate the resistance of 100 m length of a wire having a uniform cross-sectional area of 0.1 mm2 if the wire is made of manganin having a resistivity of 50 x 10-8a-m. If the wire is drawn out to three times its original length, by how many times would you except its resistance to be increased? [500 U; 9 times] 8 Electrical Technology 2. A cube of a material of side I cm has a resistance of 0.001 n between its opposite faces. If the same volume of the material has a length of 8 cm and a uniform cross-section, what will be the resistance of this length? [0.064 D] 3. A lead wire and an iron wire are connected in parallel. Their respective specific resistances are in the ratio 49 : 24. The former carries 80 per cent more current than the latter and the latter is 47 per cent longer than the former. Determine the ratio of their cross-sectional area. [2.5 : 1] 4. A rectangular metal strip has the following dimensions : x = 10cm,y =0.5 cm, Z=0.2 cm Determine the ratio of resistances Rx' R).,and Rz between the respective pairs of opposite faces. [Rz : Rl : Rz : 10,000 : 25 : 4] (Elect. Engg. A.M.Ae. S.I. June 1987) 5. The resistance of a conductor I mm in cross-section and 20 m long is 0.346 U. Determine the specific resistance of the conducting material. [1.73x 10-8D-m](Elect.Circuits-I,BangoloreUniv.199/) 6. When a current of 2 A flows for 3 micro-seconds in a coper wire, estimate the number of electrons crossing the cross-section of the wire. (Bombay University, 2000) Hint: With 2 A for 3 J.1Sec, charge transferred =6 J.1-coulombs Number of electrons crossed = 6 x 10--6/(1.6x 10-19)=3.75 X 10+13 1.8. Conductance and Conductivity Conductance (G) is reciprocal of resistance*. Whereas resistance of a conductor measures the opposition which it offers to the flow of current, the conductance measures the inducement which it offers to its flow. I I A aA From Eq. (i) of Art. 1.6, R = P A or G =P.T =I where a is called the conductivity or specific conductance of a conductor. The unit of conductance is siemens(S). Earlier,this unitwascalledmho. It is seenfromthe aboveequationthatthe conductivityof a materialis givenby G IG siemens x I metre G I. / a = - = 2 = - sIemens metre A A metre A Hence, the unit of conductivity is siemens/metre (S/m). 1.9. Effect of Temperature on Resistance The effect of rise in temperature is : (i) to increase the resistance of pure metals. The increase is large and fairly regular for normal ranges of temperatur~. The temperature/resistance graph is a straight line (Fig. 1.6). As would be presently clarified, metals have a ~ temperature co-efficient at resist~e. (ii) to increase the resistance of alloys, though in their case, the increase is relatively small and irregular. For some high-resistance alloys like Eureka (60% Cu and 40% Ni) and manganin, the increase in resistance is (or can be made) negligible over a considerable range of temperature. ' (iii) to decrease the resistance of electrolytes, insulators (such as paper, rubber, glass, mica etc.) and partial conductors such as carbon. Hence, insulators are said to possess a negative temperature-coefficient of resistance. . 1.10. Temperature Coefficient of Resistance Let a metallic conductor having a resistance of Ro at O°Cbe heated of tOCand let its resistance at this temperature be R,. Then, considering normal ranges of temperature, it is found that the increase in resistance ~ R =Rt - ~depends (i) directly on its initial resistance (ii) directly on the rise in temperature (iii) on the nature of the material of the conductor. * In a.c. circuits, it has a slightly different meaning. Electric Current and Ohm's Law 9 or R,- Ro oc R x t or R, - Ro=a Ro t ...(i) where a (alpha) is a constant and is known as the temperature coefficient of resistance of the conductor. " R,-Ro /:1R Rearrangmg Eq. (I), we get a = Rox t Rox t If Ro = I Q,t = 1°C, then a =/:1R= R,- Ro Hence, the temperature-coefficient of a material may be defined as : the increase in resistance per ohm original resistance per °C rise in temperature. From Eq. (i), we find that R, = Ro (l + at) ...(ii) - 234.5.C . 0 . -tC- -tC Fig.1.6 It should be remembered that the above equation holds good for both rise as well as fall in temperature. As temperature of a conductor is decreased. its resistance is also decreased. In Fig. 1.6 is shown the temperature/resistance graph for copper and is practically a straight line. If this line is extended backwards, it would cut the temperature axis at a point where temperature is - 234.5°C (a number quite easy to remember). It means that theoretically, the resistance of copper conductor will become zero at this point though as shown by solid line, in practice, the curve departs from a straight line at very low temperatures. From the two similar triangles of Fig. 1.6 it is seen that : !l t + 234.5 _ ( I+ LRo = 234.5 234.5) R, = Ro(I + 23~.5) or R, =Ro (l + a t) where a = lJ234.5 for copper. .. 1.11. Value of a at Different Temperatures So far we did not make any distinction between values of a at different temperatures. But it is found that value of a itself is not constant but depends on the initial temperature on which the increment in resistance is based. When the increment is based on the resistance measured at O°C, then a has the value of = Ro(1+ 40Q. Calculate the mean temperature rise of the field coil. Take the temperature coefficient of the conductor material as 0.0042 at O°e. r43.8°Cl (Elements of Elec. En!!!!.Ban!!lore Univ. 199/) 16 Electrical Technology. 1.13. Ohm's Law This law applies to electric to electric conduction through good conductors and may be stated as follows: The ratio of potential difference (V) between any two points on a conductor to the current (1) flowing between them, is constant. provided the temperature of the conductor does not change. V V In other words, I = constant or T =R where R is the resistance of the conductor between the two points considered. Put in another way, it simply means that provided R is kept constant, current is directly proportional to the potential difference across the ends of a conductor. However, this linear relationship between V and I does not apply to all non-metallic conductors. For example, for silicon carbide, the relationship is given by V= KI" where K and m are constants and m is less than unity. It also does not apply to non-linear devices such as Zener diodes and voltage-regulator (VR) tubes. Example 1.23. A coil of copper wire has resistance of 0 at 20°C and is connected to a 230-V supply. By how much must the voltage be increased in order to maintain the current consant if the temperature of the coil rises to 60°C? Take the temperature coefficient of resistance of copper as 0.00428from O°c. Solution. As seen from Art. l.l 0 Rw I + 60 x 0.00428 R = 1. ')(\ ~ {\{\{\,1')Q :. R60=90 x 1.2568/1.0856= 104.20 20 Now, current at 20°C = 230/90 = 23/9 A Since the wire resistance has become 104.20 at 60°C, the new voltage required for keeping the current constant at its previous value = 104.2 x 23/9 =266.3 V . . increasein voltagerequired=266.3 - 230 =36.3 V Example 1.24. Three resistors are connected in series across a 12-V battery. Thefirst resistor has a value of 1 0, secund has a voltage drop of 4 V and the third has a power dissipation of 12 W. Calculate the value of the circuit current. Solution. Let the two unknown resistors be R2 and R3 and I the circuit current 2 3 R2 A 1 I _ 4 .. I R3 = 12 and. 1R3=4 :. R3=4" 2' i"'-lSO, - R2 Now, 1(1 + R2+ R3) = 12 Substituting the values of I and R3, we get :2 (I+R2+iRi) = 12 or 3R/ -8R2+4=0 8 :t .J64 - 48 , 2 R2 = 6 :. R2=20 or 3"0 3 2 3 2 3 ( 2 ) 2 I R - - R2 =-x 2 =3 0 or - - =- 0 3-4 4 433 12 12 I = =2 A or I = =6 A 1+2+3 1+(2/3)+(1/3) .. .. .. 1.14. Resistance in Series When some conductors having resistances R1,R2 and R3 etc. are joined end-on-end as in Fig. 1.12,they are said to be connected in series. It can be proved that the equivalent resistance or total. resistance between points A and D is equal to the sum of the three individual resistances. Being a series circuit, it should be remembered that (i) current is the same through all the three conductors (ii) but voltage drop across each is different due to its different resistance and is given by Ohm's Law and (iii) sum of the three voltage drops is equal to the voltage applied across the three conductors. There is a progressive fall in potential as we go from point A to D as shown in Fig. 1.13. Electric Current and Ohm's Law A ~ B R2 C RJ D VI/'- I-~-I-~-I-~~ I A v t V Fig. 1.12 .. V = VI + V2+ VJ= IRI + IR2 + IRJ But V = IR where R is the equivalent resistance of the series combination. . .. IR = IR, + IR2 + IRJ or R = RI + R2+ RJ I 1 I I Also - = -+-+- G G. G2 GJ As seenfrom above, the main characteristics of a series circuit are : I. samecurrent flows through all parts of the circuit. 2. different resistors have their individual voltage drops. 3. voltage drops are additive. 4. applied voltage equals the sum of different voltage drops. 5. resistancesare additive. 6. powers are additive. Fig. 1.13 17 D -Ohm's Law 1.15. Voltage Divider Rule Since in a series circuit, same current flows through each of the given resistors, voltage drop varies directly with its resistance. In Fig. 1.14 is shown a 24- V battery connected across a series combination of three resistors. Total resistance R = R I + R2 + RJ = 12 Q According to Voltage Divider Rule, various voltage drops are : R 2 VI = V. ---1= 24x - = 4 V R 12 R2 4 V2 = V. - = 24 x _2 = 8V R 1 RJ 6 VJ = V.- =24x-=12 V R 12 1.16. Resistances in Parallel Three resistances,asjoined in Fig. 1.15aresaid to be connected in parallel. In this case (i) p.d. across all resistancesis the same (il) current in each resistor is different and is given by Ohm's Law and(iii) the total current i.. the sum of the three separatecurrents. V V V I - 1+1 +1 --+-+- - I 2 J- R R R . I 2 3 , = ~ where V is the applied voltage. R = equivalent resistanceof the parallel combination. ~ _ -Y.+ y +-Y. or 1- = -1-+ -.l +-.L R - RI R2 RJ R RI R2 RJ G = GI + G2 + G] Fig. 1.15 Now, .. Also R . 12 R . IJ RJ, I I V R2 24V RJ - Fig. 1.14 II RI A 2 VI B 4 V2 C 6 VJ D 18 Electrical Technology The main characteristics of a parallel circuit are : I. same voltage acts across all parts of the circuit 2. different resistors have their individual current. 3. branch currents are additive. 4. conductances are additive. 5. powers are additive. Example 1.25. W1u:ztis tlu?valueoftlu?unJawwnresistorR in Fig. 1.16if the voltagedrop acrosstlu? 500 0 resistoris 2.5 volts? All resistancesare in ohm (Elect.Technology,Indore Univ.April 1990) 550 50 I 550 ~ 50 A A B Fig. 1.16 Solution. By directproportion,dropon 50 0 resistance= 2.5 x 50/500= 0.25 V Drop across CMD or CD = 2.5 + 0.25 =2.75 V Dropacross5500 resistance = 12- 2.75=9.25 V I = 9.25/550 =0.0168 A, 12=2.5/500 =0.005 A I. = 0.0168 - 0.005 =0.01l8 A 0.0118 = 2.75/R; R =233n Example 1.26. Calculate the effective resistance of thefollowing combination of resistances and the voltage drop across each resistance when a P.D. of 60 V is applied between points A and B. Solution. Resistance between A and C (Fig. 1.17). 3 = 6113=20 Resistance of branch ACD = 18 + 2 =20 0 Now, there are two parallelpathsbetweenpointsA andD of resistances20 0 and 5 O. Hence, resistance between A and D =20 II5 =4 0 :. ResistancebetweenA andB =4 + 8 =120 Total circuit current =60/1 2 = 5 A Current through 5 0 resistance = 5x ;~ =4 A 5 - 5x-=IA - 25 . . P.D.across3 0 and 6 0 resistors=I x 2 =2 V P.D. across 180 resistors = I x 18=18 V P.D. across 5 0 resistors =4 x 5 = 20 V P.D. across 8 0 resistors = 5 x 8 =40 V 12 V R 500 12V B . Current in branch ACD R M~ 500 D Fig. 1.17 -Art. 1.25 Example 1.27. A circuit consists offour 100-W lamps connected in parallel across a 230-V supply. Inadvertently. a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 15000 and that of the lamps under the conditions stated is six times th~ir value then j,ur..i..o ..nrmnll" Whnt will hp thp rpndinv of the voltmeter? ., - A C .8 D u..8 B ... -., 5 / -.. .. Solution. The circuit is shown in Fig. 1.18. The wattage of a lamp is given by : W = P R = VIR .. 100 = 2302/R :. R =529 Q Resistanceof each lamp understatedconditionis = 6 x 529 = 3174 Q Equivalentresistanceofthesefourlampsconnected 230V in parallel=3174/4 =793.5 Q This resistance is connected in series with the voltmeter of 1500 Q resistance. :. total circuit resistance = 1500 + 793.5 =2293.5 Q :. circuit current = 230/2293.5 A According to Ohm's law, voltage drop across the voltmeter =1500x 230/2293.5=150 V (approx) Example 1.28. Determine the value of R and current through it in Fig. 1.19, if current through branch AO is zero. (Elect. Engg. & Electronics, Bangalore Univ. 1989) Solution. The given circuit can be redrawn as shown Fig. 1.19 (b). As seen, it is nothing else but Wheatstone bridge circuit. As is well-known, when current through branch AO becomes zero, the bridge is said to be balanced. In that case, products of the resistances of opposite arms of the bridge become equal. Electric Cu"ent and Ohm's Law .. 4 x 1.5 = R xl; R =6 Q A A 19 15000 Voltmeter Lamp Load Fig. 1.18 A Under condition of balance, it makes no difference if resistance X is removed thereby giving us the circuit of Fig. 1.19(c). Now, there are two parallel paths between points B and C of resistances (1 + 1.5) =2.5 Q and (4 + 6) = 10 Q. RBC =10 112.5 =2 Q. Total circuit resistance =2 + 2 =4 Q. Total circuit current = 10/4 =2.5 A This current gets divided into two parts at point B. Current through R is y = 2.5 x 2.5/12.5=0.5 A Example 1.29. In the unbalanced bridge circuit of Fig. 1.20 (a),find the potential difference that exists across the open switch S. Also, find the current which willflow through the switch when it is closed. . Solution. With switch open, there are two parallel branches across the 15-V supply. Branch ABC has a resistance of (3 + 12) = 15 Q and branch ABC has a resistance of (6 + 4) =10 Q. Obviously,each branch has 15 V applied across it. VB = 12x IS/IS=12 V; VD=4 x 151(6+ 4)=6 V :. p.d. across points B and D = VB - VD = 12 - 6 =6 V When S is closed, the circuit becomes as shown in Fig. 1.20 (b) where points B and D become electrically connected together. 0 I I 0 (x+y) t(x+y) 2 .J----M 10V k '" 10 V tOV 2 (a) (b) (c) Fig. 1.19 20 RAB = 3 II6 =2 Q RAC = 2 + 3 =5 Q / A Electrical Technology and RBC=4 II12=3 Q J = 15/5 =3 A 3A C (a) (b) (c) Fig. 1.20 CUlmet through ann AB =3 X 6/9 =2 A. The voltage drop over ann AB =3 x 2 =6 V. Hence, drop over ann BC =]5 - 6 =9 V. Current through BC =9/12 =0.75 A. It is obvious that at point B, the incoming current is 2 A, out of which 0.75 A flows along BC, whereas remaining 2 - 0.75 = ].25 A passes through the switch. As a check, it may be noted that current through AD =6/6 =1 A. At point D, this current is joined by 1.25 A coming through the switch. Hence, current through DC =] .25 + J =2.25 A. This fact can be further verified by the fact that there is a voltage drop of 9 V across 4 Q resistor thereby giving a current of 9/4 =2.25 A. Example 1.30. A 50-ohm resistor is in parallel with IOO-ohmresistor. Current in 50-ohm resistor is 7.2A. How will )'ou add a third resistor and what will be its value of the line-current is to be its value if the line-current is to be /2./ amp? [Nagpur Univ., Nov. 1997] Solution. Source voltage =50 x 7.2 =360 V, Current through lOO-ohm resistor =3.6 A Total current through these two resistors in parallel = 10.8 A For the total line current to be 12.1 A, third resistor must be connected in parallel, as the third branch, for carrying (] 2.1 - 10.8) =1.3 A. If R is this resistor R =360/1.3 =277 ohms Examp]e 1.31. In the circuit show in Fig. /.2/. calculate the value of the unknown resistance R and the currentflowing through it when the current in branch OC is zero. [Nagpur Univ., April 1996] Solution. If current through R-ohm resistor is / amp, AO branch carries the same current. since, current through the branch CO is zero. This also means that the nodes C and 0 are at the equal . potential. Then, equating voltage-drops, we have VAO=VAC' This means branch AC carries a current of 41;. C +"- JOy. Fig. 1.21 Thisiscurrentof 4 I alsoflowsthroughthebranchCB. Equatingthe voltage-dropsin branches nR :mctrR. J.5 x 4 J = R J. giving R =6 Q ,+ ISV D ,,+ B ]SV ,+ ]5V D A 51 2Q f 6V t 9V --1 Electric Current and Ohm's Law 21 At node A, applying KCL, a clUTentof 51 flows through the branch BA from B to A. Applying KVL around the loop BAOB, 1= 0.5 Amp. Example 1.32. Find the values of Rand Vsin Fig. 1.22. Alsofind the power supplied by the source. [Nagpur University, April 1998] Solution. Name the nodes as marked on Fig. 1.22. Treat node A as the reference node, so that VA=O. 3 Q Si~ce path ADC carries I A with a total of 4 ohms resistance, VC =+ 4 V. Since VCA+ 4, lCA=4/8 =0.5 amp from C to A. Applying KCLat node C, IBC =1.5 A from B to C. Along the path BA, 1A flows through 7-ohm resistor. VB =+ 7 Volts. VBC =7 - 4 =+ 3. This drives a current of 1.5 amp, through R ohms. Thus R =3/1.5 =2 ohms. Applying KCL at node B. IFB =2.5 A from F to B. . VFB =2 x 2.5 =5 volts, F being higher than B from the view-point of Potential. Since VB has already been evaluated as + 7 volts, V + 12 volts (w.r. to A). Thus, the source voltage V, = 12 volts. Example 1.33. In Fig. 1.23 (a), if all the resistances are of 6 ohms, calculate the equivalent resistancebetween any two diagonal points. [Nagpur Univ. April 1998] 6Q () 2Q IA F IQ D Fig. 1.22 N x y 6Q 2Q p P 6Q Y P Y 6Q Fig. 1.23(a) Fig. 1.23 (b) Fig. 1.23 (c) Solution. If X-Yare treated as the concerned diagonal points, for evaluating equivalent resistance offered by the circuit, there are two ways of transforming this circuit, as discussed below: Method 1: Delta to Star conversion applicable to the delta of PQY introducing an additional node N as the star-point. Delta with 6 ohms at each side is converted as 2 ohms as each leg of the star-equivalent. This is shown in Fig. 1.23 (b), which is further simplified in Fig. 1.23 (c). After handling series-parallel combinations of resistances, Y I 18Q Rxy Fig. 1.23 (d) Fig. 1.23 (e) Total resistance between X and Y terminals in Fig. 1.23 (c) cQmes out to be 3 ohms. Method 2: Star to Delta conversion with P as the star-point and XYQ to be the three points of concerned converted delta. With star-elements G~6 ohms each, equivalent delta-~lements will be 18 ohms, as Fig. 1.23 (d): This is included while redrawing the circuit as in Fig. 1.23 (e). After simplifvinl!:. the series-oarallel comhination re~lIlt~ intn thp fin". "n"Ulpr "" R - 1 ohm" 22 Example 1.34. For the given circuit find the current IAand lB' Electrical Technology [Bombay Univ. 1991] c -- - l IA+'A In 2n + 3n B 7V In A Fig. 1.24 Solution. Nodes A. B, C, D and reference node 0 are marked on the same diagram. IAand IBare to be found. Apply KCL at node A. From C to A, current =7 + IB At node 0. KCL is applied. which gives a current of 7 + IA through the 7 volt voltage source. Applying KCL at node B gives a current IA -IB through 2-ohm resistor in branch CB. Finally. at node A, KCL is applied. This gives a current of 7 + IBthrough I-ohm resistor in branch CA. Around the Loop OCBO. 2 (IA-IB) + 1.IA= 7 Around the Loop CABC. 1 (7 + IBJ'+ 3 IB - 2 (IA -IB) =0 After rearranging the terms, 3 IA - 21B =7 - 21A + 61B =- 7 This gives IA = 2 amp. IB=- 0.5amp. This means that IBis 0.5 amp from B to A. Example 1.35. Find RABin the circuit, given in Fig. 1.25. [Bombay Univ. 2001] A F B IOQ G IOQ Fig. 1.25 (a) Solution. Mark additional nodes on the diagram. C, D, F. G. as shown. Redraw the figure as in 1.25 (b), and simplify the circuit. to evaluate RAB'which comes out to be 22.5 ohms. 2Q C,D A 50Q ~ 25 Q :E50 Q SQ B F.G Fig. 1.25 (b) ','vvv 2Q ISQ SQ '.20Q 30Q D 40Q .. . '" 'P Electric Current and Ohm's Law Example 1.36. Find current through 4 resistance. 23 [Bombay Univ. 2001] Solution. Simplifying the series-parallel combinations, and solving the circuit, the source current is to amp. With respect to 0, VA=40, VB =40 - 16=24 volts. I) = 4 amp, hence I~=6 amp Vc = VB-12 X 1.6 =24 - 9.6 = 14.4 volts 13 = 14.4/4=3.6 amp, which is the required answer. Further 14=24 amp. 80 80 120 Fig. 1.27 120 Fig. 1.28 Tutorial Problems No. 1.3 1. Find the current supplied by the battery in the circuit of Fig. 1.27. 2. Compute total circuit resistance and battery current in Fig. 1.28. [8/3 n, 9 A] 3. Calculate battery current and equivalent resistance A of the network shown in Fig. 1.29. [15 A; 8/5 n] 4. Find the equivalent resistance of the network of Fig. 1.30 between terminals A and B. All resistance values are in ohms. [6 n] 5. What is the equivalent resistance of the circuit of Fig. 1.31 between terminals A and B? All resistances are in ohms. [4 n] B 6. Compute the value of battery current I in Fig. 1.32. All resistances are in ohm. [6 A] \ A 6 I B -=-12V 4 80 Fig. 1.31 Fig. 1.32 7. Calculate the value of current I supplied by the voltage source in Fig. 1.33. All resistance values are in ohm~ (Hint: Vo1t!l(J'P' !I~ro~~ p.!I~h rp.~i~tor i~ (, V) f1 A 1 A--1.2- ou r '3 :;60 I. . _ . 14 Fig. 1.26 F' 120 Ig. 1.29 6 [5 A] «4 £ U )6 Fig. 1.30 6 24 Electrical Technology 8. Computethe equivalentresistanceof the circuitof Fig. 1.34(a) betweenpoints(i) ab (ii) ac and (iii)be. Allresistancesvaluesarein ohm. [(i)60, (ii)4.5 0, (iii)4.5 0] 6V 2 3 A 1 B Fig. 1.33 Fig. 1.35 9. In the circuit of Fig. 1.35, find the resistance between terminals A and B when switch is (a) open and (b) closed. Why are the two values equal? [(a) 2 0 (b) 2 0] 10. The total current drawn by a circuit consisting of three resistors connected in parallel is 12 A. The voltage drop across the first resistor is 12 V, the value of second resistor is 3 n and the power dissipation of the third resistor is 24 W. What are the resistances of the first and third resistors ? [20; 60] 11. Three parallel connected resistors when connected across a d.c. voltage source dissipate a total power of 72 W. The total current drawn is 6 A, the current flowing through the first resistor is 3 A and the second and third resistors have equal value. What are the resistances of the three resistors? [40; 8 0; 80] 12. A bulb rated 110 V, 60 watts is connected with another bulb rated 11O-V, 100 W across a 220 V mains. Calculate the resistance which should be joined in parallel with the first bulb so that both the bulbs may take their rated ,power. [302.5 0] 13. Two coils connected in parallel across 100 V supply mains take 10 A from the line. The power dissipated in one coil is 600 W. What is the resistance of the other coil? [25 0] 14. An electric lamp whose resistance, when in use, is 2 n is connected to the terminals of a dry cell whose e.m.f. is 1.5 V. If the current through the lamp is 0.5 A, calculate the internal resistance of the cell and the potential difference between the terminals of the lamp. If two such cells are connected in parallel, find the resistance which must be connected in series with the arrangement to keep the current the same as before. [1 0 ; 1 V ; 0.5 0] (Elect. Technology, Indore Univ. 1978) 15. Determine the current by the source in the circuit shown below. (Bombay Univ. 2001) 30 V Fig. 1.36. (a) Hint. Series-parallel combinations of resistors have to be dealt with. This leads to the source current of 28.463 amp. G' 8 '6 2 /W' v c Fig. 1.34 Electric Current and Ohm's Law 25 16. Find the voltage of point A with respect to point B in the Fig. 1.36 (b). Is it positive with respect to B ? 5A A ~ IOV 30 t 40 c B Fig. 1.36 (b) Hint. If VA = 0, Vc = - 1.25x 3 = - 3.75V VD = -3.75-8=-11.75V VB = VD+ 15= + 3.25volts Thus, the potential of point A with respect to B is - 3.25 V. 1.17. Types of Resistors (a) Carbon Composition It is a combination of carbon particles and a binding resin with different proportions for providing desired resistance. Attached to the ends of the resistive element are metal caps which have axial leads of tinned copper wire for soldering the resistor into a circuit. The resistor is enclosed in a plastic case to prevent the entry of moisture and other harmful elements from outside. Billions of carbon composition resistors are used in the electronic industry every year. They are available in power ratings of lI8, lI4, 1/2, I and 2 W, in voltage ratings of 250,350 and 500 V. They have low failure rates when properly used. Such resistors have a tendency to produce electric noise due to the current passing from one carbon particle to another. This noise appears in the form of a hiss in a loudspeaker connected to a hi-fi system and can overcome very weak signals. That is why carbon composition resistors are used where performance requirements are not demanding and where low cost in the main consideration. Hence, they are extensively used in entertainment electronics although better resistors are used in critical circuits. (b) Deposited Carbon Deposited carbon resistors consist of ceramic rods which have a carbon film deposited on them. They are made by placing a ceramic rod in a methane-filled flask and heating it until, by a gascracking process, a carbon film is deposited on them. A helix-grinding process forms the resistive path. As compared to carbon composition resistors, these resistors offer a major improvement in lower current noise and in closer tolerance. These resistors are being replaced by metal film and metal glaze resistors. ' (c) High-Voltage Ink Film These resistors consist of a ceramic base on which a special resistive ink is laid down in a helical band. These resistors are capable of withstanding high voltages and find extensive use in cathoderay circuits, in radar and in medical electronics. Their resistances range from I ill to 100,000Mil with voltage range upto 1000 kV. (d) Metal Film Metal film resistors are made by depositing vaporized metal in vacuum on a ceramic-core rod. The resistive path is helix-ground as in the case of deposited carbon resistors. Metal film resistors have excellent tolerance and temperature coefficient and are extrememly reliable. Hence, they are very suitable for numerous high grade applications as in low-level stages of certain instruments although they are much more costlier. (Bombay University, 2000) 26 Electrical Technology (e) Metal Glaze A metal glaze resistor consists of a metal glass mixture which is appied as a thick film to a ceramic substrate and then fired to form a film. The value of resistance depends on the amount of metal in the mixture. With helix-grinding, the resistance can be made to vary from I Q to many megohms. Another category of metal glaze resistors consists of a tinned oxide film on a glass substrate. if) Wire-wound Wire-wound resistors are different from all other types in the sense that no film or resistive coating is used in their construction. They consist of a ceramic-core wound with a drawn wire having accurately-contralled characteristics. Different wire alloys are used for providing different resistance ranges. These resistors have highest stability and highest power rating. Because of their bulk, high-power ratings and high cost, they are not suitable for low-cost or high-density, limited-space applications. The completed wire-wound resistor is coated with an insulating material such as baked enamel. (g) Cermet (Ceramic Metal) The cermet resistors are made by firing certain metals blended with ceramics on a ceramic substrate. The value of resistance depends on the type of mix and its thickness. These resistors have very accurate resistance values and show high stability even under extreme temperatures. Usually, they are produced as small rectangles having leads for being attached to printed circuit boards (PCB). 1.18. Nonlinear Resistors Those elements whose V - I curves are not straight lines are called nonlinear elements because their resistances are nonlinear resistances. Their V - I characteristics can be represented by an equation of the form 1= kV = b where n is usually not equal to one and the constant b mayor may not be equal to zero. Examples of nonlinear elements are filaments of incandescent lamps, diodes, thermistors and varistors. A varistor is a special resistor made of carborundum crystals held together by a binder. Fig. 1.37(a)shows how current through a varistor increase rapidly when the applied voltage increases beyond a certain amount (nearly 100 V in the present case). 1 R mA 100V 1 o (a) (b) (c) v 0 0.4 0.8 (d) t 0 v Fig. 1.37 There is a corresponding rapid decrease in resistance when the current increases. Hence, varistors are generally used to provide over-voltage protection in certain circuits. A thermistor is made of metallic oxides in a suitable binder and has a large negative coefficient of resistance i.e. its resistance decreases with increase in temperature as shown in Fig. 1.30(b). Fig. 1.30(c) shows how the resistance of an incandescent lamp increases with voltage whereas Fig. 1.30 (d) shows the V-I characteristics of a typical silicon diode. For a germanium diode, current is related to its voltage by the relation. I = 10(eVlO.026- I) Electric Cu"ent and Ohm's Law 27 1.19. Varistor (Nonlinear Resistor) It is a voltage-dependent metal-oxide material whose resistance decreases sharply with increasing voltage. The relationship between the current flowing through a varistor and the voltage applied across it is given by the relation : i = kenwhere i = instantaneouscurrent,e is the instantaneous voltage and 11is a constant whose value depends on the metal oxides used. The value of 11for silicon-carbide-based varistors lies between 2 and 6 whereas zinc-oxide-based varistors have a value ranging from 25 to 50. The zinc-oxide-based varistors are primarily used for protecting solid-state power supplies from low and medium surge voltage in the supply line. Silicon-carbide varistors provide protection against high-voltage surges caused by lightning and by the discharge of electromagnetic energy stored in the magnetic fields of large coils. 1.20. Short and Open Circuits When two points of circuit are connectedtogetherby a thick metallic wire (Fig. 1.38),they are said to be short-circuited. Since 'short' has practically zero resistance,it gives rise to two important facts : (i) no voltage can exist across it because V =IR =I x 0 =0 (ii) currentthroughit (calledshort-circuitcurrent)is very large(theoretically,infinity) + A + ~ lsc ~B =0 lsc = lAB =mWl Series Circuit Parallel Circuit I. = 12 =13 = ......... V = V2=V3=......... VT = VI + V2 + V3 + ......... I. = II + 12 + 13 + ......... RT = RI + R2 + R3 + ......... GT = GI + G2 + G3 + ......... V. _ V2 _ V3 _ II 12 13 I = /f-T-T-...... V = (;=([=0=...... 123 I 2 3 .. RI R2 . . GI G2 Voltage DiVIder Rule VI = VT R'T V2 =VT RT Current DivIder Rule II = IT G'T 12 =IT GT Fig. 1.60 2 2 '\AI i100 WW+ 3 I 2A + 0.6A ( t) t f) 7 t Q'2 V 6 6V 0.2A 3A - + 'I i2 I I I I t - I (0) (b) (c) Fig. 1.61 Electric Current and Ohm's Law 37 4KQ A 6KQ IKQ 6KQ 3KQ -=..9V 3KQ S B Fig. 1.63 Fig. 1.64 A 3S 2S R R R v R R Fig. 1.65 Fig. 1.66 Fig. 1.67 5. Computethe values of battery cwrent I and voltage drop across6 ill resistorof Fig. 1.64when switchS is (a) closed and (b) open. All resistance values are in kilohm. [(a)3mA; 6 V; (b) 2.25 mA; OV] 6. For the parallel circuit of Fig. 1.65 calculate (i) V (ii) II (iii) 12, [(i) 20 V; (ii) 5 A; (in) - 5 A] 7. Find the voltage across terminals A and B of the circuit shown in Fig. 1.66. All conductances are in siemens (S). [5 V] 8. Prove that the output voltage Voin the circuit of Fig. 1.67 is Vl13. 9. A fault has occurred in the circuit of Fig. 1.68. One resistor has burnt out and has become an open. Which is the resistor if current supplied by the battery is 6 A ? All resistances are in ohm. [4 n] 10. In Fig. 1.69if resistance between terminalsA and B measures 1000n, which resistor is open-circuited. All conductance values are in milli-siemens (mS). [0.8 mS] A -=..12V 6 4 3 0.2 0.3 0.8 0.5 B Fig. 1.68 11. In the circuit of Fig. 1.70, find current (a) I and (b) I.. ~ +/'.,..1 1 3 r§' 0",; Fig. 1.69 [(a) 2 A; (b) 0.5 A] 2 2 200 100 Fi~. 1.70 A 4 B Fil!. 1.71 A 50 n 8 . 15 30 25 ""'- Fig. 1.62 f 4 (} 4 (} 2A\p 40A\fJ 6S 4S /2 I I B 12. Deduce the current / in the circuit of Fig. 1.71. All resistances are in ohms. [25 A] 13. Two resistors of 100 0 and 200 0 are connected in series across a 4-V cell of negligible internal resistance. A voltmeter of 200 0 resistance is used to measures P.D. across each. What will the voltage be in each case? [1 V across 100 0 ; 2 V across 200 0] 14. Using series-parallel combination laws, find the resistance between terminals A A and B of the network shown in Fig. 1.72. [4R] 15. A resistance coilAB of 100Q resistance is to be used as a potentiometer and is connected to a supply at 230 V. Find, by calculation, the position of a tapping point C between A and B such that a B current of 2 A will flow in a resistance of 50 Q connected across A and C. [43.4 Q from A to C] (London Univ.) 16. In the circuit shown in Fig. 1.73, calculate (a) current / (b) current /1and (c) VAB.All resistances are in ohms. [(a) 4 A (b) 0.25 A (c) 4 V] 38 Electrical Technology R 2R Fig. 1.72 ~~1~ ~~1~ 17. In the circuit given in Fig. 1.74, calculate (a) current through the 25 Q resistor (b) supply voltage V. All resistances are in ohms. [(a) 2 A (b) 100 V] 18. Using series and parallel combinations for the electrical network of Fig. 1.75, calculate (a) current flowing in branch AF (b) p.d. across branch CD. All resistances are in ohms. [(a) 2 A (b) 1.25 V] 13 11 '"' + 24 V 30V dc S Supply 9 Fig. 1.75 Fig. 1.76 19. Neglecting the current taken by volemters VIand V2in Fig. 1.76, calculate (a) total current taken from the supply (b) reading on voltmeter VI and (c) reading on voltmeter V2[(a)15 A (b) 14 V (c) 16 V] A . /I 0 + S:;> '\. "f '\. F I t . I I V ! $6 SA . " 0 ... B Electric Current and Ohm's Law 39 20. Find the equivalent resistance between tenninals A and B of the circuit shown in Fig. 1.77. Also. find the value of cumnts II' 12, and 13' All resistances are in ohm. [8 n; II =2A; 12=0.6 A; 13= 0.4 A] 21. In Fig. 1.78, the IO n resistor dissipates 360 W. What is the voltage drop across the 5 n resistor? [30 V] 22. In Fig. 1.79, the power dissipated in the 10 n resistor is 250 W. What is the total power dissipated in the circuit ? [850 W] 25 60 Fig. 1.77 + + 5 -+ E [ 8V 2 6 12 5 E 10 10 Fig. 1.78 Fig. 1.79 Fig. 1.80 23. What is the value of E in the circuit of Fig. 1.80? All resistances are in ohms. R 2 [4 V] 24. Find the equivalent resistance Ra_b at the tenninals a - b of the networks shown in Fig. 1.81. [(0) 0 (b) 0 (c) R (d) 2 n] 25. Find the equivalent resistance between tenninals a and b of the circuit shown in Fig. 1.82 (a). Each resistance has a value of 1 n. [5111n] a b (a) (b) Fig. 1.82 26. Find the equivalent resistance between tcnninals a and b of the Circuitshown in Fig. 1.82 (b). Each resistorhas a valueof 1n. r5/12n] ao , . - a I R Ra a b >R> 1 I R b I I Rb bO (a) (b) (c) (d) Fig. 1.81 40 Electrical Technology 27. Two resistors of value 1000Q and 4000 Q are connected in series across a constant voltage supply of 150 V. Find (a) p.d. across 4000 ohm resistor (b) calculate the change in supply current and the reading on a voltmeter of 12,000 Q resistance when it is connected across the larger resistor. [(a) 120 V (b) 7.5 mA; 112.5 V] 1.28. Relative Potential It is the voltage of one point in a circuit with respect to that of another point (usually called the reference or common point). Consider the circuit of Fig. 1.83 (a) where the most negative end-point C has been taken as the reference. With respect to point C, both points A and B are positive thoughA is more positive than B. The voltage of point B with respect to that of C i.e. VBC=+ 30 V. Similarly, VAC=+ (20 + 30) =+ 50 V. + A 20 V In Fig. 1.83 (b), the most positive end point ~ A has been taken as the reference point. With I respect to A, both B and C are negative though · fAc C is more negative than B. VBA=-20 V, VCA=-(20 + 30) =-50 V In Fig. 1.83 (c), mid-point B has been taken as the reference point. With respect to B, A is at positive potential whereas C is at a negati~ p0- tential. Hence, VAB=+ 20 V and VCB=- 30 V (of course, VBC=+ 30 V) It may be noted that any point in the circuit can be chosen as the reference point to suit our requirements. This point is often called ground or earth because originally it meant a point in a circuit which was actually connected to earth either for safety in power systems or for efficient radio reception and transmission. Although, this meaning still exists, yet it has become usual today for 'ground' to mean any point in the circuit which is connected to a large metallic object such as the metal chassis of a transmitter, the aluminium chassis of a receiver, a wide strip of copper plating on a printed circuit board, frame or cabinet which supports the whose equipment. Sometimes, reference point is also called common point. The main advantage of using a ground system is to simplify our circuitry by saving on the amount of wiring because ground is used as the return path for may circuits. The three commonly-used symbols for grounnd are shown in Fig. 1.84. Example 1.48. In Fig. 1.85. calculate the values of (i) VAF(ii) VEAand (iii) VFB" Solution. It should be noted that VAFstands for 12V · the potential of point A with respect to point F. The A0---1 7 B ~ E l--o F easist way of finding it is to start from the reference 24V point F and go to point A along any available path ~J 4 V and calculate the algebraic sum of the voltages met on the way. Starting from point F as we go to point C D A, we come across different battery voltages. Tak- 8V . ing the sign convention given in Art. 1.28, we get Fig. 1.85 (i) VAF = - 24 + 4 + 8 - 6 + 12 = - 6 V The negative sign s~ows that point A is negative with respect to point F by 6 V. (ii) Similarly, VEA = -12+6-8-4=-18V (iii) Starting from point B, we get VFB=6 - 8 - 4 + 24 = 18 V. Since the result is positive it means that point F is at a higher potential than point B by 18 V. B + 30 V C ~ I--vBc~ -I Reference (a) + A .... it Reference 20V I\IW' - B . 30V /\II.AI' c_- (b) + A - 20V I\IW'- 30V ~ it Reference c _ - (c) Fig.1.83 Fig.1.84 Electric Current and Ohm's Law 41 Example 1.49. In Fig. 1.86 compute the relative potentials of points A. B. C. D and E which (i) point A is grounded and (ii) point D is grounded. Does it affect the circuit operation or potential difference between any pair of points? Solution. As seen, the two batteries have been connected in series opposition. Hence, net circuit voltage = 34 - 10 =24 V Total circuit resistance = 6 + 4 + 2 =12 Q Hence, the circuit current = 24/12 = 2 A Drop across 2 Q resistor = 2 x 2 =4 V Drop across 4 Q resistor =2 x 4 =8 V Drop across 6 Q resistor = 2 x 6 =12 V OV A 34 1 Y B 2 n C - I '2A A 34V II B 2n C 2A 2A - 6n IOV IOV 2A 4n 2A 2A 4n D E D E - ~~1.~ ~~1.~ (i) Since point B is directly connected to the positive tenninal of the battery whose negative tenninal is earthed, hence VB = + 34 V. Since there is a fall of 4 V across 2 Q resistor, Ve= 34 - 4 =30 V As we go from point C to D i.e. from positive tenninal of IO-Vbattery to its negattve tenninal, there is a decrease in potential of 10 V. Hence, Vo =30 - 10=20i.e. point D is 20 V above the groundA. Similarly, VE = V0 - voltage fall across 4 Q restors =20 - 8 =+ 12 V Also VA = VE - fall across 6 Q resistor =12 (2 x 6) =0 V (ii) In Fig. 1.87, point D has been taken as the ground. Starting from point D, as we go to E there is a fall of 8 V. Hence, VE = - 8 V. Similarly, VA =- (8 + 12) =- 20 V. As we go from A to B, there is a sudden increase of 34 V because we are going from negative terminal of the battery to its positive tenninal. :. VB = - 20 + 34 =+ 14 V Ve = VB - voltage fall across 2 Q resistor = \4 - 4 = + 10 V. It should be so because C is connected directly to the positive terminal of the 10 V battery. Choice of a reference point does not in any way affect the operation of a circuit. Moreover, it also does not change the voltage across any resistor or between any pair of points (as shown below) because the ground current ig=O. Reference Point A VC.4 = Ve- VA= 30-0=+ 30V; VCE=Ve- VE=30-12 =+ 18V VBO= VB-Vo=34-20=+14V Reference Point D VeA = Ve- VA =10- (-20)= + 30 V; VCE=Ve- VE= 10-(-8) =+ 18 V VBO = VB- Vo= 14-0= + 14 V . Example 1.50. Find the voltage V in Fig. 1.88 (a). All resistances are in ohms. Solution. The given circuit can be simplified to the final form shown in Fig. 1.88 (d). As seen, current supplied by the the battery is I A. At point A in Fig. 1.88 (b), this current is divided into two equal parts of 0.5 A each. 42 Electrical Technology Obviously, voltage V represents the potential of point B with respect to the negative terminal of the battery. Point B is above the ground by an amount equal to the voltage drop across the series combination of (40 + 50) =90n. V = 0.5 x 90 =45 V. A IA 100 100 100 100V c (c) (d) 1.29. Voltage Divider Circuit A voltage divider circuit (also called potential divider) is a series network which is used to feed other networks with a number of different voltages and derived from a single input voltage source. Fig. 1.89 (a) shows a simple voltage I I 0 I I 0 A divider.circuit which provides two output R Vi ~ voltages VI and V2. Since no load is con- ) nected across the output terminals, it is called an unloaded voltage divider. As seen from Art. 1.15. VI = V RI and V2 =V .~ R) + R2 RI + R2 The ratio VIV is also known as voltage-ratiotransferjunction. - As seen, V2= R2 = 1 (a) V RI + R2 1+ R/R2 Fig. 1.89 The voltage divider of Fig. 1.89 (b) can be used to get six different volta~es : VCG = V3. VBC = V2. VAS = VI' VBG =(V2+ V3), VAC= (VI + V2)and VAG= V Example 1.51. Find the values of different voltages that can be obtainedfrom a 12-V battery with the help of voltage divider circuit of Fig. 1.90. " OA Solution. R = RI + R2 + R3 = 4 + 3 + 1=8 Q I 4 Q~RJ ~ Drop across RJ = 12x 4/8=6 V .. VB = 12- 6 =6 V above ground Drop across R2 = 12 x 3/8 =4.5 V .. Vc = VB-4.5=6-4.5=1.5 Drop across R3 = 12x 1/8= 1.5V Differentavailableloadvoltagesare : (i) VAB = VA - VB =12 - 6 =6 V ...L- B v v c OV - - (b) B J2 c D ---1..- - Fig. 1.90 +IOOV IA A . I 60 1100 V T 6O + - 100 oov 100 BV V.r-iB 40 C I I 50 AAA to C -=- (a) (b) Fig. 1.88 Electric Current and Ohm's Law 43 (ii) VAC = 12 - 1.5= 10.5 V (iii) VAD = 12 V (iv) VBC =6 - 1.5=4.5 V (v) VCD = 1.5 V Example 1.52. What are the output voltages of the unloaded voltage divider shown in Fig. 1.91 ? What is the direction of current through AB ? Solution. It maybe remembered that both VIand V2are with respect to the ground. R = 6 + 4 + 2 =120 VI = drop across R2 = 24 x 4/12 = + 8 V V2 = drop across R3=- 24 x 2/12 =- 4 V It should be noted that point B is at negative potential with respect to the ground. _ Current flows from A to B i.e. from a point at a higher potential to a . B point at a lower potential. Fig. 1.91 Example 1.53. Calculate thepotentials of point A, B, Cand D in Fig. 1.92. What would be the new potential values if connections of 6-V battery are reversed ? All resistances are in ohm. Solution. Since the two batteries are connected in additive series, total voltage around the circuit is =12 =6 =18 V. The drops across the three resistors as fo~nd by the voltage divider rule as shown in Fig. 1.92 (a) which also indicates their proper polarities. The potential of any point in the circuit can be found by starting from the ground point G (assumed to be at OV)and going to the point either in clockwise direction or counter-clockwise direction. While going around the circuit, the rise in potential would be taken as positive and the fall in potential as negative. (Art. 2.3). Suppose we start from point G and proceed in the clockwise direction to point A. The only potential met on the way is the battery voltage which is taken as positive because there is a rise of potential since we are going from its negative to positive terminal. Hence, VAis + 12 V. VB = 12 - 3 =9 V; Vc =12 - 3 - 6 =3 V Similarly, VD = 12-3-6-9=-6V. It is also obvious that point D must be at - 6 V because it is directly connected to the r~gative terminal of the 6-V battery. We would also find the potentials of various points by starting from point G and going in the counter-clockwise direction. For example, VB = - 6 + 9 + 6 =9 V as before. A + .. -=-24 V R2 40 A + l'2v 3V I'I - & B Grr J: G OV + 2 6VC C 6V + 6V + 3 9V 3 3V - D D (a) (b) Fig. 1.92 Electrical Technology The connections of the 6 - V battery have been reversed in Fig. 1.92(b). Now, the net voltage around the circuit is 12- 6 =6 V. The drop over the 1Q resistor is =6 x 1/(1+ 2 + 3) =1 V; Drop over 2 Q resistor is =6 x 2/6 =2 V. Obviously, VA=+ 12 V. VB=12 - I =11 V, Ve =12 - 1 - 2 =9 V. Similarly, Vo = 12 - 1 - 2 - 3 =+ 6 V. Example 1.54. Using minimum number of components, design a \'oltage divider which can deliver 1 Wat 100 V, 2 W at - 50 Vand A 100mA B 10mA 100 V 1;6"W at - 80 V. The voltage source has I 1+ an internal re:>'istanceof 200 Q and sup- 200 II L1 plies a current of 100 mA, What is the ~ 100 V open-circuitvoltageof the voltage source? I All resistances are in ohm. OV Solution. From the given load condi- V I + tions, the load currents are as follows : - ILl = 1/100=10 mA, IL2 = 2/50 =40 mA, IL3= 1.6/80=20 mA For economising the number of com- RJ I ponents, the internal resistance of 200 Q I L3 can be used as the series dropping resis- 100mA 3 20 mA tance. The suitable circuit and the ground D C connection are shown in Fig. 1.93. Fig. 1.93 Applying Kirchhoff's laws to the closed circuit ABCDA, we have V-200x 100 x 10-3-100-80=0 or V=200V II = 100 - 10 = 90 mA :. RI = 100 VI90 mA = 1.111d1 13 = 100 - 20 = 80 mA; voltage drop across R3 =- 50 - (- 80)=30 V R3 = 30 V/80 mA =375 Q 12+ 40 = 80 :. 12=40 mA; R2=50 V/4OmA =1.25 Id1 Example 1.55. Fig. 1.94 shows a transistor with proper voltages established across its base. collector are emitterfor properfunction. Assume that there is a voltage drop VBEacross the base-emitter junction of 0.6 Vand collector current Ie is equal to collector current IE Calculate (a) VI (b) V2and VB(c) V4and VE(d)1Eand Ide) VJ(f) Ve (g) VeE All resistances are given in kilo-ohm. Solution. (a) The 250 k and 50 k resistors form a voltage-divider bias network across 20 V supply. :. VI =20 x 250/300=16.7 V (b) V2 =20 - 16.7=3.3V The voltage of point B with respect to ground is V2=3.3 V (c) VE = V2 - VBE = 3.3 - 0.6 = 2.7 V. Also V4 =2.7 V (d) IE = Vi2 = 2.7 V/2 k = 1.35 mA. It also equals Ie- (e) V3 =drop across collector resistor =1.35 mA x 8 k =10.8 V if) Potential of point C is Ve =20 - 10.8 =9.2V (g) VCE= Ve-VE=9.2-2.7=6.5V 44 .. hz- 50V 80V J40mA -80V +() 20 V Ie ~~250 B J'2~ 50 2 - - Fig. 1.94 Tutorial Problems No. 1.5 1. Find the relative potentials (i) VAD(ii) VDC(iii) VBDand (iv) VACin Fig. 1.95. [(i) 10 V (ii) - 20 V (iii) 10 V (iv) - 30 V] Electric Current and Ohm's Law 2. Calculatetherelativepotentialof pointA withrespectto that toB in Fig. 1.%. C + 20V A 45 ~~1~ ~~1~ 3. Give the magnitude and polarity of the following voltages in the circuit of Fig. 1.97 (I) VI (ii) V2 (iil) V3(iv) V3-2 (v) VI_2 (vi) VI_3. [-75 V, - 50 V, 125 V, 175 V, - 25 V, - 200 V] 4. Fig. 1.98shows the equivalent circuit of a digital-to-analog (DIA) converter. What is the value of the output voltage Vo? + OBJECTIVE TESTS-l ~g. 1.98 1. A good electric conductor is one that (a) has low conductance (b) is always made of copper wire (c) produces a minimum voltage drop (d) has few free electrons 2. Two wires A and B have the same cross-section and are made of the same material. RA= 600 Q and R8=100Q. The number of times A is longer than B is (a) 6 (b) 2 (c) 4 (d) 5 3. A coil has a resistance of 100 Q at 9O.C. At l00"C, its resistance is 101 Q. The temperature coefficient of the wire at 90.C is (a) 0.01 (b) 0.1 (c) 0.001 (d) 0.001 4. Which of the following material has nearly zero temperature-coefficient of resistance ? (a) carbon (b) porcelain (c) copper (d) manganin 5. Which of the following material has a negative temperature coefficient of resistance ? (a) brass (b) copper (c) aluminium (d) carbon 6. A cylindrical wire, 1m in length, has a resistance of 100 Q. what would be the resistance of a wire made from the same material ifboth the length and the cross-sectional area are doubled ? (a) 200 Q (b) 400 Q (c) 100 Q (d) 50 Q 7. Carbon composition resistors are most popular because they . (a) cost the least (b) are smaller (c) can withstand overloads (d) do not produce electric noise 2 4 -=-24 V lA 6 I: 5Q T 2T V. -=-200V IOOH 25 Q f3 3-1- Fig. 1.97 8K 2K - =-12V ...., 4K lK Vo - 46 8. A unique feature of a wire-wound resistor is its (a) low power rating (b) low cost (c) high stability (d) small size 9. If, in Fig. 1.99, resistor Rz becomes opencircuited, the reading of the voltmeter will become (a) zero (c) 50 V (b) 150 V (d) 200 V 20 20 20 20 200 V Fig. 1.99 10. WhateverthebatteryvoltageinFig. 1.100,it is certainthat smallestcurrentwill flow in the resistanceof ohm. (a) 300 (c) 200 (b) 500 (d) 100 300 500 200 Fig. 1.100 11. Which of the following statement is TRUE both for a series and a parallel d.c. circuit? (a) powers are additive (b) voltages are additive (c) currents are additive (d) elements have individual currents 12. The positive terminal of a 6-V battery is connected to the negative terminal of a l2-V battery whose positive terminal is grounded. The potential at the negative terminal of the 6-V battery is-volt. Electrical Technology ~) +18 ~) -12 (c) - 12 (d) + 12 13. In the above question, the potential at the positive terminal of the 6-V battery is volt. (a) +6 (b) -6 (c) - 12 (d) + 12 14. A 100-W, llO-V and a 50-W, 1l0-V lamps are connected in series across a 220-V d.c. source. If the resistance of the two lamps are assumed to remain constant, the voltage across the 1000Wlamp is volt. (a) 110 (b) 73.3 (c) 146.7 (d) 220 15. In the parallel circuit of Fig. 1.10I, the value of Vois volt. (a) 12 (c) 0 (b) 24 (d) - 12 2 2 - - Fig.1.101 16. In the series circuit of Fig. 1.102, the value of Vois volt. (a) 12 (c) 0 (b) - 12 (d) 6 Fig. 1.102 17. In Fig. 1.103,there is a drop of 20 V on each resistor. The potential of point A would be volt. (a) + 80 (b) - 40 (c) + 40 (d) - 80 2 ) Vo f2V -- Electric Cu"ent and Ohm's Law A B .=.80 V D C Fig. 1.103 18. From the voltmeter reading of Fig. 1.104. it is obvious that (a) 3 Q resistor is short-circuited (b) 6 Q resistor is short-circuited (c) nothing is wrong with the circuit (d) 3 Q resistor is open-circuited 4 .=.24 V 3 6 Fig. 1.104 19. With reference to Fig. 1.105. which of the following statement is true ? (a) E and R( fonn a series circuit (b) R( is in series with R3 (c) R( is in series with R3 (d) there is no series circuit E Fig. 1.105 PTZ: q't>1 v'L q'O, J'n J'9 P'61 P"ZJ p'S SH~MSNV 47 20. Which of the following statements is correct concerning the circuit of Fig. 1.106. (a) R2and R3 fonn a series path (b) E is in series with RI (c) R. is in parallel with R3 (d) RI' R2 and R3 fonn a series circuit. R2 E Fig. 1.106 21. What is the equivalent resistance in ohms between points A and B of Fig. 1.107? All resistances are in ohms. (a) 12 (b) 14.4 (c) 22 (d) 2 P'81 v'n p.t> A 6 Fig. 1.107 J'LI v'OI P'f P'91 P'6 v', J'SI J'8 J'I 2 DC NETWORK THEOREMS 2.1. Electric Circuits and Network Theorems There are certain theorems, which when applied to the solutions of electric networks, wither simplify the network itself or render their analytical solution very easy. These theorems can also be applied to an a.c. system, with the only difference that impedances replace the ohmic resistance of d.c. system. Different electric circuits (according to their properties) are difined below: I. Circuit. A circuit is a closed conducting path through which an electric current either flows or is inteneded flow. 2. Parameters. The various elements of an electric circuit are called its parameters like resistance, inductance and capacitance. These parameters may be lumped or distributed. 3. Liner Circuit. A linear circuit is one whose parameters are constant i.e. they do not change with voltage or current. 4. Non-linear Circuit. It is that circuit whose parameters change with voltage or current. 5. Bilateral Circuit. A bilateral circuit is one whose properties or characteristics are the same in either direction. The usual transmission line is bilateral, because it can be made to perform its function equally well in either direction. 6. Unilateral Circuit. It is that circuit whose properties or characteristics change with the direction of its operation. A diode rectifier is a unilateral circuit, because it cannot perform rectification in both directions. 7. Electric Network. A combination of various electric elements, connected in any manner whatsoever, is called an electric network. 8. Passive Network is one which contains no source of e.m.f. in it. 9. Active Network is one which contains one or more than one source of e.m.f. 10. Node is a junction in a circuit where two or more circuit elements are connected together. II. Branch is that part of a network which lies between two juncions. 12 Loop. It is a close path in a circuit in which no element or node is epcountered more than once. 13. Mesh. It is a loop that contains no other loop within it. For example, the circuit of Fig. 2.1 (a) has even branches, six nodes, three loops and two meshes whereas the circuit of Fig. 2.1 (b) has four branches, two nodes, six loops and three meshes. It should be noted that, unless stated otherwise, an electric network would be assumed passive in the following treatment. We will now discuss the various network theorems which are of great help in solving complicated networks. Incidentally, a network is said to be completely solved or analyzed when all volt-. ages and all currents in its different elements are determined. 48 DC Network Theorems 49 2 3 v-=- R C ~LC> 6 5 (a) 4 (b) Fig. 2.1 There are two general approaches to network analysis : (i) Direct Method Here, the network is left in its original form while determining its different voltages and currents. Such methods are usually restricted to fairly simple circuits and include Kirchhoff slaws, Loop analysis, Nodal analysis, superposition theorem, Compensation theorem and Reciprocity theorem etc. (ii) Network Reduction Method Here, the original network is converted into a much simpler equivalent circuit for rapid calculation of different quantities. This method can be applied to simple as well as complicated networks. Examples of this method are: Delta/Star and StarlDelta conversions. Thevenin's theorem and Norton's Theorem etc. 2.2. KirchhoWs Laws * These laws are more comprehensive than Ohm's law and are used for solving electrical networks which may not be redily solved by the latter. Kirchhoffs laws, two in number, are particularly useful (a) in determining the equivalent resistance of a complicated network of conductors and (b) for calculating the currents flowing in the various conductors. The two-laws are : 1. KirchhoWs Point Law or Current Law (KCL) It states as follows: in any electrical network, the algebraic sum of the currents meeting at a point (orjunction) is zero. Put in another way, it simply means that the total current leaving a juncion is equal to the total current entering that junction. It is obviously true because there is no accumulation of charge at the junction of the network. ' Consider the case of a few conductors meeting at a point A as in Fig. 2.2 (a). Some conductors have currents leading to point A, whereas some have currents leading away from point A. Assuming the incoming currents to be positive and the outgoing currents negative, we have II + (-12) + (-13) + (+ 14) + (-Is) =0 m ~+4-~-~-4=O m ~+4=~+~+4 or incomingcurrents =outgoing currents Similarly, in Fig. 2.2 (b) for node A +I + (-II) + (-12) + (-13) + (-I~ =0 or 1=II + 12+ 13+ 14 We can express the above conclusion thus: 1: 1= 0 at a junction * After Gustave Robert Kirchhoff (1824 - 1887), an outstanding German Physicist. 50 Electrical Technology (a) (b) Fig. 2.2 -'. Kirchhoff's Mesh Law or Volta2e Law (KVU It sraresas follows : the algebraic sum of the products of currents and resistances in each of the conductors in any closed path (or mesh) in a network plus the algebraic sum of the e.mjs. in that path is zero. In other words, 1:IR + 1:e.mj. = 0 ...round a mesh It should be noted that algebraic sum is the sum which takes into account the polarities of the voltage drops. The basis of this law is this : If we.start from a particular junction and go round the mesh till we come back to the starting point, then we must be at the same potential with which we started. Hence, it means that all the sources of e.m.f. met on the way must necessarily be equal to the voltage drops in the resistances, every voltage being given its proper sign, plus or minus. 2.3. Determination of Voltage Sign In applying Kirchhoff's laws to specific problems, particular attention should be paid to the algebraic signs of voltage drops and e.m.fs., otherwise results will come out to be wrong. Following sign conventions is suggested : (a) Sign of Battery E.M.F. A rise in voltage should be given a + ve sign and afall in voltage a -ve sign. Keeping this in mind, it is clear that as we go from the -ve terminal of a battery to its +ve terminal (Fig. 2.3), there is a rise in potential, hence this voltage should be given a + ve sign. If, on the other hand, we go from +ve tenninal to -ve tenninal, then there is afall in potential, hence this voltage should be preceded current current .. .. A + V _ B A _ V +B ~~ R . . motIon Fall in Voltage -V= -IR ~~ . Rise in Voltage +E . Fallin Voltage -E ..motion Rise in Voltage + V = + IR Fig. 2.3 Fig. 2.4 by a -ve sign. It is important to note that the sign of the battery e.m.! is independent of the direction of the current through that branch. (b) Sign of IR Drop Now, take the case of a resistor (Fig. 2.4). If we go through a resistor in the same direction as the current, then ther is a fall in potential because current flows from a higher to a lower potential.. Hence, this voltage fall should be taken -ve. However, if we go in a direction opposite to that of the current, then there is a rise in voltage. Hence, this voltage rise should be given a positive sign. It is clear that the sign of voltage drop across a resistor depends on the direction of current through that resistor but is independent of the polarity of any other source of e.m.! in the circuit under consideration. DC Network Theorems 51 Consider the closed path ABCDA in Fig. 2.5. As we travel around the mesh in the clockwise direction, different voltage drops will have the follwing signs : I)R_ is - ve (fall in potential) "' A IJ 12R2 is - ve (fall in potential) 13R3 is + ve (rise in potential) 14R4 is - ve (fall in potential) E2 is - ve (fall in potential) E) is + ve (rise in potential) Using Kirchhoff's voltage law, we get -I)R) - 1~2 - 13R3 - 14R4 - E2 + E) =0 or I,R) + 12R2 - 13R3 + 14R4 =E) - E2 /D R:J 2.4. Assumed Direction of Current Fig. 2.5 In applying Kirchhoff's laws to electrical networks, the question of assuming proper direction of current usually arises. The direction of current flow may be assumed either clockwise or anticlockwise. If the assumed direction of current is not the actual direction, then on solving the quesiton, this current will be found to have a minus sign. If the answer is positive, then assumed direction is the same as actual direction (Example 2.10). However, the important point is that once a particular direction has been assumed. the same should be used throughout the solution of the question. Note. It should be noted that Kirchhoff's laws are applicable both to d.c. and a.c. voltages and currents. However, in the case of alternating currents and voltages, any e.m.f. of self-inductance or that existing across a capacitor should be also taken into account (See Example 2.14). B c 2.5. Solving Simultaneous Equations Electric circuit analysis with the help of Kirchhoff's laws usually involves solution of two or three simultaneous equations. These equations can be solved by a systematic elimination of the varia~les but the procedure is often lengthy and laborious and hence more liable to error. Determinants and Cramer's rule provide a simple and straight method for solving network equations through manipulation of their coefficients. Of course, if the number of simultaneous equaitons happens to be very large, use of a digital computer can make the task easy. 2.6. Determinants The symbol I ~ ~ I is called a determinant of the second order (or 2 x 2 ~eterminant) because it contains two rows (ab and cd) and two columns (ac and bd). The numbers a, b, c and d are called the elements or constituents of the determinant. Their number in the present case is 22=4. The evaluation of such a determinant is accomplished by cross-multiplicaiton is illustrated below : Ll = I :x; I =ad - bc The above result for a second order determinant can be remembered as upper left times lower right minus upper right times lower left a) bl c) The symbol Iaz b2 c2 repre8ents a third-order determinant having 32 I =9 elements. It may . a3 b3 c3 be evaluated (or expanded) as under: 52 Electrical Technology 1. Multiply each element of the first row (or alternatively, first column) by a determinant obtained by omitting the row and column in which it occurs. (It is called ~or determinant or just minor as shown in Fig. 2.6). Minorof a. Minorof a1 Minorof a3 Fig. 2.6 2. Prefix + and - sing alternately to the tenns so obtained. 3. Add up all these tenns together to get the value of the given determinant. Considering the farstcolumn, minors of various elements are as shown in Fig. 2.6. Expanding in tenns of farstcolumn, we get I b2 c2 1 _ I bl c. I + I ha cl I ~ = al b a2 b a3 b 3 ~ 3 ~ 2 ~ = a. (b2c3- b3c2) - a2 (blc3 - b3c.) + a3 (blc2 - b2cl) ...(i) Expanding in tenns of the first row, we get ~ = a I b2 c2 1 _ b I a2 c2 1 + c I ~ b2 I b3 c3 I ~ c3 I ~ b3 1 = al (b2c3 - b3c2) - bl (~c3 - a3c2) + cl (~b3 - a3b2) whichwillbe foundto be the sameas above. I 7 -3 -4 Example 2.1. Evaluate the detenninant 1 - 3 6 - 2 . -4 -2 11 Solution. We will expand with the help of 2st column. I 6 -2 1 1 -3 -4 1 1 -3 -4 D = 1 7 _ 2 II - (- 3) _ 2 II + (- 4) 6 - 2 =7 [(6 x II) - (- 2 x - 2)]+ 3 [(- 3 x II) - (- 4 x - 2)]- 4 f.(-3 x - 2) - (- 4 x 6)] = 7 (66 - 4) + 3 (- 33 - 8) - 4 (6 + 24) =191 2.7. Solving Equations with Two Unknowns Suppose the two given simultaneous equations are ax+by = c dx+ey =/ Here, the two unknown are x and y, a, b, d and e are coefficients of these unknowns whereas c and/are constants. The procedure for solving these equations by the method of determinants is as follows : 1. Writethetwoequationsinthematrixformas [~ :] [; ] =Lr] Example 2.2. Solve thefollowing two simultaneous equations by the method of determinants : 4ij - 3i2 = 1 3ij-5i2 = 2 Solution. The matrix form of the equations is [j =: ~] [:~] =[~ ] 1 4 -3 1 Ll = 3 -5 =(4x-5)-(-3x3)=-1l 1 1 -3 \ Lli = 2 -5 =(lx-5)-(-3x2)=1 Llz= I j ~I =(4 x 2) - (1x 3)=5 . Lli _ 1 _ 1. . _ Llz_ 5 'I = ~-=Ti --Ii' 'z-~ --0 .. 2.8. Solving Equations With Three Unknowns Let the three simultaneous equaitons be as under: ax+by+ez = d ex + fy + gz = h jx+ky+lz = m The above equations can be put in the matrix form as under : [~ {r] [~] = [!] The valueof commondeterminantis givneby a bel Ll = Ie f g I =a(fl-gk)-e(bl-ek)+j(bg-Cf) j k I The determinant for x can be found by replacing coefficients of x in the original matrix by the constants. d b e I Lli = Ih f g =d (fl- gk)- h(bl- ek) + m(bg - ef) m k I Similarly, determinant for y is given by replacing coefficients of y with the three constants. I a d e Llz = ~ h g I=a(hl- mg) - e(dl- me) + j (dg - he) I] m I .. DC Network Theorems 53 2. The common determinant is givne as Ll = [ ]=ae-bd 3. For finding the determinant for x, replace the rei b I coefficients of x in the original matrix by the Lli = I I =(ee- bl> constants so that we get determinant Lligiven by tf-.J e 4. For fmd thedeterminant for y, replacecoefficients la e I =(af-ed) of y by the constants so that we get Llz = i b 5. ApplyCramer'sruleto get the valueof x andy I fl x =Lli =ee - bf and y =Llz af - cd Ll ae-bd Ll ae-bd 54 Electrical Technology In the same way, determinant for z is given by a b d Ll3 = I~ f h I=a(fm-hk)-e(bm-dk)+j(bh-df) ] k m Ll Ll Ll As per Cramer's rule x = -t, y = ;, z= ; Example 2.3. Solve thefollowing three simultaneous equaitons by the use of determinants and. Carmer's rule i]+3i2+4i3 = 14 i] + 2i2 + i3 = 7 2i] + i2 + 2i3 = 2 Solution. As explained earlier, the above equations can be written in the form [ 1 3 4 ] [ ~t ] _ [ 14 ] 121'2 _ 7 2 I 2 i3 2 f l 3 4 ] Ll = I 2 I =1(4 -I) - I (6 - 4) + (3 - 8) =- 9 L2 I 2 [ 14 3 4 ] LlI = 7 2 1 =14(4 -1) - 7 (6- 4) + 2(3- 8)=18 2 I 2 [ I 14 4 ] Ll2 = 1 7 I =1(14-2)-1(28-8)+2(14-28)=-36 2 2 2 [ 1 3 14 ] Ll3 = 1 2 7 =1(4-7)-1(6-14)+2(21-28)=-9 2 I 2 According to Cramer's rule, Lll 18 . Ll2 -36 . Ll3 -9 it = "A= -9 =- 2A; '2=~=~= 4A; '3=~= -9 = lA Example 2.4. What is the voltage Vsacross the open switch in the circuit of Fig. 2,7 ? Solution. We will apply KVL to find Vs' Starting from point A in the clockwise direction and using the sign convention given in Art. 2.3, we have , , , I . 1 / , t o V I" 3° 1 " / I . D I I A / ~~ , ! \ Fig. 2.7 +Vs+ 10 - 20 - 50 + 30 = 0 .. A C 16A 40 V 3 F 4A L 2 E Fig. 2.8 D Vs= 30 V DC Network Theorems 55 Example 2.5. Find the unknown voltage VJ in the circuit of Fig. 2.8. Solution. Initially, one may not be clear regarding the solution of this question. One may think of Kirchhoff's laws or mesh analysis etc. But a little thought will show that the question can be solved by the simple application of Kirchhoff's voltage law. Taking the outer closed loop ABCDEFA and applying KVL to it, we get - 16 x 3 - 4 x 2 + 40 - VI = 0; :. VI=-16 V The negative sign shows there is a fall in potential. Example 2.6. Using Kirchoff's Current Law and Ohm's Law,find the magnitude and polarity ofvoltge V in Fig.2.9 (a). Directions A T A of the two current sources are as shown. L b. .1 H1 Solution. et us ar Itran y choose the directions of Ii' 12and 13 and polarity of V as shown in Fig. 2.9.(b). We will use the sign convention for currents as given in Art. 2.3. Applying KCL to node A, we have -II + 30 + 12- 13- 8 = 0 or II - 12 + 13 = 22 Applying Ohm's law to the three resistive branches in Fig. 2.9 (b), we have V V V II = 2,13=4,12,=-(; Substituting these values in (i) above, we get B (a) B (b) Fig. 2.9 ...(i) (Please note the - ve sign.) V ( - V ) V 2" - (; +4' = 22 or V =24 V .. I( = V/2 =24/2 =12 A, 12=- 24/6 =- 4 A,13= 24/4= 6 A The negative sign of 12indicates that actual direction of its flow is opposite to that show in Fig. 2.9 (b). Actually, 12,flows from A to B and not from B to A as shown. Incidentally, it may be noted that all currents are outgoing except 30A which is an incoming current. Example 2.7. For the circuit shown in Fig. 2.10,find VCEand VAG' (F.Y. Engg. Pune Univ. May 1988) Solution. Consider the two battery circuits of Fig. 2.10 separately. Current in the 20 V battery circuitABCD is 20 (6 + 5 + 9) =lA. Similarly, current in the 40 V battery curcuit EFGH is = A 40/(5 + 8 + 7) =2A. Voltage drops over different resistors can be found by using Ohm's law. For finding VCEi.e. voltage of point C with respect to point E, we will start from point E and go to C via points H and B. We will find the algebraic sum of the voltage drops met on the way from point E to C. Sign convention of D the voltage drops and battery e.m.fs. would be the same as discussed in Art. 2.3. 6 E 8 F 20V 5 5 40V 9 7 C Fig. 2.10 G 56 Eledrical Technology .. VCE = - 5 x 2 + 10 - 5 x 1=- 5V The negativesignshowsthatpoint C is negativewithrespectto pointE. VAG = 7 x 2 + 10 =6 x 1=30 V. The positive sign shows that point A is at a positive potential of 30 V with respect to point G. Example 2.8. Determine the currents in the unbalanced bridge circuit of Fig. 2.JJ below. Also. determine the p.d. across BD and the resistancefrom B to D. Solution. Assumed current directions are as shown in Fig. 2.11. Applying Kirchhoffs Second Law to circuit DACD, we get - x - 4z + 2X=0 or x - 2y + 4z =0 ...(1) Circuit ABCA gives - 2(x - z) + 3 (y + z) + 4z = 0 or 2x - 3y - 9z=0 ...(2) CircuitDABED gives Hx+y) -x - 2(x - z) - 2 (x + y) + 2 = 0 or 5x + 2y - 2z =2 ...(3) Multiplying (1) by 2 and subtracting (2) from it, we get - y + 17z = 0 ...(4) Similarly, multiplying (1) by 5 and subtracitng (3) from it, we have A C 2~.h!:=2 E Fig. 2.11 - 12y + 22z = - 2 or - 6y + llz =- 1 Eliminating y from (4) and (5), we have 91z =1 or z=1/91 A From (4); y =17/91 A. Putting these values ofy and z in (1), we get x =30/91 A Current in DA =x =30/91 A Current in DC =y = 17/91 A . 30 1 29 Current 10. AB = x - z=-- - - = A 91 91 91 17 1 18 CB = y+z=-+-= A 91 91 91 30 17 47 - x+ y=-+-= A - 91 91 91 CurrentinAC = z = 1/91A Internal voltage drop in the cell =2 (x + y) =2 x 47/91 =94/91 V . 9488* .. P.O. across pomts D and B = 2 - 91= 91 V Equivalent resistance of the bridge between points D and B p.d. between points Band D 88/91 88 = - =-= 1 87 n (approx) currentbetweenpointsB andD 47/91 47 . Solution By Determinants The matrix from the three simultaneous equaitons (1), (2) and (3) is [~ =~ =iJm=[g] [ 1 - 2 Ii = 2 -3 5 2 Current in Current in external circuit (x+y) ...(5) -~ J =1(6+ 18)- 2 (4 -8) +5 (18+ 12)=182 -2 * P.D. between D and B =drop across DC + drop across CB =2 x 17191+ 3 x 18191= 88/91 V. DC Network Theorems 57 [ 0 - 2 4 ] ~I = 0 -3 -9 =0(6+18)-0(4-8)+2(18+12)=60 2 2-2 [ 1 0 4 ] [ 1 - 2 0 ] ~2 = 2 0 - 9 = 34,~3 = 2 - 3 0 =2 5 2 -2 5 2 2 _ ~I =~_30 = 34 =17 =2= 1 A x - x ~ 182- 91 A, Y 182 91 A, z 182 91 Example 2.9. Determine the branch currents in the network of Fig. 2.12 when the value of each branch resistance is on ohm. (Elect. Technology, Allahabad Univ. 1992) Solution. Let the current directions be as shown in Fig. 2.12. Apply Kirchhoff s Second law to the closed circuit ABDA, we get 5 - x - z + y = 0 or x - y + z =5 Similarly, circuit BCDB gives - (x - z) + 5 + (y + z) + z= 0 or x - y - 3z = 5 Lastly, from circuit ADCE4, we get -y-(y+z)+lO-(x+y)= 0 or x + 3y + z = 10 From Eq. (i) and (ii), we get, z =0 Substitutingz=0 either in Eq. (i) or (ii) and in Eq. (iii), we get x-y = 5 x + 3y = 10 Subtracting Eq. (v) from (iv), we get - 4y =- 5 or y =5/4 =1.24 A Eq. (iv) gives x =25/4 A =6.25 A Current in branch AB =current in branch BC =6.25 A Current in branch BD =0; current in branch AD =current in branch DC =1.25 A; current in branch CE4 =6.25 + 1.25 =7.5 A. Example 2.10. State and explain Kirchhoff's laws. Determine the current supplied by the battery in the circuit shown in Fig. 2.12 A. (Elect. Engg. I, Bombay Univ. 1987) , Solution. Let the current distribution be as shown in the figure. Considering the close circuit ABCA and applying Kirchhoffs Second Law, we (x+y) 100n A v 100n have .. - lOOx- 300z + 500y =0 or x - 5y + 3z = 0 (i) Similarly, considering the closed loop BCDB, we have - 300z- lOO(y+ z) + 5OO(x- z) = 0 or 5x - y - 9z = 0 ...(ii) Taking the circuit ABDE4, we get - l00x - 500(x - z) + 100 - loo(x + y) =0 or 7x + y - 5z =1 ...(iii) ...(i) B ...(ii) ...(iii) ...(iv) ...(v) (x+y) D (x+y) tOY Fig. 2.12 T booy -LE z (x+y) D Fig. 2.12 A C (y+z) 58 Electrical Technology The value of x, y and z may be found by solving the above three simultaneous equations or by the method of determinants as given below : Putting the above three equations in the matrix form, we have [~ ~J ~!][n=[~] [ 1 - 5 3 ] [ 0 - 5 3 ] A = 5 - 1 - 9 =240, Al = 0 - 1 - 9 =48 7 1 -5 1 1-5 [ 1 0 3 ] [ 1 -5 0 ] A2 = 5 0 - 9 = 24,A3= 5 - 1 0 =24 7 1 -5 7 1 1 48 1 24 1 24 1 .. x = 240 = "5 A; y =240 = 10 A ; z =240 = 10 A Current supplied by the battery is x + y =1/5 + 1/10 =3/10 A. Example 2.11. Two batteries A and Bare connected in parallel and load of 10 Q is connected across their terminals. A has an e.m.f of 12 Vand an internalresistanceof 2 Q .. B hasan e.m.j.of 8 Vand A an internal resistance of 1 Q. Use Kirchhoff's laws to determine the values and directions of the currents flowing in each of the batteries and in the external resistance. Also determine the potential difference across the external resistance. x 12V 2 B C y 8V ID (x+y) 10 E (F.Y.Engg. Pone Univ.May 1989) . Solution. Applying KVL to the closed circuit Fig. 2.13 ABCDAof Fig.2.13,we get - 12 + 2x - ly + 8 = 0 or 2x - y =4 ...(i) Similarly, from the closed circuit ADCEA, we get - 8 + ly + 10 (x + y) = 0 or lOx + lly =8 ...(ii) From Eq. (i) and (ii), we get x =1.625 A and y =- 0.75 A The negative sign of y shows that the current is flowing into the 8-V battery and not out of it. In other words, it is a charging current and not a discharging current. · Current flowing in the external resistance =x + y =1.625 - 0.75 = 0.875 A P.D. across the external resistance =10 x 0.875 =8.75 V Note. To confIrm the correctness of the answer, the simple check is to fInd the value of the external voltage available across point A and C with the help of the {wo parallel branches. If the value of the voltage comes out to be the same, the the answer is correct, otherwise it is wrong. For example, VCBA =- 2 x 1.625 + 12 =8.75 V. From the second branch VeDA=1 x 0.75 + 8 =8.75 V. Hence, the answer found above is correct. Example 2.12. Determine the current x in the 4-Q resistance of the circuit shown in Fig. 2.13 (A). Solution. The given circuit is redrawn with assumed distribution of currents in Fig. 2.13 A (b). Applying KVL to different closed loops, we get Circuit EF ADE - 2y + IOz + (x - y - 6) = 0 Circuit ABCDA 2 (y + z + 6) - 10 + 3 (x - y - z - 6) - IOz =0 or 3x - 5y - 14z =40 Circuit EDCGE - (x - y - 6) - 3(x - y - z - 6) - 4x + 24 =0 or 8x - 4y - 3z=48 From above equations we get x =4.1 A Example 2.13. Applying Kirchhoff's laws to different loops in Fig. 2.14,find the values of VI and V2. + - + - 30 V -IS V Solution. Starting from point A and I I B applying Kirchhoff's voltage law to loop No.3, we get - V3+ 5 = 0 or V3= 5 V S . f . A d I . JOV tartmg rom pomt an app ymg Kirchhoff's voltage law to loop No.1, we get 10 - 30 - V, + 5 =0 or V, =-15 V The negative sign of V, denotes that its polarity is opposite to that shown in the figure. Starting from point B in loop No.3, we get -(-15)-V2+(-15)=0 or V2=O Example 2.14. In the network of Fig. 2.J5, the different currents and voltages are as under: . 5 - 2r. 3 . d 4 - 2r 12= e ,14 = Sin t an v] = e Using KCL,find voltage vI' Solution. According to KCL, the algebraic sum of the currents meeting at juncion " is zero Le. i, + i2 + i3 + (- i4) = 0 it + i2 + i3- i4 = 0 ...(i) Now, current through a capacitor is givne by i =C dv/dt . - C. dVJ _., .!£.(4 ,- 2r) = - 16e- ~r .. 13 - dt - _. dl e or x - 3y + IOz=6 + +Ji3- A + 5VFig. 2.14 ...(i) ...(ii) ...(iii) + C Fig. 2.15 60 Electrical Technology Substituting this value in Eq (I) above, we get . 5 -21 16 -21 3 . 'I + e - e - SIDt = 0 or ;1 = 3 sin t + lle-2I The voltage vI developed across the coil is vI = L ~~ =4. ;t (3sin t + lle-2I) = 4 (3 cos t- 22e - 21) =12 cos t - 88e- 21 Example 2.15. In the network shown in Fig. 2.16, vI =4V, v4 =4 cas 2t and i3 =2e- 1/3. Detennine i2. .. - 4 - v2- 4e-l13+ 4 cos 2t= 0 or v2= 4 cos 2t - 4e-t13- 4 N . dV2 d -1/3 ow, '2 = C - =8-(4cos 2t -4e -4) dt dt . 8 ( 8 . 2t 4 -,13 ) 64 . 2t 32 -1/3 '2 = - SID + 3" e =- SID + "3 e Example 2.16. Use nodal analysis to detennine the voltage across 5 resistance and the current in the I2 V source. [Bombay University 2001] 9A Solution. Applying KVL to closed mesh ABCDA, we get - vI - v2 + v3 + v4 = 0 = L d~=6 !£ (2e-1/3) v3 dt' dt = _ 4e- 1/3 Now .. 9A + 'i _ ~ 4V B Fig. 2.16 Fig. 2.17 (a) Fig. 2.17 (b) Solution. Transform the 12-volt and 4-ohm resistor into current-source and parallel resistor. Mark the nodes O. A. B and C on the diagram. Salf-and mutual conductance terms are to be wirtten down next. At A, G(IQ= 1/4 + 1/2 + 1/4 = 1 mho At B, Gbb= 1/2 + 1/5 + 1/100 = 0.71 mho At C, Gee= 1/4 + 1/5 + 1/20 = 0/50 mho Between A and B, Gab=0.5 mho, Between B and C, Gbe=0.2 mho, BetweenA and C, Gac= 0.25mho. 840 40 AI . 8 ?O A 20 8 50 ... C C 40 1000 200 1000 f200 3AQ) f 0 DC Network Theorems 61 Current Source matrix: At node A, 3 amp incoming and 9 amp outgoing currents give a net outgoing current of 6 amp. At node C, incoming current =9 amp. At node B, no current source is connected. Hence, the c"""nt-source matrix is: [- g] The potentials of three nodes to be [ found 1are :_V~:/8'_ ~~25 ] [ VA ] [ _ 6 ] - 0.5 0.71 - 0.20 VB = 0 - 0.25 - 0.20 0.5 Vc 9 For evaluatingVA'VB'Vc'followingstepsare required. 1 -0.5 -0.25 ~ =I - 0.5 0.71 - 0.201= Ix (0.710.5 - 0.04) + 0.5 (- 0.25 - 0.05) - 0.25 (0.1 + 0.71x 0.25) - 0.25 - 0.20 0.5 = 0.315 - 0.15- 0.069375=0.095625 -6 -0.5 -0.25 ~a = I- 0.5 0.71 - 0.201=+ 0.6075 9 - 0.20 + 0.5 1 - 6 - 0.25 ~b = I - 0.5 0 - 0.201=1.125 - 0.25 9 0.50 1 -0.5-6 - 0.5 0.71 0 1=2.2475 - 0.25 - 0.20 9 I VA =~j~ =+0/6075/0.095625 =6.353 volts VB =~J~ =1.125/0.095625 =11.765 volts Vc =~/~ =2.475/0.95625=25.882volts Hence, voltage across 5-ohm resistor =Vc - VB =14.18 volts. Obviously. B is positive w.r. to A. From these node potentials, current through 100-ohm resistor is 0.118 amp; (i) current through 20 ohm resistor is 1.294 amp. (ii) Current through 5-ohm resistor =14.18/5=2.836 amp. (iii) Current through 4-ohm resistor between C and A =19.53/4 =4.883 amp Check: Apply KCL at node C Incoming current =9 amp, from the source. Outgoing currents as calculated in (i), (ii) and (iii) above = 1.294 + 2.836 + 4.883 ==9 amp (iv) Current through 2-ohm resistor =(VB - VA)/2=2.706 amp, from B k>A. (v) Current in A-O branch =6.353/4 =1.588 amp ~c = 4Q A 2Q 1.412 amp 4.883 amp 2.706 amp 4Q D I o ?t2 V o Fig. 2.17 (c) Equivalent Fig. 2.17 (d) Actual elements In Fig. 2.17 (c), the transformed equivalent circuit is shown. The 3-amp current source (0 toA) 62 Electrical Technology and the current of 1.588amp inA-O branch have to be interpreted with reference to the actual circuit, shown in Fig. 2.17 (d), where in a node D exists at a potential of 12 volts w.r. to the reference node. The 4-ohm resistor between D and A carries an upward current of {(l2 - 6.353)/4 =} 1.412 amp, which is nothing but 3 amp into the node and 1.588 amp away from the node, as in Fig. 2.17 (c), at node A. The current in the 12-V source is thus 1.412 amp. Check. KCL at node A should give a check that incoming currents should add-up to 9 amp. 1.412 + 2.706 + 4.883 ==9 amp Example 2.17. Determine current in 5-Q resistor by anyone method. (Bombay University 2001) IOY ~r+ 200~ 30 8Y 20 A c 20 ~ 50 +11- Fig. 2.18 (b) A 8YI- ),__ 20 A C 20 D I- ' Yrn 12Y Fig. 2.18 (c) 10 3Q At9A 20 D . C X y It DC Network Theorems 63 x By observation, Resistance-elements of 2 x 2 matrix have to be noted. Raa = 3, Rbb =5, Rab= I = I+~~I . 1 8 -I II 3 -1 1 'a = 10 6 + -I 6 =58/17=3.412amp ib = I-i 1~1+(17)=38/17=2.2353amp VXY = VTH= 12 + 2ia+ 2ib= 23.3 Volts, with y positive w.r. to X. RTHcan be evaluated from Fig. 2.18 (c), after transforming delta configuration at nodes B-D-C to its equivalent star, as shown in Fig. 2.18 (d) Further simplification results into : Rxy =RTH =1.412 ohms Hence, Load Current =Vnf(RL + RTH) =23.3/6.412 D =3.643 amp. Fig. 2.18 (d) This agrees with result obtained ealier. Example 2.18 (a). Determine the voltages 1 and 2 of the network in Fig. 2.19 (a) by nodal analysis. (Bombay University, 2001) I 3 - 1 II ~a - I 6 'b I B y 10 2 20 10 20 o Fig. 2.19 (a) Solution. Write the conductance matrix for the network, with nodes numbered as 1, 2, 4 as shown. This gives VI =AI/A =2.50/1.25 =2 Volts And V2 =~/A =2.50/1.25 =2 Volts It means that the 2-ohm resistor between nodes 1 and 2 does not carry CUlTent. gll = 1+ 0.5 =2 mho,g22= I + 0.5 = 1.5mho, g33 = 1 mho,gl2= 0.5 mho,g23= 0, g\3= I mho 2 - 0.5 - 1\ I 0 - 0.5 - 1 A = I- 0.5 1.5 0 =1.25, AI= 2 1.5 0 I= 2.5 - 1 0 1.0 1 0 1 2 0 -1 A2 = I- 0.5 2 o I=2.5 -1 1 1.0 DC Network Theorems 65 2.9. Independent and Dependent Sources Those voltage or current sources, which do not depend on any other quantity in the circuit, are called independent sources. An independent d.c. voltage source is shown in Fig. 2.20 (a) whereas a time-varying voltage source is shown in Fig. 2.20 (b). The positive sign shows that terminal A is positive with respect to terminal B. In other words, potential of terminal A is v volts higher than that of terminal B. v OA OA OA v (t) (a) (b) ~2 = Similarly, Fig. 2.20 (c) shows an ideal constant current source whereas Fig. 2.20 (d) depicts a time-varying current source. The arrow shows the direction of flow of the current at any moment under consideration. A dependent voltage or current source is one which depends on some other quantity in the curcuit which may be either a voltage or a current. Such a source is represented by a diamondshaped symbol as shown in Fig. 2.21 so as not to confuse it with an independent source. There are four possible dependent sources : I. Voltage-dependent voltage source [Fig. 2.21 (a)] 2. Current-dependent voltage source [Fig. 2.21 (b)] 3. Voltage-dependent current source [Fig. 2.21 (c)] 4. Current-dependent current source [Fig. 2.21 (d)] Such sources can also be either constant sources or time-varying sources. Such sources are often met in electronic circuits. As seen above, the voltage or current source is dependent on the and is proportional to another current or voltage. The constants of proportionality are writeen as a, r, g and~. The constants a and ~ have no unis, r has the unit of ohms and g has tbe unit of siemens. i (1) R I . OR I oR (c) _ (d) Fig. 2.20 2 - 0.5 - I I I 0 - 0.5 - 1 - 0.5 1.5 0 =1.25, l = 2 1.5 0 I=2.5 - I 0 1.0 I 0 I +0 0 j 0 +0 V ( )av " < }ri v -0 r r -0 (a) (b) (c) (d) Fig. 2.21 66 Electrical Technology Independent sources actually exist as physical entities such as a battery, a d.c. generator and an alternator etc. But dependent sources are parts of models that are used to reperesent electrical properties of electronic devices such as operational amplifiers and transistors etc. Example 2.19. Using Kirchhoff's current law. find the values of the currents iI and i2 in the curcuit of Fig. 2.22 (a) which contains a current-dependent current source. All resistances are in ohms. Solution. ApplyingKCLto nodeA, we get 2 - il + 4 il - i2 = 0 or - 3il + i2=2 By Ohm's law, i( =vl3 and i2 = v/2 Substituting these values above, we get - 3(vI3) + vl2 = 2 or v =- 4 V .. il = - 4/3 A and i2=- 4/2 =- 2 A . The value of the dependent current source is =4il =4 x (- 4/3) =- 16/3A. A A tJ2A 3 2 t J2A 3 2 Fig. 2.22 Since i( and izcome out to be negative, it means that they flow upwards as shown in Fig. 2.22(b) and not downwards as presumed. Similarly, the current of the dependent source flows downwards as shown in Fig. 2.22 (b). It may also be noted that the sum of the upwards currents equals that of the downward currents. (a) (b) Example 2.20. By applying Kirchhoff's current law. obtain the values of v, il and i2 ill the circuit of Fig. 2.23 (a) which contains a voltage-dependent current source. Resistance values are in ohms. Now, Solution. Applying KCL to node A of the circuit, we get 2-i(+4v-iz = 0 or il+iz-4v=2 i( = v/3 and i2= v/6 v v -4 - + - - 4v = 2 or v = - V 367 - 4 . - 2 - 4 -16 -A and lz=-A and 4v=4x-=-V 21 21 7 7 A .. .. A t )2A 3 + 1~.4vA v(t' 6 t }2A 3 + , v1t4VA 6 (a) Fig. 2.23 (b) DC Network Theorems 67 Since il and i2come out to be negative and value of current source is also negative, their directions of flow are opposite to those presumed in Fig. 2.23 (a). Actual current directions are shown in Fig. 2.23 (b). Example 2.21. Apply Kirchhoff's voltage law. to find the values of current i and the voltage drops vJand v2in the circuit of Fig. 2.24 which contains a current-dependent voltage source. What is the voltage of the dependent source? All resistance values are in ohms. Solution. .. Applying KVL to the circuit of Fig. 2.24 and starting from point A, we get - vI + 4i- v2+ 6 =0 or vI- 4i + v2=6 vI = 2i and v2=4i 2i - 4i + 4i = 6 or i =3A vI = 2 x 3 =6V and v2=4 x 3 =12 V 4 vI 4iv ~ v2~4 2A f 2 + v Now, .. A + 6V D C Fig. 2.24 Voltage of the dependent source =4i =4 x 4 =12 V Example 2.22. In the circuit shown in Fig. 2.25. apply KCL tofind the value of ifor the case when (a) v =IV (b) v =4 V (c) v =6 V. The resistor values are in ohms. Solution. (a) When v =4 V, current through 2 Q resistor which is connected in parallel with the 2 v source =2/2 =IA. Since the source current is 2 A, i =2 - I =1 A. (b) When v = 4V, current through the 2Q resistor =4/2 =2 A. Hencei = 2 - 2 =0 A. (c) When v = 6 V, current through the 2Q resistor =6/2 =3 A. Since current source can supply only 2 A, the balance of I A is supplied by the voltage source. Hence, i =- I A i.e. it flows in a direction opposite to that shown in Fig. 2.25. Example 2.23. In the curcuit of Fig. 2.26. apply KCL + to find the value of current i when (a) k =2 (b) K =3 and (c) K =4. Both resistances are in ohms. Solution. Since 6 (} and 3 Q resistors are connected in parallel across the 24-V battery, iI = 24/6 = 8 A. Applying KCL to node A, we get i - 4 + 4 K - 8 =0 or i =12 - 4 K. (a) When K =2, i =12 - 4 x 2 =4 A (b) WhenK=3,i=12-4x3=OA (c) WhenK= 4, i = 12- 4 x 4 =-4 A It means that current i flows in the opposite direciton. Example 2.24. Find the current i and also the power and voltage of the dependent source in Fig. 2.72 (a). All resistances are in ohms. Fig. 2.25 A 6 · 3 Fig. 2.26 68 Ekchical Technology Solution. The two current sources can be combined into a single source of 8 - 6=2A. Thetwo parallel 4 Q resistances when combined have a value of 2 Q which, being in series with the 10 Q resistance, gives the branch resistance of 10 + 2 =12 Q. This 12 Q resistance when combined with the other 12 Q resistance gives a combination resistance of 6 Q. The simplified circuit is shown in Fig. 2.27 (b.) ApplyingKCLto nodeA, we get 0.9i+ 2 - i- V/6 = 0 or 0.6i =12 - v Also v =3i :. i =10/3 A. Hence, v = 10 V. The power furnished by the current source =v x 0.9 i =10 x 0.9 (10/3) =30 W. Example 2.25. Byusing voltage-dividerrule,calculatethe voltagesVxand v"in the net work shown in Fig. 2.28. 7' 5 Solution. As seen, 12 V drop in over the series combination of 1, 2 and 3 Q resistors. As per voltage-divider ruel v.. = drop over 3 Q = 12 x 3/6=6 V The voltage of the dependent source = 12 x 6 =72 V The voltage v" equals the drop across 8 Q resistor connected' across the voltage source ~f 72V. 3 4 6 2 + 4 + V.r~3 12V Again using voltge-divider rule, drop over 8 Q resistor = 72 x 8/12 = 48 V. Hence, vy= - 48 V. The negative sign has been given because positive and negative signs of vy are actually opposite to those shown in Fig. 2.28. cf A Example 2.26. Use KCL tofind the value ofv in t the curcuit of Fig. 2.29. Solution. Let us start from ground and go to point a andfindthe valueof voltageva' Obviously,5 + v = vaor v = va- 5. Applying KCL to point, we get 6 - 2 v + (5 - va)/l = 0 or 6 - 2 (va5) + (5 - va>= 0 or va= 7 V Hence, v = va- 5 = 7 - 5 = 2 V. Since it turns out to be positive, its sign as indicated in the figure is correct. OV - Fig. 2.29 -A + I I nA 3 !12 09 6A t v (n (f)"JO 2A(t ) 3 6 4 I -I (a) (b) Fig. 2.27 Fig. 2.28 a . . + v + + )2V 5 DC Network Theorems 69 Example 2.27.(a) BasicElectric Circuits by Cwminghan. Find the value of current i2 supplied by the voltatagei 1 controlled current source (VCCS) shown in Fig. 2.30. 2 + Solution. ApplyingKVLto theclosedcircuitABCD,we have - 4 + 8 - vI =0 :. vI=4 V The current supplied by VCCS is 10 VI=10 x 4 =40 A. Since i2 flows in an opposite direction to this current, hence i2=- 40A. Example 2.27. (b). Find the voltage drop v2across the current-controlled voltage source (CCVS)shown in Fig. 2.28. Solution. Applying KCL to point A, we have 2 + 6 - il or i. =8 16 V D C Fig. 2.30 A. Application of KVL to the closed circuit on the right hand side gives 5 i. - v2 =0 or v2 =5 i. =5 x 8 =40 V. A Fig. 2.31 Fig. 2.32 Example 2.28. Find the values of i/, V/, Vx and vahin the network of Fig. 2.32 with its terminals a and b open. Solution. It is obvious that il =4 A. Applying KVL to the left-hand closed circuit, we get - 40 + 20 - vI =0 or vI=- 20 V. Similarly, applying KVL to the second closed loop, we get v. - Vx+ 4vI - 50 =0 or Vx= 5 vI - 50 =- 5 x 20 - 50 =- 150 V Again applying KVL to the right-hand side circuit containing v00'we get 50 - 4vI - 10 vab =0 or voo=50 - 4 (- 20)- 10=120 V . Example 2.29 (a). Find the current i in the circuit of Fig. 2.33. All resistances are in ohms. Solution. The equivalent resitance of the two parallel paths across point a is 3 II(4 + 2) =2 Q. Now, applying KVL to the closed loop, we get 24 - v - 2v - 2i =O. Since v =2i, we get 24 - 2i2(2i)- 2i =0 or i =3 A. 2 + _ a a Fi2. 2.33 Fil!. 2.34 I \.J d +y 1- \.J a + il +1 ) tJ2Am6A n 5i.0 v2l,J 040 V cb It vab 1+ 4A I 'y!JV. - b +v2v . 4 A J' I , i2 3 + 24V 3$ 1 v15 2 Example 2.29. (b) Determine the value of current i2 and voltage drop v across 15 Q resistor in Fig. 2.34. Solution. It will be seen that the dependent current source is related ot i2. Applying KCL to node a, we get 4 - i + 3i2 - i2 =0 or 4 - i,+3 i2 =O. Applying Ohm's law, we get i, = vl5 and i2 =v115. Substituting these values in the above equation, we get 4 - (vI5) + 2 (vI15) =0 or v =60 V and i2=4 A. Example2.29(c). In thecurcuitof Fig.2.35,find thevalues of i andv. All resistanceare in ohms. Solution. It may be noted that 12 + v = vaor v = va- 12. ApplyingKCLto nodea, we get 70 0- v v v -12 ~+--~ =0 ~ v=4V 2 4 2 a Hence, v =4 - 12 =- 8 V. The negative sign shows that its polarity is opposite to that shown in Fig. 2.35. The current flowing from the point a to ground is 4/2 =2 A. Hence, i = - 2 A. Tutorial Problems No. 2.1 1. ApplyKCLto find!hevalueof / in Fig. 2.36. ",,~~~ '" " ' , , " C " , \ , , I ' I I ,A R '5A , " ~ ' ' , " " - -- ~,." ------.... + J) Fig. 2.36 2. Applying Kirchhoffs voltage law, find VI and V2in Fig. 2.37. 3. Find !he values of currents /2 and /4 in !he network of Fig. 2.38. I -L ~ =6A 2A 9A Electrical Technology a 2 + v + 2 12 OV - Fig. 2.35 [8 A] +15 Y- + lOY + ~ B Fig. 2.37 [VI =10 V; V2 - 5 V] [12 =4 A; 14 =5 A] I I Fig. 2.38 Fig. 2.39 4. Use Kirchhoffs law,to find!hevaluesof voltagesVI andV2in thenetworkshownin Fig.2.39. [V 1 =2 V ; V2 =5 V] 5. Find !he unknown currents in !he curcuits shown in Fig. 2.40(a). [II =2 A ; 12 =7 A] DC Network Theorems lOA 4A' 8A 5A 3A (a) 6. Using Kirchhoffs current law, find the values of the unknown currents in Fig. 2.40 (b). [I) =2 A; 12=2 A; 13=4 A; 14=10 A] 7. In Fig. 2.41, the potential of point A is - 30 V. Using Kirchhoffs voltage law, find (a) value of V and (b) powerdissipatedby 5 Q resistance.Allresistancesarein ohms. [100V; 500 W] 8. Using KVL and KCL, find the values of V and I in Fig. 2.42. All resistances are in ohms. [80 V; -4 A] 9. Using KCL, find the values VAD'II' 12and 13in the circuit of Fig. 2.43. All resistances are in ohms. [VAB =12 V ; I) =2/3 A; 12 =1 A; 13 =4/3 A] 10. A bridge network ABCD is arranged as follows: Resistances between terminals A-B, B-C, C-D, D-A, and B-D are 10,20, 15, 5 and 40 ohms respectively. A 20 V battery of negligible internal resistance is connected between terminals A and C. Determine the current in each resistor. [AB = 0.645A; BC = 0.678A; AD = 1.025A; DB = 01033A; DC = 0.992A] II. Two batteries A and B are connected in parallel and a load of 10Q is connected across their terminals. A has an e.m.f. of 12 V and an internal resistance of 2 Q ; B has an e.mJ. of 8 V and an internal resistance of I Q. Use Kirchhoffs laws to determine the values and directions of the currents flowing in each of the batteries and in the external resistance. Also determine the p.d. across the external resistance. [IA = 1.625A (discharge),IB= 0.75A (charge); 0.875A; 8.75V] 12. The four arms of a Wheastone bridge have the following resistances; AB = 100,BC= 10,CD= 4, DA = 50 ohms. A galvanometer of 20 ohms resistance is connected across ED. Calculate the current through the galvanometer when a potential difference of 10 volts is maintained across AC. [0.00513A][Elect.Tech.Lond. Univ.] 13. Findthe voltageVdain the networkshownin Fig.2.44(a)if R is 10Q and(b) 20 Q. [(a) 5 V (b) 5 V] 14. In the network of Fig. 2.44 (b), calculate the voltage between points a and b i.e. Vab. [30 V] (Elect. Engg. I, Bombay Univ. 1979) 71 _A I h 12A h 6A 14 I 7A IA I I f"",--8A Fig. 2.40 (b) 5 + 20 V- IAA A I V-L- 6 3 I I I I t t/2 I 13 30V. -30v1 TV T60V ,- If} I8 I2 ( ) 9 G I I I B 0 Fig. 2.41 Fig. 2.42 Fig. 2.43 72 3 5 2 30V to 30V~ (a) 5V (b) Electrical Technology 4 b + 6 + to -4 II 20V Fig. 2.44 [Hint: In the above two cases, the two closed loops are independent and no CU1Tentpasses between them]. 15. A battery having an E.M.F. of 110 V and an internal resistance of 0.2 Q is connected in parallel with another battery having an E.M.F. of 100 V and internal resistance 0.25 Q. The two batteries in parallel are placed in series with a regulating resistance of 5 Q and connected across 200 V mains. Calculate the magnitude and direction of the current in each battery and the total current taken from the supply mains. [IA =11.96 (discharge); IB =30.43 A (charge) : 18.47 A] (Elect Technology, Sumbhal Univ. May 1978) 16. Three batteries P, Q and R consisting of 50,55 and 60 cells in series respectively supply in paralle a common load of 100 A. Each cell has a e.mJ of 2 V and an internal resistance of 0.005 Q. Determine the current supplied by each battery and the load voltage. [1.2 A; 35.4A : 65.8A: 100.3V](BasicElectricity.BombayUniv.1980) 17. Two storge batteries are connected in parallel to supply a load having a resistance of 0.1 Q. The opencircut e.mJ. of one battery (A) is 12.1 V and that of the other battery (B) is 11.8 V. The internal resistances are 0.03 Q and 0.04 Q respectively. Calculate (i) the current supplied ot the lead (ii) the current in each battery (iii) the terminal voltage of each battery. [(i) 102.2A (ii) 62.7A (A). 39.5A (B) (iii) 10.22V] (LondonUniv.) 18. Two storage batteries, A and B, are connected in parallel to supply a 30 load the resistance of which is 1.2Q. Calculate (i) the current in this 1000and (ii) the CU1Tentsupplied by each battery if the open-circuit e.m.f. of A is 12.5V and that of B is 12.8V, the internal resistance of A being 0.05 Q an that of B 0.08 Q. [(i)10.25A (ii) 4 (A), 6.25A (B)] (LondonUniv.) 19. The circuit of Fig. 2.45 contains a voltage-dependent voltage source. Find the current supplied by the battery and power supplied by the voltage source. Both resistances are in ohms. [8A ; 1920W] 20. Find the equivalent resistance between terminals a and b of the network shown in Fig. 2.46. [2 QJ a 12V + 2v + + v 15 Fig. 2.45 Fig. 2.46 Fig. 2.47 Fig. 2.48 21. Find the value of the voltage v in the network of Fig. 2.47. [36 V] DC Network Theorems 73 22. Determine the current i for the network shown in Fig. 2.48. [- 40 A] 23. State the explain Kirchhoffs current law. Determine the value of Rs and RI, in the network of Fig. 2.49 if V2=Vl2 and the equivalent resistance of the network between the terminals A and B is 100 n. [Rs =100/3 n. Rp =400/3 n] (Elect. Engg. /. Bombay Univ. /978) 24. Four resistance each of R ohms and two resistances each of S ohms are connected (as shown in Fig. 2.50) to four terminasl AB and CD. A p.d. of V votls is applied across the terminals AB and a resistance of Z ohm is connected across the terminals CD. Find the value of Z in terms of S and R in order that the current at AB may be VIZ. Find also the relationship that must hold between R and S in order that the p.d. at the points EF be V/2. [Z =JR (R+ 2S); S =4R] A E C A R R E R R C S z F Fig. 2.50 D 2.10. Maxwell's Loop Curent Method This method which is particularly well-suited to coupled circuit soluions employs a system of loop or mesh currents instead of branch currents (as in Kirchhoffs laws). Here, the currents in different meshes are assigned continuous paths so that they do not split at a juncion into branch currents. This method eliminates a great deal of tedious work involved in the branch-current method and is best suited when energy sources are voltage sources rather than current sources. Basically, this method consists of writing loop voltage equations by Kirchhoff's voltage law in terms of unknown loop currents. As will be seen later, the number of independent equations to ~ reducesfromb by Kirchhoffs lawsto b -(j- 1)forthe loopcurrentmethodwhereb is the number of branches andj is the number of junctions in a given network. RI ~ R3 Fig. 2.51 shows two batteries E) and E2 B .... E .... G connected in a network consisting of five resistors. Let the l~op currents for the three meshes be II' 12and 13' It is obviE ous that current through R4 (when con2 sidered as a part of the first loop) is (/) - 12)and that through Rs is (12-13), However, when R4 is considered part of the second loop, current through it is (/2 - H I). Similarly, whe Rs is considered part of the third loop, current through it is (13 -12), Applying Kirchhoff's voltage law to the three loops, we get, E) -IIR) - R4 (I) -12) = 0 or I) (R) + R,J -12 R4 - E) =0 -I~2 - Rs (/2 -13) - R4 (/2-II) =0 A Ej D C F Fig. 2.51 ...loop 1 Similarly, c: <:> v v2 I V S II N B B Fig. 2.49 74 Electrical Technology or 12R4 -12 (R2 + R4 + RS) + I3RS = 0 ...loop 2 Also -I3R3 - E2 - RS (13-C12) = 0 or I2Rs-13 (R3+ Rs) - E2=0 ...loop 3 The above three equations can be solved not oniy to find loop currents but branch currents as well. 2.11. Mesh Analysis Using Matrix Form Consider the network of Fig. 2.52, which contains resistances and independent voltage sources and has three meshes. Let the three mesh currents be designated as II' 12 and 13and all the three may be assumbed to flow in the clockwise direcion for obtaining symmetry in mesh equations. Applying KVL to mesh (i), we have EI -IIRI - R3 (/1 -13) - R2 (/1 -12) =0 or (RI + R2 + R3) II - R/2 - Rl3 =EI ...(i) Similarly, from mesh (ii), we have E2 - R2 (12-II) - Rs (/2 -13) -I2R4 =0 or - R/I + (R2 + R4 + Rs) 12 - Rsl3 =E2 ...(if) Applying KVL to mesh (iii), we have E3 -I3R7 - Rs (13-12) - R3 (13-II) -13 R6 =0 or - Rll - RsI2 + (R3 + Rs + R6 + R7) 13=E3 ...(iii) It should be noted that signs of different items in the above three equations have been so changed as to make the items containing self resistances positive (please see further). The matrix equivalent of the above three equations is [ +(RI+R2+R3) -R2 -R3 ][ 11 ] [ EI ] - R2 + (R2+ R4+ Rs) - Rs 12 = E2 -R3 -Rs +(R3+RS+R6+R7) 13 E3 It would be seen.that"thefirst item is the first row i.e. (RI + R2+ R3)represents the self resistance of mesh (i) which equals the sum of all resistance in mesh (i). Similarly, the second item in the first row represents the mutual resistance between meshes (i) and (ii) i.e. the sum of the resistances common to mesh (i) and (ii). Similarly, the third item in the first row represents the mutual-resistance of the mesh (i) and mesh (ii). The item E.. in general, represents the algebraic sum of the voltages of all the voltage sources acting around mesh (i). Similar is the case with E2 and E3' The sign of the e.m.fs is the same as discussed in Art. 2.3 fe. while going along the current, if we pass from negetive to the positive terminal of a battery, then its e.m.f. is taken postive. If it is the other way around, then battery e.mf. is taken negative. In general, let RII =self-resistance of mesh (i) R22 =self-resistance of mesh (ii) i.e. sum of all resistances in mesh (ii) R33 =Self-resistance of mesh (iii) i.e. sum of all resistances in mesh (iii) RI2 =R21 =- [Sum of all the resistances common to meshes (i) and (ii)] * R23 =R32 =- [Sum of all the resistances common to meshes (ii) and (iii)]* Fig. 2.52 * Although, it is easier to take all loop currents in one direction (Usually clockwise), the choice of direcion for any loop current is arbitary and may be chosen independently of the direction of the other loop currents. DC Network Theorems 75 R31 = RI3 = - [Sum of all the resistances common to meshes (i) and (iii)] * UsingIh~sesymbols,the generalizedformof the abovematrixequivalentcan be writtenas [ RII RI2 RI3 ][ /I ] [ EI ] ~I R22 ~3 12 = E2 R31 R32 R33 13 E3 If there are m independent meshes in any liner network, then the mesh equations can be written in the matrix form as under : [ RII RI2 RI3 ... RIm ] [ II ] [ EI :::~::::Z ;~~: ] The above equaitons can be written in a more compact form as [Rm][1m]= [Em]' It is known as Ohm's law in matrix form. In the end, it may be pointed out that the directions of mesh currents can be selected arbitrarily. If we assume each mesh current to flow in the clockwise direction, then (i) All self-resistances will always be postive and (ii) all mutual resistances will always be negative. We will adapt this sign convention in the solved examples to follow. The above main advatage of the generalized form of all mesh equations is that they can be easily remebered because of their symmetry. Moreover, for any given network, these can be written by inspection and then solved by the use of determinants. It eliminates the tedium of deriving simultaneous equations. Example. 2.30. Write the impedance matrix of the network shown in Fig. 2.53 and find the value of current /3. =:; J ? / 3 (Network Analysis A.M.I.E. Sec. B.W. 1980) Solution. Different items of the mesh-resistance matrix [Rm] are as under: Ru = 1 + 3 + 2 = 6 Q ; R22 = 2 + I + 4 = 7 Q ; R33 = 3 + 2 + 1 = 6 Q ; RI2 = R21 = - 2 Q ; R23 = R32 = - 1 Q ; RI3 = R31 = - 3 Q ; EI = + 5 V ; E2 = 0 ; E3 = O. The mesh equations in the matrix form are [ RII RI2 RI3 ][ /I ] [ ~ ] [ 6 -2 -3 ][ /1 ] [ 5 ] R21 Rn R23 12 = E2 or = 2 _7 ~ 1 12 = 0 R31 R32 ~3 13 E3 3 1 6 13 0 [ 6 - 2 -3 ] A = -2 7 -1 =6(42-1)+2(-12-3)-3(2+21)=147 -3 -I 6 [ 6 - 2 5 ] A3 = -2 7 0 =6+2(5)-3(-35)=121 -3 -1 0 121 13 = A/A = 147= 0.823A +1 3 j) 2 sv 2 1 :IV 4 Fig. 2.53 In general,if the twocurrentsIhroughthecommonresistanceflowin the samedirecion,thenthe mutual resistance is taken negative. One the other hand, if the two currents flow in the same direcion, mutual resistance is taken as positive. * 76 Electrical Technology Example 2.31. Determine the current supplied by each battery in the circuit shown in Fig. 2.54. (Electrical Engg. Aligarh Univ. 1989) Solution. Since there are three meshes, let the three loop currents be shown in Fig. 2.51. 50 40 5 V For loop I we get 20 - 51)- 3 (I) - 12)- 5 =0 For loop 2 we have - 412 + 5 - 2 (12- 13) + 5 + 5 - 3 (12- I) =0 or 31) - 912 + 213=- 15 Similarly, for loop 3, we get - 813- 30 - 5 - 2(13- 12)=0 or 212- 1013=35 . Eliminating I) from (i) and (ii), we get 6312- 1613 =165 Similarly, for 12from (iii) and (iv), we have 12 =5421299A From (iv), 13 =- 1875/598 A Substituting the value of 12in (i), we get I) =765/299 A Since 13turns out to be negative, actual directions of flow of loop currents are as shown in Fig. 2.55. ...(i) ...(ii) ...(iii) ...(iv) Fig. 2.55 Discharge current of BI = 765/299A , Charging current of B2 = I) - 12=220/299 A Discharge current of B3 = 12+ 13=2965/598 A . Discharge current of B4 = 12 =545/299 A; Discharge current of Bs =1875/598 A Solution by Using Mesh Resistance Matrix. The different items of the mesh-resistance matrix [Rm] are as under: RIl =5 + 3 =8 Q; R22 =4 + 2 + 3 =9 Q; R33 =8 + 2 =10 Q R)2 =R2) =- 3 Q ; R13=R3) =0 ; R23=R32 =- 2 Q E) =algebraic sum of the voltages around mesh (i) = 20 - 5 = 15 V E2 = 5 + 5 + 5 = 15 V ; E3 = - 30 - 5 = - 35 V B4 I rQ rQ 130V 20 V I) 12 5V h IBs B2T 5 V B3..... Fig. 2.54 5V -! B4 I rQ rQ s 20 V I) h 5 V h T30V B2T 5 V B3..... . DC Network Theorems 77 Hence, the mesh equaitons in the matrix fonn are [ RlI R2" R3' Example 2.32. Determine the current in the 4-0. branch in the circuit shown in Fig. 2.56. (Elect. Technology, Nagpur Univ. 1992) Solution. The three loop currents are as shown in Fig. 2.53 (b). For loop 1, we have - 1 (I, - 12) - 3 (II - 13) - 41, + 24 =0 or 81, - 12 - 313=24 For loop 2, we have 12 - 212 - 12 (12- 13) - 1 (12- I,) = 0 or I, - 1512+ 1213= - 12 Similarly, for loop 3, we get - 12 (13- 12) - 2/3 - 10 - 3 (/3 -I,) =0 or 3/, + 12/2 - 1713 =10 Eliminating 12from Eq. (i) and (ii) above, we get, 1191,- 5713=372 Similarly, eliminating 12from Eq. (ii) and (iii), we get, 571,- 11113 =6 From(iv) and (v)we have, I, =40,950/9,960 =4.1 A Solution by Determinants The three equations as found above are 81, -12 - 313 = 24 II - 15/2+ 1213 = - 12 311+ 1212- 1713 = 10 [ 8 -1 -31 [ X ] [ 24 ] Their matrix fonn is 1 -15 12 J Y = -12 3 12 -17 z 10 [ 8 -1 -3 ] [ 24-1 ~= 1 -15 12 =664, ~I = -12 -15 3 12 -17 10 12 .. II =~,/~ =2730/664 =4.1 A ...(i) ...(ii) ...~iii) ...(iv) ...(v) -3 ] 12 =2730 -17 8 -3 0 = 1-3 9 -21=8(90-4)+3(-30)=598 o -2 10 , 15 -3 0 ,= 115 9 -21=15(90-4)-15(-30)-35(6)=1530 - 35 - 2 10 I 8 15 0 2 = -3 15 - 21=8(150-70) + 3(150+0)=1090 o - 35 10 8 -3 151 3 = 1-3 9 151=8(-315+30)+3(105+30)=-1875 o - 2 - 35 , _ 1530_ 765 . _ 2 _ 1090 545 . _ 3 _ -1875 I, = - 598 - 299A,I2 -t;- 598 - 299A,13--598A Fig.2.56 Solution by Using Mesh Resistance Matrix For the network of Fig. 2.53 (b), values of self resistances, mutual resistances and e.m.f s can be written by more inspection of Fig. 2.53. RII =3 + 1 + 4 =8 Q ; R22 =2 + 12 + 1 =15 Q ; R33 =2 + 3 + 12 =17 Q R12 =R21 =- I; R23 =R32 =- 12 ; R13 =R31 =- 3 EI =24 V ; E2=12V ; E3=- 10V . The matrix form of the above three equations can be written by inspection of the given network as under :- [ RII RI2 R13 ][ II ] [ EI ] [ 8 -1 -3 ][ 11 ] [ 24 ] ~I R22 ~3 12 = £2 or -1 15 -12 12 = 12 R31 R32 R33 13 E3 -3 -12 17 13 -10 ~ = 8 (255 - 144) + 1(- 17 - 36) - 3 (12 + 45) =664 [ 24 - 1 - 3 ] ~I = 12 15 -12 =24 (255-144) -12(-17 - 36) ~ 10(12+ 45)=2730 -10 -12 17 II = ~I =2730 =4.1 A . ~ 664 .. It is the same answer as found above. Tutorial Problems No. 2.2 1. Findtheammetercurrentin Fig. 2.57by usingloopanalysis. [In A](NetworkTheoryIndoreUniv.1981) 10 Fig. 2.57 Fig. 2.58 Fig. 2.59 2. Using mesh analysis, determine the voltage across the 10 ill resistor at terminals a-b of the circuit . shown in Fig. 2.58. [2.65 V](Elect. Technology, Indor Univ. April 1978) 3. Apply loop current method to find loop currents II' 12and 13in the circuit of Fig. 2.59. [II =3.75 A, 12 = 0,13 = 1.25 A] 78 ElectricalTechnology 2 2 2 2 2vLl 12V \0 12\0 12 10V T T IIOV . , 3 N----I I 24V 4 (a) IK I J) 10VT /' "1-L' I \ }IVV Y 2) 11 Tsov DC Network Theorems 79 2.12. Nodal Analysis With Sources The node-equation method is based directly on Kirchhoffs current law unlike loop-current method which is based on Kirchhoff s voltage law. However, like loop current method, nodal method Node R Node R also has the advantage that a minimum numt 2 2 3 ber of equations need be written to determine I I"A-/2 fBI-h the unknown quantities. Moreover, it is par- + ~ + ticularly suited for networks having many 15 parallel circuits with common ground con- £2 T nected such as electronic circuits. For the application of this method, every juncion in the network where three or more branches meet is regarded a node. One of these is regarded as the reference node or datum node or zero-potential node. Hence the number of simultaneous equations to be solved becomes (n - 1) where n is the number of independent nodes. These node equations often become simplified if all voltage sources are converted into current sources (Art. 2.12). (i) First Case Consider the circuit of Fig. 2.60 which has three nodes. One of these i.e. node 3 has been take in as the reference node. VArepresents the potential of node 1 with reference to the datum node 3. Similarly, VBis the potential difference between node 2 and node 3. Let the current directions which have been chosen arbitarily be as shown. For node 1, the following current equation can be written the help of KCL. II = 14+ 12 Now IIRI = EI - VA :. II =(EI - VA)IRI Obviously, 14 = ViR4 Also, 1~2 = VA- VB (": VA > VB) .. 12 = (VA - VB)IR2 Substituting these values in Eq. (i) above, we get, EI-VA VA VA-VB = -+ RI R4 R2 Simplifying the above, we have VA( 1...+ J + ...!.. ) - VB - 5.. - 0 RI R2 R4 R2 RIThe current equation for node 2 is 15 =12 + 13 VB VA -VB E2 -VB - = + ~ R2 R3 or V ( ...!...+ ...! +...!.. ) - VA - E2 =0 BR2 R3 R5 R2 R3 Though the above nodal equations (ii) and (iii) seem to be complicated, they employ a very simple and systematic arrangement of terms which can be written simply by inspection. Eq. (ii) at node I is represented by 1. The product of node potential VAand (lIRI + lIR2 + lIR4) i.e. the sum of the reciprocals of the branch resistance connected to this node. . 2. Minus the ratio of adjacent potential VBand the interconnecting resistance R2. 3. Minus ratio of adjacent battery (or generator) voltage-EI and interconnecting resistacne RI' 4. All the above set to zero. Reference Node Fig. 2.60 ...(i) ...(ii) or ...(iii) ...(iv) 80 Electrical Technology Same is the case with Eq. (iii) which applies to node 2. Rg Node 2 ..-- Vc ReferenceNode Fig. 2.61 Using conductances instead of resistances, the above two equations may be written as VA(G1 + G2+ G~ - VBG2- E1G1=0 VB (G2 + G3 + GS) - VAG2- E2G3 =0 To emphasize the procedure given above, consider the circuit of Fig. 2.61. ...(iv) ...(V) . ( 1 1 1 1 ) Vc VB EI_ The three node equatIons are VA If +R + If + R - R - R - If - 0 (node 1) 12 S"8 2"81 ( 1 1 1 ) VA VB Vc - + -+- =0 (node 2) Rz R3 R6 R2 R3 ( 1 1 1 1 ) Vc VA E4 VB - +- +- +- =0 (node3) R3 R4 R7 Rg R3 Rg R4 After finding different node voltages, various currents can be calculated by using Ohm's law. (ii) Second Case Now, consider the case when a third battery of e.m.f. E3 is connected between nodes 1 and 2 as shown in Fig. 2.62. It must be noted that as we travel from node 1to --£ node 2, we go from the -ve terminal of E3 to its +ve T 1 terminal. Hence, according to the sign convention given in Art. 2.3, E3must be taken aspositive. However, if we travel from node 2 to node 1, we go from the +ve to the -ve terminal of E3' Hence, when viewedfrom node 2, E3 is taken negative. For node 1 Node 2 ~ £, Reference Node Fig. 2.62 .. ...(i) Now, or DC Network Theorems 81 It is exactly the same expression as given under the First Case discussed above except for the additional tenn involving E3' This additional tenn is taken as +EiR2 (and not as - EiR2) because this third battery is so connected that when viewed from mode I, it represents a rise in voltage. Had it been connected the other way around, the additional tenn would have been taken as -EiR2' For node 2 12+ 13-Is =0 or 12+ 13=Is - as perKCL V +E -V E -V V Now as before I = A 3 B I = 2 B I =...1!... , , 2 ~'3 R3's Rs . ~+~-~+~-~=~ .. ~ R3 Rs O . lify. V ( I + 1 + 1 ) E2 VA E3 0 ( 00 ) n sunp mg, we get B R2 R3 Rs - R3 - R2 - ~ = .nII ~s seen, the additional tenns is -EiR2 (and not + EiR2) because as viewed from this node, E3 represents afall in potential. It is worth repeating that the additional tenn in the above Eq. (i) and (ii) can be either +EiR2 or -EiR2 depending on whether it represents a rise or fall of potential when viewed from the node under consideration. Example 2.33. Using Node voltage method.find the current in the 30. resistancefor the network shown in Fig. 2.63. (Elect. Tech. Osmania Univ. Feb. 1992) Solution. As shown in the figure node 2 has 3 2 V. (1) 2 been taken as the reference node. We will now find the value of node voltage VI' Using the technique developed in Art. 2.10, we get V. ( !+!. +1. ) -.1.- ( ~ ) =0 15222 5 The reason for adding the two battery voltages of 2 V and 4 V is because they are connected in additive series. Simplifying above, we get VJ= .8/3 V. The current flowing through the 3 0. resistance . 6 - (8/3) 2 towards node 1 IS= (3 + 2) - 3"A Alternatively . 6-~ 4 VI -+--- = 0 5 2 2 12 - 2VI + 20 - 5VI = 0 7VI = 32 6-~ 4-~ ~ -+- = - 5 2 2 12-2V(+20-5V( = 5.V( 12VI = 32; VI =8/3 Example2.34. Frameandsolvethenodeequationsof thenetworkof Fig.2.64. Hence,find the totalpowerconsumedby thepassiveelementsof the network.(Elect.CircuitsNagpur Univ.1992) Solution. The nodeequationfor node I is ( 1 ) V2 15 ~ 1+1+- =0 0.5 0.5 I 2 2 4V 4V - Fig. 2.63 Also 82 Electrical Technology or 4VI - 2V2=15 Similarly, for node 2, we have \II ( 1+1+1- ) - V2 - 20 =0 2 0.5 0.5 1 or 4VI -7V2 = - 40 ...(ii) .. V2= 11 volt and VI= 37/4 volt Now, I = 15-37/4 =23 A~ 575A .I _11-37/4 =35A I I 4 . '2 05 . 20 - 11 14 =5.75 + 3.5 =9.25 A; 13 = 1 =9 A ; Is =9 - 3.5 =5.5 A The passive elements of the net- J. 6 E 2 work are its five resistances. Total I . . . . A 3 powerconsumedby them is =5.752X 1 + 3.52x 0.5 + 92 x 1 + 9.252 x 1 + 5.52x 2 =266.25 L.... Example 2.35. Find the branch EIT 6 V currents in the curcuit of Fig. 2.65 by using (i) nodal analysis and (ii) loop analysis. ...(i) CD 10 10 IS V 20V Datum Node Fig. 2.64 B 4 13 w., 4. R3 ~ IOY 4 Is Reference Node Solution. (i) Nodal Method The equation for node A can be writtenby inspectionas explaine in Art. 2-12. Fig. 2.65 V ( .l..+-1-+-1- ) _ EI _ VB + E3 =0 A RI R2 R4 RI R2 R2 Substituting the given data, we get. V ( 1+1+1 )_&-VB +~=o or 2 VA- VB";'-3 A6 2 3622 For node B, the equation becomes . V ( -1 + -1- + -1- ) _5: VA _ E3 =0 B R2 R3 Rs R3 R2 R2 V ( 1.+1+~ ) _1O_ VA-2=0 .. B244 422 From Eq. (;) and (i;), we get, 4 17 VA = -VV B =-V 3' 3 I = EI- VA_ 6 - 4/3 _ 2. A I RI 6 9 6 1= VA+E3-VB (4/3)+5-(17/3)_!A 2 ~. 2 3 I = E2 -VB _10-17/3 _ 13 A 3 R3 4 12 _ VA=4/3= iA I =VB =1713= 17 A 14 - R4 3 9' s Rs 4 12 ...(i) .. ...(ii) A 6 B5V 2 c 4 JOV F E Fig. 2.66 DC Network Theorems (ii) Loop Current Method Let the direction of flow of the three loop currents be as shown in Fig. 2.66. Loop ABFA : 83 or - 6/1 - 3(/1 -12) + 6 = 0 3/1-/2 = 2 ...(i) Loop BCEFB : + 5 - 2/2 - 4(/2 -13) - 3 (/2 -II) 3/1 - 9/2 + 4/3 . or = 0 = -5 ...(ii) Loop CDEC : - 4/3- 10- 4 (/3-/2) = 0 or 2/2 - 4/3 =5 ...(iiO The matrixfonn of the abovethree simultaneousequationsis [ 3 - 1 0 ] [ X ] [ 2 ] [ 3 -I O } 3 -9 4 = Y = -5 ;~= 3 -9 4 84-12-0=72 o 2 -4 z 5 0 2-4 I 2 I 0 1 1 3 2 0 1 1 3 -I 2 ~I = - 5 - 9 4 =56; ~2 = 3 - 5 4 =24; ~3 = 3 - 9 - 5 I=- 78 I 5 2 -4 0 5 -4 0 2 5 II =6.1/6.=56n2 =7/9 A; 12 = Aj6. =24n2 =1/3 A 13 = ~/6. =- 78n2 =- 13/12 A The negative sign of 13shows that it is flowing in a direcion opposite to that shown in Fig. 2.64 i.e. it flows in the CCW direction. The actual directions are as shown in Fig. 2.67. The various branch currents are as under: 7 I 4 lAB =/1 = 7/9 A; IBF =/1 -/2 =9"-'3= 9"A .. A 6 B 5V 2 c 4 1 . 1 13 17 IBC=12= 3"A, ICE=12+ 13=3"+12= 12 A 6V 13 I DC = 13 = 12 A Solution by Using Mesh Resistance Matrix From inspection of Fig. 2.67, we have R11 = 9; R22=9; R33 =8 . RI2 = R21=-3Q;~23=R32=-4Q;R13=R;1 =OQ R13 ][ /~ ] 1 = 6 [ ~ ] :E2= [ 5~;E~;-I~ ] V [ /I ] [ 6 ] Rn 12 = E2 or -3 _ 9 -4 12 = _ 5 R33 13 E3 0 4 8 13 10 I 9 -3 0 6. = -3 9 -41=9(72-16)+3(-24)=432 o -4 8 6 -3 0 1 5 9 - 4 =6(72-16) - 5(- 24)-10(12)=336 -10 -4 8 I 9 6 0 ~ = 1 -3 5 -41=9(40-40)+3(48)=144 o -10 8 ~ J JOV F E Fig. 2.67 6.1 = 84 Electrical Technology 9 -3 6 ~ = 1-3 9 51=9(-90-90)-3(30+24)=-468 o -4 -10, I) = J1,il!.= 336/432 = 7/9 A 12 = 1!./1!.= 144/432 = 1/3 A 13 = I!.ll!.= -- 468/432 = - 13/12 A These are the same values as found above. 2.13. Nodal Analysis with Current Sources Consider the network of Fig. 2.68 (a) which has two current sources and three nodes out of which I and 2 are independent ones whereas No.3 is the reference node. The given circuit has been redrawn for ease of understanding and is shown in Fig. 2.68 (b). The current directions have been taken on the assumption that I. both V) and V2are positive with respect to the reference node. That is why their respective curents flow from nodes I and 2 to node 3. 2. V) is positive with respect to V2because current has been shown flowing from node 1 to node 2. A positive result will confirm out assumption whereas a negative one will r . dicate that actual directionis oppositeto that assumed. 1 ~ 2 I ~ 2 ~~ t t 3 (b) Fig. 2.68 We will now apply KCL to each node and use Ohm's law to express branch currents in terms of node voltages and resistances. Node 1 3 (a) 13- 12- 14 = 0 or 13=12 + 14 V2 V. - V. 14 = 11 and 13 =~ - as before "2 - R3 \-I - V2 V2 ( I I ) \-I -.- = 12 + 11 or V2 -+- --=-1) '3 "2. ~ R3 R3 The above two equations can also be written by simple inspection. For example, Eq. (i) is represented by Now, .. ...(ii) I) - 12- 13 = 0 or I) =12+ 13 Now 1-\-1 and I - V)-V2 2 - 3-- R) R3 I - \-I \-I- V2 or V. (1.. + -L )_ V2 =I) .. ) --+ ...(i) R) R3 )R) R3 R3 Node 2 DC Network Theorems 85 1. product of potential VI and (l/RI + 1- R3) i.e. sum of the reciprocals of the branch resistances connected to this node. 2. minus the ratio of adjoining potential V2and the interconnecting resistance R3' 3. all the above equated to the current supplied by the current source connected to this node. This current is taken positive if flowing into the node and negative if flowing out of it (as per sign convention of Art. 2.3). Same remarks apply to Eq. (ii) where 12has been taken negative because it flows away from node 2. In tenns of branch conductances, the above two equations can be put as VI (GI + G3) - V2G3 = II and V2 (G2 + G3) - VIG3 =-/2 Example 2.36. Use nodal analysis method to find currents in the various resistors of the circuit shown in Fig. 2.69 (a). Solution. The given circuit is redrawn in Fig. 2.66'(b) with its different nodes marked 1,2,3 and 4, the last one being taken as the reference or datum node. The different node-voltage equations are as under : ...(i) ...(ii)] ...(iii) Fig. 2.69 The matrix fonn of the above three equations is [ 11 - 5 -1 ] [ X ] [ 280 ] 5 -17 10 y = 0 1 10 - 13.5 z 20 111 -5 -1 ~ = I 5 -17 101=1424.5-387.5-67=970 I 1 10 - 13.5 i 280 - 5 - 1 ~I = 0 - 17 10 I=34, 920, ~2 = , 20 10 -13.5 1 11 - 5 280 I ~3 = 5 -17 C ==15,520 I 1 10 20 (a) 11 280 - 1 5 0 10 I=19,400 1 20 -13.5 Node 1 V. (1 + 1 +.!...)_ V2_ V3 = 8 1.2 2 10 2 10 or 11VI - 5V2- V3- 280 = 0 Node2 V.2 2 5 2 1 (1.+1+1)- VI _ V3 = 0 or 5VI - 17V2+ 10V3 = 0 Node 3 e I)V2 VI V3 '4+1+10 -Y-1O =-2 or VI + 10V2-13.5 V3- 20 = 0 13 1 12 2 14 r12 5 J. 86 Electrical Technology V = ~I =34,920 =36 V V =~z =19,400 =20 V V =~3 =15,520 = 16 V I ~ 970 ' z ~ 970 ' 3 ~ 970 It is obvious that all nodes are at a higher potential with respect to the datum node. The various currents shown in Fig. 2.69 (b) can now be found easily. II = Vl2 =36/2 = 18 A Iz = (VI - Vz)/2=(36 - 20)/2 =8 A 13 = (VI - V3)/1O=(36 - 16)/10=2 A It is seen that total current, as expected, is 18 + 18 + 2 =28 A 14 = (Vz - V3)/1 =(20 - 16)/1 =4 A Is = V/5=20/5=4A,/6=Vi4=16/4=4A Example 2.37. Using nodal analysis. find the different branch currents in the circuit of Fig. 2.70 (a). All branch conductances are in siemens (i.e. mho). Solution. Let the various branch currents be as shown in Fig. 2.70 (b). Using the procedure detailed in Art. 2.11, we have First Node VI (1 + 2) - V2 x 1- V3 = - 2 or 3VI - V2- 2V3=- 2 ...(i) Second Node Vz(1 + 4) - V x 1 = 5 or VI - 5V2=- 5 ...(ii) Third Node V3 (2 + 3) - VI x 2 = - 5 or 2VI - 5V3=5 2S ...(iii) 2S 4S 4S 2A 3S - - (a) Fig. 2.70 Solving for the different voltages, we have 3 7 8 VI = - 2 V, Vz= 10 V and V3= - 5"V II = (VI - Vz) x 1 = (- 1.5 - 0.7) x 1 = - 2.2 A Iz = (V3- VI) x 2 = [- 1.6 - (- 1.5)] x 2 = - 0.2 A 14 = Vzx 4 = 4 x (7/10) = 2.8 A 13 = 2 + 2.8 = 4.8 A As seen, II and Iz flow in directions opposite to those originally assumed (Fig. 2.71). (b) . Fig. 2.71 2S Izlo.2A to.2A IS -3 I 2.2AI: SA 4.8A b2A 48 38 t13 14=2.8A DC Network Theorems 87 Example 2.38. Find the current I in Fig. 2.72 (a) by changing the two voltage sources into their equivalent current sources and then using Nodal method. All resistances are in ohms. Solution. The two voltage sources have been converted into their equivalent current souces in Fig. 2.72 (b). The circuit has been redrawn as shown in Fig. 2.72 (c) where node No.4 has been __15A. Q5A 5A 4V 4V Q) (c) CD I,@ CD G) (b) Fig. 2.72 taken as the reference node or common ground for all other nodes. We will apply KCL to the three nodes and taken currents coming towards the nodes as positive and those going away from them as negative. For example, current going away from node No. 1is (VI - V2)/l and hence would be taken as negative. Since 4 A current is coming towards node No. 1. it would be taken as poisitive but 5 A current would be taken as negative. Node 1: (VI~ 0) _ (VI~ V2) _ I cD ~ + + '30V ( ) 3~ ~ ~ 30V 3 'I '1 ~ 3 v2 ~ sA() 3 - I - 3 5A - 3 (b) Fig. 2.99 Let us now keep 5 A source alive and 'kill' the other two independent sources. KCL to node 1, we get, from Fig. 2.99 (c). 6 + v2 v2 (vzl3 - v2) --5--+ -0 or v =-6V 6 3 2 - 2 Let us now 'kill' 30 V source and 5 A source and find v3 due to 20 V source only. The two parallel resistances of 6 Q and 3 Q can be combined into a single resistance of 2 Q. Assuming a circulating current of i and applying KVL to the indicated circuit, we get, from Fig. 2.100. (a) (e) Again applying 2 20V + + ~ v3 3~: ~ 3 - 2i- 20 - 2i- I (- 2i): 0 or i = 6 A 3 Fig. 2.100 Hence, according to Ohm's law, the component of v that corresponds to 20 V source is v3 =2 x 6 =12 V. :. v =vI + v2 + v3 =6 - 6 + 12 =12 V. Example 2.50. Using Superposition theorem,find the current through the 40 W resistor of the curcuit shown in Fig. 2.101 (a). (F.Y. Engg. Pune Univ. May 1990) Solution. We will first conside when 50 V battery acts alone and afterwards when 10-V battery is alone is the circuit. When 10-V battery is replaced by short-circuit, the circuit becomes as shown in Fig. 2.101 (b). It will be seen that the right-hand side 5 Q resistor becomes connected in parallel with 40 Q resistor giving a combined resistance of 5 1140=4.44 Q as shown in Fig. 101 (c). This 4.44 Q resistance is in series with the left-hand side resistor of 5 Q giving a total resistance of (5 + 4.44) =9.44 Q. As seen there are two resistances of 20 Q and 9.44 Q connected in parallel. In Fig. 2.101 (c) current I =50/9.44 =5.296 A. At point A in Fig. 2.101 (b) there are two resistances of 5 Q and 40 Q connected in parallel, hence, current I divides between them as per the current-divider rule. If II is the current flowing through the 40 Q resistor, then 98 20 5 5 50V 40 10V 50V Electrical Technology 20 5 A 5 I A 5 20 4.44 40 50 V (b) (c) Fig. 2.101 5 5 II = Ix-=5.296x-=0.589A. 5 + 40 45 In Fig. 2.lO2 (a), lOV battery acts alone because 50-V battery has been removed and replaced by a short-circuit. As in the previous case, there are two parallel branches of resistances 20 Q and 9.44 Q across the 10-V battery. Current I through 9.44 Q branch is 1= lO/9.44= 1.059 A. This current divides at point B between 5 Q resistor and 40 Q resistor. Current through 40 Q resistor 12=1.059 x 5/45 =0.118 A. According to the Superposition theorem, total current through 40 Q resistance is =I) + 12=0.589 + 0.118 =0.707A. Example 2.51. Solvefor the power delivered to the 10 Q resistor in the circuit shown in Fig. 2.103 (a). All resistance are in ohms. (Elect. Science - I, Allahabad Univ. 1991) Solution. The 4-A source and its parallel resistance of 15 Q can be converted into a voltage source of (15 x 4) =60 V in series with a 15 Q resistances as shown in Fig. 2.lO3 (b). Now,we will use Superpositiontheoremto findcurrentthroughthe lOQ resistances. When 60- V Sourceis Removed _ AAAA_.__ 5 A 5 When60 - V batteryis removed i the total resistance as seen by 2 V bat- I teryis=l+lOlI(l5+5)=7.67Q. ~1O ~15 I The battery current = 2/7.67 A = 2 '--:-/ I ~ 10 0.26 A. At point A, this current is di- Vvided into two parts. The current passingthrough the 10Q resistor from A to B is (a) 20 20 5 5 B I 4.44 10V 5 40 10V (a) (b) Fig. 2.102 15 60 V (a) B (b) Fig. 2.103 I) = 0.26 x (20/30) =0.17 A When 2-V Battery is Removed Then resistance seen by 60 V battery is =20 + lO III =20.9 Q. Hence, battery current =60/20.9 =2.87 A. This current divides at point A. The current flowing through lO Q resistor from A to B is 12 = 2.87 x 1/(1 + lO)=0.26 A Total current through lO Q resistor due to two batteries acting together is =II + 12=0.43 A. Power delivered to the lO Q resistor =0.432 x lO = 1.85 W. \ DC Network Theorems 99 Example 2.52. Compute thepower dissipated in the 9-W resistor of Fig. 2.104 by applying the Superposition principle. The voltage and current sources should be treated as ideal sources. All resistances are in ohms. Solution. As explained earlier, an ideal constant-voltage sources has zero internal resistances whereas a constant-current source has an infinite internal resistance. (i) When Voltage Source Acts Alone This case is shown in Fig. 2.104 (b) where constant-current source has been replaced by an open-circuit i.e. infinite resitance (Art. 2.16). Further circuit simplification leads to the fact that total resistances offered to voltage source is =4 + (121115) =32/3 Q as shown in FIg. 2.104 (c). A ..~. B' A ..~. B Fig. 2.104 Hence current =32 + 32/3 =3 A. At point A in Fig. 2.104 (d), this current divides into two parts. The part going alone AB is the one that also passes through 9 Q resistor. /' = 3 x 12/(15 + 12) = 4/3 A (ii) When Current Source Acts Alone As shown in Fig. 2.105 (a), the voltage source has been replaced by a short-circuit (Art 2.13). Further simplification gives the circuit of Fig. 2.015 (b). 6 A 6 A I' = 2A 12 I. 4A( t) 9 12 1+ Open 9 Circuit n 32 V - (a) (b) A 6 A r A . . 3A <1 I 4 20 9 12 I. 15 I, 3 32V I I Y32V I ( 132 V B B C B (c) (d) (e) 12 <-4 Q)4A 9 3 4AC! ) 9 Short Circuit I I I I (a) (b) Fig. 2.105 EkctrU:al Technology The 4 -A current divides into two equal parts at point A in Fig. 2.105 (b). Hence 1=4/2 =2 A. Sincebothr and f' flowin the samedirection,totalcurrentthrough9-0 resistoris 1=r +f' = (4/3)+ 2 = (10/3)A Power dissipated in 9 0 resistor =P R =(10/3)2 x 9 =100 W Example 2.53(a). With the help of superposition theorem, obtain the value of current / and voltage Voin the circuit of Fig. 2./06 (a). Solution. We will solve this question in three steps. First, we will find the value of 1 and Vo when current source is removed and secondly, when voltage sources is removed. Thirdly, we would combine the two values of 1and Voin order to get their values when both sources are present. First Step As shown in Fig. 2.106 (b), current source has been replaced by an open-circuit. Let the values of current and voltage due to 10 V source be 11and VOl' As seen II =0 and VOl=10V. Second Step As shown in Fig. 2.106 (c), the voltage source has been replaced by a short circuit. Here 12 =- 5 A and V02=5 x 10 =50 V. I Ion 100 Isn Isn Third Step By applyingsuperpositiontheorem,we have I = /1 +I2=0+(-5)=-5A Vo = VOl+ V02= I~ + 50 = 60 V Example 2.53(b). Using Superposition theorem,find the value of the output voltage Voin the curcuit of Fig. 2.107. Solution. As usual, we will break down the problem into three parts involving one source each. (a) When 4 A and 6 V sources are kiUed* As shown in Fig. 2.108 (a), 4 A source has been replacedby an open circuit and 6 V source by a shortcircuit. Using the cunent-divider rule, we find current iI throughthe 2 0 resistor=6 x I/(l + 2 + 3) 6A =I A :.' VOl=I x 2 =2 V. (b) When 6 A and 6 V sources are killed As shown in Fig. 2.108 (b), 6 A sourceshas been replacedby an open-circuitand 6 V source by a shortcircuit. The current i2 can again be found with the (e) * The process of setting of voltage source of zero is called killing the sources. +T . .. 'II\N n+ +1 + IOV SAm Vo ( )IOV Vol _1 (a) (b) Fig. 2.106 +T f) §In 6v1 Fig.2.107 DC Network TheoremS 101 3 t )6A 10 20 3 6V B ~ ~ ~ ~~~ , According to Superposition theorem, we have Vo = VOl + V02 + V03 =2 + 4 - 4 =2 V Example 2.54. Use Superposition theorem. tofind the voltage V in Fig. 2.109 (a). 40 4 1 12 V 40 4 SC 1--0 A A + + 15 V 10 V 15V 10 O.C B B (a) Fig. 2.109 (b) Solution. The given circuit has been redrawn in Fig. 2.109 (b) with 15 - V battery acting alone while the other two sources have been killed. The 12 - V battery has been replaced by a short-circuit and the current source has been replaced by an open-circuit (O.C) (Art. 2.19). Since the output terminals are open, no current flows through the 4 Q resistor and hence, there is no voltage drop across it. Obviously VI equals the votIage drop over 10 Q resistor which can ,be found by using the voltage-divider rule. VI = 15 x 10/(40 + 50) =3 V Fig. 2.110 (a) shows the circuit when current source acts alone, while two batteries have been killed. Again, there is no current through 4 Q resistor. The two resistors of values 10Q and 40 Q are in parallelacrossthe currentsource. Theircombinedresistancesis 10II40 =8 Q 40 4 SC 40 ~ 4 12V o A ~AAr--r-- 1---0 A + T I + SC 10S ( t)2.5A ~ ISC 10$ O.C. V3 B B (a) Fig. 2.110 (b) 102 Electrical Technology .. V2 = 8 x 2.5 =20 V with point A positive. Fig. 2.110 (b) shows the case when 12 -V battery acts alone. Here, V3 =- 12 V*. Minus sign has been taken because negative tenninal of the battery is connected to point A and the positive tenninal to point B. As per the Superposition theorem, V = VI + V2 + V3=3 + 20 - 12 =11 V Example 2.55. Apply Superposition theorem to the circuit of Fig. 2.107 (a) for finding the voltage drop V across the 5 Q resistor. Solution. Fig. 2.111 (b) shows the redrawn circuit with the voltage source acting alone while the two current sources have been 'killed' i.e. have been replaced by open circuits. Using voltagedivider principle, we get VI =60 x 5/(5 + 2 + 3) =30 V. It would be taken as positive, because current through the 5 Q resistances flows from A to B, thereby making the upper end of the resistor positive and the lower end negative. Fig. 2.111 Fig. 2.112 (a) shows the same circuit with the 6 A source acting alone while the two other sources have been 'killed'. It will be seen that 6 A source has to parallel circuits across it, one having a resistance of2 Q and the other (3 + 5) =8 Q. Using the current-divider rule, the current through the 5 Q resistor =6 x 2/(2 + 3 + 5) =1.2 A. -6A J r'I D.C. * Because Fig. 2.110 (b) resembles a voltage source with an inlemal resislance =4 + 10 II40 = 12 Q and which is an open-circuil a.c. / I A i5 3 6 + D a.c. 5 A 1 60V B B (a) (b) A + 3 cb 53 $3 D.C. 5 V2 2A :J I t sc B B (a) (b) Fig. 2.112 DC Network Theorems 103 .. Vz=1.2 x 5 =6 V. It would be taken negative because current is flowing from B to A. i.e. point B is at a higher potential as compared to point A. Hence, Vz=- 6 V. . Fig. 2.112 (b) shows the case when 2-A source acts alone, while the other two sources are dead. As seen, this current divides equally at point B, because the two parallel paths have equal resistances of 5 Q each. Hence, V3=5 x 1=5 V. It also would be taken as negative becuase current flows from B toA. Hence, V3=- 5 V. Using Superposition principle, we get V = VI + Vz + V3 =30 - 6 - 5 =19 V Example 2.56. (b) Determine using superposition theorem. the voltage across the 4 ohm resistor shown in Fig. 2.113 (a) [Nagpur University, Summer 2000] 2 ohm 8 ohm 13 F B ~ 4 ohm 4 ohm 2 ohm 2 ohm r=O...L+ lOy. G Fig. 2.113 (a) Fig. 2.113 (b) Solution. Superposition theorem needs one source acting at a time. Step I: De-acting current source. The circuit is redrawn after this change in Fig. 2.113 (b) 10 10 II = 4 x (8 + 2) =2 + 40 =2.059 amp 2 + 4 + (8 + 2) 14 2.059 x 10 =1.471. . . Iz = 14 amp, In downward direction Step n: De-activate the voltage source. The circuit is redrawn after the change, in Fig. 2.113 (c) With the currents marked as shown. Id = 2Ie relating the voltage drops in Loop ADC. A. .2.?~~. D8 ohm Ib F H c 4 ohm i5 Amp 2 ohm + IOY o G H Fig. 2.113 (c) c Thus Ib=3 Ie' Resistance of parallel combination of 2x4 2 and 4 ohms = 2 + 4 =1.333 Q Resistance for flow of Ib = 8 + 1.333 =9.333 Q 104 Electrical Technology The 5-amp cUITentfrom the sources gets divided into Ib (= 3 Ie) and la' at the node F. Ib = 3 Ie= _ _ 2.~ _ _ _ X 5 =0.8824 Ie = 0.294 amp, in downward direction. Step m. Apply superposition theorem, for finding the total current into the 4-ohm reistor =Current due to Current source + CUITent due ~o Voltage source =0.294 + 1.471=1.765 amp in downward direction. Check. In the branch AD, The voltage source drives a current from A to D of 2.059 amp, and the cUITentsource drives a current of Id (= 2Ie) which is 0.588 amp, from D to A. The net current in branch AD .. = 2.059 - 0.588 = 1.471 amp With respect to 0, A is at a potential of + 10 volts. Potential of D with respect to 0 == (netcurrentin resistor)x 4 = 1.765 x 4 =+ 7.06 volts Between A and D, the potential difference is (10 - 7.06)volts Hence, the cUITenthro' this branch ...eqn. (a) = 10 -7.06 =1.47 amp from A to D ...eqn (b) . This is the same as eqn. (a) and hence checks the result, obtained previously. Example 2.57. Find the currentflowing in the branch XY of the circuit shown in Fig. 2.114 (a) by superposition theorem. [Nagpur University, April 1996] Solution. As shown in Fig. 2.114 (b), one source is de-activated. Through series-parallel combinations of resistances, the cUITentsdue to this source are calculated. They are marked as on Fig. 2.114 (b) 20 x 10 20 --./'II\M 2.67A X-r4A 1.33A 1.33 A 11.33 A + -- 20V T10 30 30 10 30 -=20 V . 60 ~20V 1.33 A 2.67 A y 20 Fig. 2.114 (b) y Fig. 2.114 (a) 20 X 4 A ~ 1.33A -=-20 V 30 <:60 4A 2.67A ...LY Fig. 2.114 (c) 1.33A DC Network Theorems 105 In the next step, sei:ond source is de-activated as in Fig. 2.114 (c). Through simple senes parallel resistances combinations, the currents due to this source are marked on the same figure. According to the super-position theorem, the currents due to both the sources are obtained after adding the individual contributions due to the two sources, with the fmal results marked on Fig. 2.114 (a). Thus, the current through the branch XYis 1.33 A from Y to X. Example 2.58. Find the currents in all the resistors by Superposition theorem in the circuit shown in Fig. 2.115 (a). Calculate the power consumed. [Nagpur University, Nov. 1996] Solution. According to Superposition theroem, one source should be retained at a time, deactivating remaining sources. Contributions due to individual sources are fmally algebraically added to get the answers required. Fig. 2.115 (b) shows only one source retained and the resultant currents in all branches/elements. In Fig. 2.115 (c), other source is shown to be in action, with concerned currents in all the elements marked. To get the total current in any element, tow component-currents in Fig. 2.115 (b) and Fig. 2.15 (c) for the element are to be algebraically added. The total currents are marked on Fig. 2.115 (a). A In In 0.7147A 2n 0.7143 D Fig. 2.115 (a) All resistors are in ohms A In 3n 0.7143A Fig. 2.115 (c) Power loss calculations. (l) from power consumed by resistors : 222 Power =(0.7147 x 4) + (3.572 x 2)+(2.875 x 8) =92.86 watts (ij) From 'Source-power. Power =10 x 3.572 + 20 x 2.857 =92.86 watts Tutorial Problems No. 2.4. 1. Apply'the principle-of Superposition to the network shown in Fig. 2.116 to find out the current in the 10 n resistance. [0.464 A] (F.Y. Engg. Pune Univ. May 1987) 2. Find the current through the 3 n resistance connected between C and D Fig. 2.117. [1 A from C to D] (F.Y. Engg. Pune Univ. May 1989) B 2.8n C 3.S72A ..£857 A A . B .n 1.5 + 20 V .. :ti..!..0v ;o 10.2n 2n o.sn l 10V 0 sn D I 12.143 A ( 1.429A o NW\Jsn Fig. 2.115 (b) B 3n 20 20V 1.429A 12.143 A 0 .. 's'0 D 3. Using the Superposition theorem, calculate the magnitude and direction of the current through each resistor in the circuit of Fig. 2.118. [11=6n A; 12=IOn A; 13=16n A] 4. ForthecircuitshowninFig.2.119find the cur- 480 rent in R =8 Q resistance in the branch AB using superposition theorem. [0.875A] (F.Y.Engg.PuneUniv.May 1988) 5. Apply superposition principle to conpute current in the 2-Q resistor of Fig. 2.120. All resis- -t28V 480 tors are in ohms. [lab =5 A] 6. Use Superpositiontheorem to calculate the voltage drop across the 3 Q resistor of Fig. 2.121. All resistance values are in ohms. [18 V] 106 3Q 1.50 c 30 30 9V 3 0 4.5 V Fig. 2.116 D Fig. 2.117 a 2 b t J24A 4 6 6A( t + Fig. 2.120 Electrical Technology 18V D Fig. 2.118 A 60 80 80 40 50 120 14V B Fig. 2.119 6 , 2 3 Fig. 2.121 7. With the help of Superposition theorem, compute the current lab in the circuit of Fig. 2.122. All resistances are in ohms. [lab= - 3 A] 10 5 a b 10 5 12A( t 60A ( t 30V a 20V b 50 ISO t )20A Fig. 2.122 Fig. 2.123 8. Use Superposition theorem to find current labin the circuit of Fig. 2.123. All resistances are in ohms. [100 A] 9. Find the current in the 150 resistor of Fig. 2.124 by using Superposition principle. Numbers represent resistances in ohms. [2.8 A] 10. Use Superposition principle to find current in the 10-0 resistor of Fig. 2.125. All resistances are in ohms. [1 A] DC Network Theorems 107 11. State and explain Superposition theorem. For the circuit of Fig. 2.126. (a) determine currents II' 12and 13when switch S is in position b. (b) using the results of part (a) and the principle of superposition, determine the same currents with switch S in position a. [(a) 15 A, 10 A, 25 A (b) 11 A ,16 A, 27 A] (Elect. Technology Vikram Univ. 1978) 30 25 15 It h 30V 15 25V 10 2 _120 V I~ 20V 4 10 50 V 60V Fig. 2.124 Fig. 2.125 Fig. 2.126 2.1iI'."fhevenin Theorem -/ It provides a mathematical technique for replacing a given network, as viewed from two output terminals, by a single voltage source with a series resistance. It makes the solution of complicated networks (particularly, electronic networks) quite quick and easy. The application of this extremely useful theorem will be explained with the help of the following simple example. R. C A .~}. C A .~!. C A r r r E E D B (b) Fig. 2.127 Suppose, it is required to find current flowing through load resistance RL'as shown in Fig. 2.127 (a). We will proceed as under: 1. Remove RLfrom the circuit terminals A and B and redraw the circuit as shown in Fig. 2.127 (b). Obviously, the terminals have become open-circuited. · 2. Calculate the open-circuit voltage Vocwhich appears across terminals A and B when they are open i.e. when RL is removed. As seen, Voc=drop across R2=lR2whereI is thecircuitcurrentwhenA andB are open. I = _; :. Vue= lR2 = R £;2 [ris the internal resistance of battery] .+ 2+r .+ 2+r It is also called 'Thevenin voltage' Vth. 3. Now, imagine the battery to be removed from the circuit, leaving its internal resistance r behind and redraw the circuit, as shown in Fig. 2.127 (e). When viewed inwards from terminals A and B, the circuit consists of two parallel paths: one containing R2 and the D (a) B D (e) B * After the French engineer M.L. Thevenin (1857-1926) who while working in Telegraphic Department published a statement of the theorem in ]893. 108 Electrical Technology other containing (Rt + r). The equivalent resistance of-the network, as viewed from these terminals is given as ~(Rt + r) R = R211(Rt + r) = D2 + (Rt + r This resistance is also called, *Thevenin resistance RSh(though, it is also sometimes written as Rj or Ro). Consequently, as viewed from terminals A and B, the whole network (excluding Rt) can be reduced to a single source (called Thevenin's source) whose e.m.f. equals Vo<(or Vsh) and whose internal resistance equals RSh(or Rj) as shown in Fig. 2.128. Thevenin 4. RL is now connected back across terminals A and B Source from where it was temporarily removed earlier. Current flowing through RL is given by V,h 1- - - Rth + RL It is clear from above that any network of resistors and voltage sources (and current sources as well) when viewed from any points A and B in the network, can be replaced by a single voltage source and a single resistance* in series with the voltage source. After this replacement of the network by a single voltage source with a series resistance has been accomplished, i~is easy to find current in any load resistance joined across terminals A and B. This theorem is valid even for those linear networks which have a nonlinear load. Hence, Thevenin' s theorem, as applied to d.c. circuits, may be stated as under : The current flowing through a load resistance RL connected across any two tenninals A and B of a linear, acti~e bilateral network is given by VocII(Rj + RJ where Vocis the open-circuit voltage (i.e. voltage across the two tenninals when RL is removed) and Rj is the internal resistance of tM network as viewed back into the open-circuited network from tenninals A and B with all voltage sources replaced by their internal resistance (if any) and current sources by infinite resistance. 2.19. How to Thevenize a Given Circuit? I A B Fig. 2.128 " 1. Temporarily remove the resistance (called load resistance RJ whose current is required. 2. Find the open-circuit voltage Vocwhich appears across the two terminals from where resistance has been removed. It is also called Thevenin voltage Vth. 3. Compute the resistance of the whose network as looked into from these two terminals after all voltage sources have been removed leaving behind their internal resistances (if any) and current sources have been replaced by open-circuit i.e. infinite resistance. It is also called Thevenin resistance Rthor Tj' 4. Replace the entire network by a single Thevenin source, whose voltage is Vthor Vocand whose internal resistance is Rthor Ri. 5. Connect RL back to its terminals from where it was previously removed. 6. Finally, calculate the current flowing through RLby using the equation. I = VtJ(Rth + RL) or I = V,)(Rj + RL) Example 2.59. Convert the circuit shown in Fig. 2.129 (a), to a single voltage source in series with a single resistor. . (AMIE Sec. B, Network Analysis Summer 1992) * Or impedance in the case of a.c. circuits. DC Network Theorems Solution. Obviously, we have to fmd equivalent Thevenin circuit. For this purpose, we have to calculate (i) V,hor VABand (ii) R'hor RAB' With tenninals A and B open, the two voltage sources are connected in subtractive series because they oppose +. each other. Net voltage around the circuit is (15 - 10) =5 V and total F resistance is (8 + 4) = 12 Q. Hence circuit current is =5/12 A. Drop across 4 Q resistor =4 x 5/12=5/3 V with the polarity as shown in Fig. 2.129 (a). .. VAB = V,h =+ 10 + 5/3 =35/3V. Incid~ntly, we could also fmd VABwhile going along the parallel route BFEA. Drop across 8 Q resistor =8 x 5/12 =10/3 V. VAB equal the algebraic sum of voltages met on the way from B to A. Hence, VAB=(- 10/3) + 15 =35/3 V. As shown in Fig. 2.129 (b), the single voltage source has a voltage of 35/3 V. For findign R'h' we will replace the two voltage sources by short-circuits. In that case, R'h =RAB =4118=8/3Q. c A 109 A tOv 35/3 V B B Fig. 2.129 (b) Example 2.60. State Thevenin's theorem and give a proof. Apply this theorem to calculate the current through the 4 Q resistor of the circuit of Fig. 2.130 (a). (A.M.I.E. Sec. B Network Analysis W. 1989) Solution. As.shown in Fig. 2.130 (b), 4 Q resistance has been removed thereby open-circuiting the terminals A and B. We will now fmd VABand RABwhich will give us V,hand R'hrespectively. The potential drop across 5 Q resistor can be found with the help of voltage-divider rule. Its value is =15 x 5/(5 + 10) =5 V. 10 4 10 A B .I.q. A BI) + + + 15V + 5~5V 6V £ (c) 15 V 5 6V (b) Fig. 2.130 For finding VAB'we will go from ponit B to point A in the clockwise direction and find the algebraic sum of the voltages met on the way. .. VAB = - 6 + 5 =- 1 V. It means that pointA is negative with respect to point E, or point B is at a higher potential than point A by one volt. In Fig. 2.130 (c), the two voltage source have been shortcircuited. The resistance of the network as viewed from points A and B is the same as viewed from points A and C. (a) A 4 IV 1 B Fig. 2.131 no Electrical Technology .. RAB = RAC =5 1110=10/3 Q Thevenin's equivalent source is shown in Fig. 2.131 in which 4 Q resistor has been joined back across tenninals A and B. Polarity of the voltage source is worth noting. I 3 I = (10/3) + 4 =22 =0.136 A .. From E to A Example 2.61. With reference to the network of Fig. 2.132 (a), by applying Thevenin's theorem find thefollowing: (i) the equivalent e.m.j of the network when viewedfrom terminals A and B. (ii) the equivalent resistance of the network when looked intofrom terminals A and.B. (iii) current in the load resistance RL of 15 Q. (Basic Circuit Analysis, Nagpur Univ. 1993) Solution. (i) Current in the network before load resistance is connected [Fig. 2.132 (a)] = 24/(12 + 3 + l) =1.5 A :. voltage across tenninals AB = Voc = Vth =12 x 1.5=1.8 V Hence, so far as tenninals A and B are concerned, the network has an e.m.f. of 18 volt (and not 24 V). (ii) There are two parallel paths between points A and B. Imagine that battery of 24 V is removed but not its internal resistance. Then, resistance of the circuit as looked into from point A and B is [Fig. 2.132 (c)] R; = Rth =12 x 4/(12 + 4) =3 Q (iii) When load resistance of 15 Q is connected across the tenninals, the network is reduced to the structure shown in Fig. 2.132 (d). Fig. 2.132 1 = Vtl,l(Rth+RL)= 18/(15+ 3) = 1 A Example 2.62. Using Thevenin theorem, calculate the currentflowing through the 4 Q resistor of Fig. 2.133'(a). Solution. (i) Finding Vth If we remove the 4-Q resistor, the circuit becomes as shown in Fig. 2.133 (b). Since full 10 A current passes through 2 Q resistor, drop across it is 10 x 2 =20 V. Hence, VB=20 V. Hence, VB= 20 V with respect to the common ground. The two resistors of 3 Q and 6 Q are connected in series across the 12 V battery. Hence, drop across 6 Q resistor =12 x 6/(3 + 6) =8 V. . . VA = 8 V with respect to the common ground* .. Vth = VBA =VB- VA =20 - 8 =12 V-with B at a higher potential * Also. VA=12- drop across 3-0 resistor =12- 12x 3/(6+ 3) = 12- 4 = 8 V 3 A 3 A 3 A I A I r I 3 24V 12 15 rv . "1 r 1 ,, 1 15 r=1n r=1n LL 110 v . 1 B B B B (a) (b) (c) (d) DC Network Theorems \ 111 (ii) Finding Rth Now, we will find Rthi.e. equivalent resistnace of the network as looked back into the opencircuited terminals A and B. For this purpose, we will replace both the voltage and current sources. Since voltage source has no internalresistance, it would be replacedby a short circuit i.e. zeroresistance. However, current source would be removed and replaced by an 'open' i.e. infinite resistance (Art. 1.18). In that case, the circuit becomes as shown in Fig. 2.133 (c). As seen from Fig. 2.133 (d), Fth= 6 II3 + 2 = 4 Q. Hence, Thevenin's equivalent circuit consists of a voltage source of 12V and a series resistance of 4 Q as shown in Fig. 2.134 (a). When 4 Q resistor is connected across terminals A and B, as shown in Fig. 2.134 (b). 1= 12/(4 + 4) = 1.5 A-from B to A Example 2.63. For the circuit shown in Fig. 2.135 (a), calculate the current in the 10 ohm resistance. Use Thevenin's theorem only. (Elect. Science-I Allahabad Univ. 1992) Solution. When the 10 Q resistance is removed, the circuit becomes as shown in Fig. 2.135 (b). (a) (b) Fig. 2.133 I A I;~1:48 Now, we will find the open-circuit voltage VAB=Vth. For this purpose, we will go from point B to point A and find the algebraic sum of the voltages met on the way. It should be noted that with termnals A and B open, there is no voltage drop on the 8 Q resistance. However the two resistances of 5 Q and 2 Q are connected in series across the 20-V battery. A~ per voltage-divider rule, drop on 2 Q resistance =20 x 2/(2 + 5) =5.71 V with the polarity as shown in figure. As per the sign convention of Art. (c) (d) ((a) B A 4 2V (b) B Fig. 2.134 8 A 5 2 B B (c) A 10 I B Fig. 2.136(0) 8 'I A D 8 I 2 wi 5_ 20VU 1 20 B D C (a) (b) Fig. 2.135 112 I Electrical Technology VAS = V,h =+ 5.71 - 12=- 6.29V The negative sign shows that point A is negative with respect to point B or which is the same thing, point B is positive with respect to point A. For finding RAS=Rth'we replace the batteries by short-circuits as shown in Fig. 2.128 (c). :. RAS = Rth =8 + 2 II5=9.43 Q Hence, the equivalent Thevenin' s source with respect to tenninals A and B is an shown in Fig. 2.136. When 10 Q resistance is reconnected across A and B, current through it is 1= 6.24/9.43 + 10) = 0.32A. Example 2.64. Using Thevenin's theorem, calculaie the p.d. across terminals A and B in Fig. 2.137 (a). Solution. (i) Finding VDC First step is to remove 7 Q resistor thereby open-circuiting terminals A and B as shown in Fig. 2.137 (b). Obviously, there is no current through the 1 Q resistor and hence no drop across it. Therefore VAS=VDC= VCD'As seen, current I flows due to the combined action of the two batteries. Net voltage in the CDFE circuit =18 - 6= 12V. Totalresistance= 6 + 3 = 9 Q. Hence,1= 12/9= 4/3 A .. VCD = 6 V + drop across 3 Q resistor =6 + (4/3) x 3 =10 V* VDC= V'h= 10V. (ii) Finding Rj or R'h As shown in Fig. 2.137 (c), the two batteries have been replaced by short-circuits (SC) since their internal resistances are zero. As seen, Rj =R'h=1 + 3 II6 =3 Q. The Thevenin's equivalent circuit is as shown in Fig. 2.137 (d) where the 7 Q resistance has been reconnected across terminals A and B. The p.d. across this resistor can be found with the help of Voltage Divider Rule (Art. 1.15). I A 7Q (d) Example 2.65. Use Thevenin's theorem to find the current in a resistance load connected between the terminals A and B of the network shown in Fig. 2.138 (a) ifthe load is (a) 2 Q (b) J Q. (Elect. Technology, Gwalior Univ. 1987) Solution. For findingopen-circuitvoltageVDC or VthacrossterminalsA andB, we must first find current 12flowing through branch CD. Using MaxweU's loop current method (Art. 2.11), we have from Fig. 2.131 (a). - 2I, - 4 (I,- 12)+ 8 = 0 or Also - 2 12- 2 12- 4 - 4 (12- I,) = 0 or From these two equations, we get 12=0.25 A As we go from point D to C, voltage rise =4 + 2 x 0.25 =4.5 V Hence, VCD=4.5 or VAS=V,h =4.5 V. Also, it may be noted that point A is positive with respect topointB. 3I,- 2 12=4 I, - 212=1 I. = I>!-dropacross6 n resIstor=18 - (4/3) x 6 = 10 V E 6 C I 6 I .A A A 7 j"ViJ:-6V+ :! f r. S.C 10 d (b) . Fig. 2.165 For finding the Thevenin's resistance with respect to tenninals a and b, we would first 'kill' the independent voltage source as shown in Fig. 2.165 (b). However, the dependent current source cannot be 'killed'. Next, we will connect a current source of I A at terminals a and b and find the value of vab. Then, Thevenin' s resistance Rth=vat/I. It will be seen that current flowing away from node A i.e. from point c to d is = vat/IOO.Hence,i = - vo/lOO.ApplyingKCLto nodeA, weget vab 9 ( vab ) vab --+ -- --+1 = 0 or v =5V 100 100 10 ab :. Rth= 5/1 = 5 a. Hence, Thevenin's equivalent source is as shown in Fig. 2.165 (c). Example 2.85. Find the Thevenin's equivalent circuit with respect to terminals a asnd b of the network shown in Fig. 2.166 (a). All resistances are in ohms. Solution. It should be noted that with terminals a and b open, potential of node A equals vab' Moreover, v = vab' Applying KCL to node A, we get - 5 - ~a; + 1~ [( V3b + 150) - Vab]= 0 or Vab= 75 V 10 I~V+ A 30 10 A 30 a Fig. 2.166 For finding Rth'we will connect a current s.ourceof iA* across tenninals a and b. It should be particularly noted that in this case the potential of node A equals (vab- 30 i). Also, v =(vab - 30 i) = potential of node A, Applying KCL to node A, we get from Fig. 2.166 (b). i - (vab.~_~?-'l+ 1.. [( vab - 30 i ) - (V b - 30 i) ] =0 - 15 10 3 a .. 4 vab = 150 i or vat/i = 75/2 a. Hence, Rth = vat/i = 75/2 n. The Thevenin's equivalent circuit is shown in Fig. 2.166 (c). 2.21. Reciprocity Theorem It can be stated in the following manner : In any linear bilateral network, if a source of e.m.f. E in any branch produces a current I in any * We could also connect a source of 1 A as done in Ex. 2.83. DC Network Theorems 80 c 20 A oa + 6A, I 10:(: Voc d b (a) 125 a <:>--, I oa + IA vabci ,.... 3V b b (c) - a I a + I ;L cbSA + "'1 + TI Y75V b v 3 (c) b (a) (b) b 126 Electrical Technology other branch, then the same e.m.f. E acting in the second bamch would produce the same current I in thefirst branch. In other words, it simply means that E and I are mutually transferrable. The ratio Ell is known as the transferresistance (or impedance in a.c. systems). Another way of stating the above is that the receiving point and the sending point in a network are interchangebale. It also means that interchange of an ideal voltage sources and an ideal ammeter in any network will not change the ammeter reading. Same is the case with the interchange of an ideal current source and an ideal voltmeter. Example 2.86. In the netwrok of Fig. 2.167 (a),find (a) ammeter current when battery is at A and ammeter at B and (b) when battery is at B and ammeter at point A. Values of various resitances are as shown in diagram. Also, calculate the transfer resistance. Solution. (a) Equivalent resistance between .f.. C 3 points C and B in Fig. 2.167 (a) is = 12 x 4/16 =3 0. +E . . AT36V .. Total CircUltreistance = 2+3+4=90. .. Batterycurrent = 36/9=4 A .. Ammetetercurrent = 4 x 12/16 =3 A. (b) Equivalentresistancebetweenpoints C andD in Fig.2.167(b)is = 12 x 6/18 =4 0. Total circuit resitance = 4 + 3 + 1 =8 0. Battery current = 36/8 =4.5 A .. Ammeter current = 4.5 x 12/18 =3 A Hence, ammeter current in both cases in the same. Transfer resistance = 36/3 =12 Q. Example 2.87. Calculate the currents in the various branches of the network shown in Fig. 2.168 and then utilize the principle of Superpositon and Reciprocity theorem together to find the value of the current in the I-volt battery circuit when an e.mj of 2 votls is added in branch BD opposing theflow of original current in that branch. Solution. Let the currents in the various branches be as shown in the figure. Applying Kirchhoff s second law, we have For loop ABDA ;- 211- 813+ 612=0 or II - 312+ 413=0 For loop BCDB, - 4 (/1 - 13) + 5 (/2 + 13)+ 813=0 or 411- 512- 17f3 =0 For loop ABCEA, - 211- 4(/1 - 13)- 10(/1 + 12) + I =0 or 1611+ 1012":,,,413 = 1 Solving for II' 12 and 13, we get II =0.494 A; 12 =0.0229 A; 13=0.0049 A 2 12 4 4 (a) D Fig. 2.167 B B (1\+'2) D E Ion 'I-- 'IV Fig. 2.168 D E Ion I-- IV Fig. 2.169 C 3 12 (b) ...(i) ...(ii) ...(iii) DC Network Theorems 127 .. Current in the 1 volt battery circuit is II + 12=0.0723A. The new circuit having 2 -V battery connected in the branch BD is shown in Fig. 2.169. According to thePrinciple of Superposition,the new current in the 1-voltbattery circuit is due to the superposition of two currents; one due to 1 - volts battery and the other due to the 2 - volt battery when each acts independently. The current in the external circuit due to 1 - volt battery when 2 - battery is not there, as found above, is 0.0723 A. Now, according to Reciprocity theorem; if 1 - volt battery were tansferred to the branch BD (where it produced a current of 0.0049 A), then it would produce a current of 0.0049 A in the branch CEA (where it was before). Hence, a battery of 2 -V would produce a current of (- 2 x 0.0049) =- 0.0098 A (by proportion). The negative sign is used because the 2 - volt battery has been so connected as to oppose the current in branch BD, .. new current in branch CEA =0.0723- 0.0098=0.0625 A Tutorial Problems No. 2.5 1. Calculate the current in the 8-W resistor of Fig. 2.170 by using Thevenin's theorem. What will be its value of connections of 6-V battery are reversed? [0.8 A ; 0 A] 2. Use Thevenin's theroem to calculate the p.d. across terminals A and B in Fig. 2.171. [1.5 V] 2 A A ?II 5 A Fig. 2.170 Fig. 2.171 Fig. 2.172 3. Compute the current flowing through the load resistance of 10Q connected across terminals A and B in Fig. 2.172 by using Thevenin' s theorem. 4. Find the equivalent Thevenin voltage and equivalent Thevenin resistance respectively as seen from open-circuited terrniansA and B to the circuits shown in Fig. 2.173. All resistances are in ohms. A 4 B (c) 30 (j) Fig. 2.173 [(a) 8 V, 6 Q; (b) 120V, 6 Q; (c) 12V, 6 Q; (d) 12V, 20Q; (e)- 40 V, 5 Q; if) - 12V, 30 Q] 6 16 r 8 1 3 5 10 10 10 - 6 6V T6V T6V T4.5 5 0 B B B A 4 6 12VI B (a) 10 15 24V A 10 B (d) A 4 3 6 IOOV -i (b) A 128 Electrical Technology 5. FindThevenin'sequivalentof thecircuitsshownin Fig.2.174betweentenninalsA andB. RR R RR V,R+VR RR [(a) V,h=1 L.L + V L- ; R,h= L.L (b) V,h= I 2 2 I ;R,h =-LL ~+~ ~+~ ~+~ ~+~ ~+~ (c) V,h =- IR; R'h =Rt (d) V,h =- VI - IR, R'h =R (e) Not possible] ~ ~ ~ Fig. 2.174 6. The four arms of a Wheatstone bridge have the following resistances in ohms. AB = 100,BC= 10,CD= 5,DA = 60 A galvanometer of 15 ohm resistance is connected across BD. Calculate the current through the galvanometer when a ptential difference of 10 V is maintained across AC. [Elect. Engg.A.M.Ae.S.I.Dec.1991][4.88mA] 7. Find the Thevenin equivalent circuit for the network shown in Fig. 2.175. [(a) 4 V; 80 (b) 6 V; 3 a (e") OV; 2/50] a b ~ ~ ~ Fig. 2.175 8. Use Thevenin's theorem to find current in the branchAB of the network shown in Fig. 2.176. [1.84 A] 4 A 3 2 1 3 -=-10V 2 2V 12 A B 4V 4 B Fig. 2.176 Fig. 2.177 A . . oA I , V ( IV, ( 1f2 ,B I I oB (a) (b) A J - A A B B a I T"",,"'- I a 2Vab 2 b Y2V 'j 6 2 b DC Network Theorems 129 9. In the network shown in Fig. 2.177 find the current that would flow if a 2-Q resistor were connected between points A and B by using. (a) Thevenin's theorem and (b) Superposition theorem. The two batteries have negligible resistance. [0.82 A] 10. State and explain Thevenin's theorem. By applying Thevenin's theorem or otherewise, find the current through the resistance R and the voltage across it when connected as shown in Fig. 2.178. [60.49A, 600.49V] (Elect. and Mech. Technology, Osmania Unvi. Dec. 1978) 5 A 3 10 20 15 B Fig. 2.178 Fig. 2.179 11. State and exaplin Thevenin's theroem. For the circuit shown in Fig. 2.179, determine the current through RLwhen its value is 50 Q. Find the value of RL for which the power drawn from the source is maximum. (Elect. Technology 1, Gwalior Univ. Nov. 1979) 12. Find the Thevenin's equivalent circuit for terminal pair AB for the netwrok shown in Fig. 2.180. [VI/,= - 16 V and Rth = 16 Q] 15 A 2 5 2 A -=-20Y 10 IOY 4 12Y 6 6 4 B B Fig. 2.180 Fig. 2.181 Fig. 2.182 13. For the circuit shown in Fig. 2.181, determine current through RLwhen it take$ values of 5 and 10 Q. [0.588 A, 0.408 A] (Network Theorem and Fields, Madras Univ. 1980) 14. Determine Thevenin's equivalent circuit which may be used to represent the network of Fig. 2.182 at the terminals AB. [Vth =4.8 V. Rth =2.4 Q] 15. For the circuit shown in Fig. 2.183 find Thevenin's equivalnet circuit for terminal pair AB. [6 V, 6 Q] A 6 3 3 o B 6 Fig. 2.184 5:; $15 c:-10 R=IO 1 30Y TSOY m RL 2 30 4 6 ,IT 6V 12Y Fig. 2.183 130 Electrical Technology 16. ABCDis a rectanglewhoseoppositesideAB andDCrepresentresistancesof 6 Q each,whileAD and BC represent3 Q each. A batteryof e.m.f.4.5 V and negligibleresistancesis connectedbetween diagonal points A and C and a 2 -Q resistance between B and D. Find the magnitude and direction of thecurrentin the2-ohmresistorbyusingThevenin'stheorem.The positiveterminalis connectedto A. (Fig.2.184) [0.25A from D to B] (BasicElectricityBombayUniv.Oct.1977) 2.22. Delta/Star* Transformation In solving networks (having considerable number of branches) by the application of Kirchhoff s Laws, one sometimes experiences great difficulty due to a large number of simultaneous equations that have to be solved. However, such complicated network can be simplified by successively replacing delta meshes by equivalent star system and vice versa. Suppose we are given three resistance RI2' R23and R31connected in delta fashion between terminals 1,2 and 3 as in Fig. 2.185 (a). So far as the respective terminals are concerned, these three given resistances can be replaced by the three resistances RJ' R2and R3 connected in star as shown in Fig. 2.185 (b). These two arrangements will be electrically equivalent if the resistance as measured between any pair of terminals is the same in both the arrangements. Let us find this condition. 3 2 3 Fig. 2.185 . First, take delta connection: Between terminals 1 and 2, there are two parallel paths; one having a resistance of RI2 and the other having a resistance of (R12+R31). . R x (R + R ) .. ResIstance between terminals 1 and 2 is = nl2 (~3 ;1 12+ 3+ 31 Now, take star connection: The resistance between the same terminals 1 and 2 is (R1+ R2). As terminal resistances have to be the same R + R _ R12~ ~~~!!"_R31) .. 1 2- RI2 + R23 + R31 Similarly, for terminals 2 and 3 and terminals 3 and I, we get ~3 x (R31+ R12) R2 + R3 = -.-. RI2 + R23 + R31 R31 x (R12 + R23) R +R - -- -- 3 1 - RI2 + R23 + R31 Now, subtracting (ii) from (i) and adding the result to (;;i), we get R R R R R R RI=--1L JI_ ; R2= 1Ll.L - and R3= 31 23 RI2+ R23+ R31 RI2+ ~3 + R31 RI2+ R23+ R31 (0) (b) ...(i) ...(ii) and ...(iii) * In Electronics, star and delta circuits are generally referred to as T and 1tcircuits respectively. DC Network Theorems 131 How to Remember? It is seen from above that each numerator is the product of the two sides of the delta which meet at the point in star. Hence, it should be remembered that: resistance of each arm of the star is given by the product of the resistances of the two delta sides that meet at its end divided by the sum of the three delta resistances. 2.23. StarlDelta Transformation . This tarnsformation can be easily done by using equations (i), (ii) and (iii) given above. Multiplying (i) and (ii), (ii) and (iii), (iii) and (i) and adding them together and then simplifying them, we get How to Remember? The equivalent delta resistance between any two terminals is given by the sum of star resistances between those terminals plus the product of these two star resistances divide by the third star resistances. Example 2.88. Find the input resistance of the circuit between the points A and B of Fig 2.186(a). (AMIE Sec. B Network Analysis Summer 1992) Solution. For finding RAB'we will convert the delta CDE of Fig. 2.186 (a) into its equivalent star as shown in Fig. 2.186 (b). Res =8 x 4/18 =16/9 n; RES' =8 x 6/18 =24/9 n; RDS =6 x 4/18 = 1219n. The two parallel resistances between S and B can be reduced to a single resistance of 35/9 n. A .~.. c .1 A~" C .1 A~" C . (' ~. ~ ~ Fig 2.186 As seen from Fig. 2.186 (c), RAB= 4 + (16/9) + (35/9) =87/9n. Example 2.89. Calculate the equivalent resistance between the terminalsA and B in the netY.'rok shown in Fig. 2.187 (a). (F.Y. Engg. Pone Univ. May 1987) Solution. Thegivencircuitcanbe redrawnas shownin Fig.2.187(b). Whenthe deltaBCDis converted to its equivalent star, the circuit becomes as shown in Fig. 2.187 (c). Each arm of the delta has a resistance of 10 n. Hence, each arm of the equivalent star has a resistance= lOx 10/30= 10/3n. Asseen,threearetwoparallelpathsbetweenpointsA andN, each having a resistance of (10 + 10/3) =40/3 n. Their combinedresistance is 20/3 n. Hence, RAB = (20/3)+ 10/3= 10n. Electrical Technology \U 10 (c) B Example 2.90. Calculate the current flowing through the 10 Q resistor of Fig. 2.188 (a) by using any method. (NetworkTheory, Nagpur Univ. 1993) Solution. It will be seen that there are two deltas in the circuit i.e.ABC andDEF. They have been converted into their equivalent stars as shown in Fig. 2.188 (b). Each ann of the delta ABC has a resistance of 12 Q and each ann of the equivalent star has a resistance of 4 Q. Similarly, each ann of the delta DEF has a resistance of 30 Q and the equivalent star has a resistance of 10 Q per ann. The total circuit resistance between A and F =4 + 48 II24 + 10 =30 Q. Hence I =180/30 =6 A. Current through 10 Q re~istoras given by current-divider rule =6 x 48/(48 + 24) =4 A. Example 2.91. A bridge network ABCD has arms AB, BC, CD and DA of resistances 1, 1, 2 and 1 ohm respectively. If the detector AC has a resistance of 1 ohm, determine by star/delta transformation, the network resistance as viewedfrom the battery terminals. (Basic Electricity, Bombay Univ. 1980) A A O.SO o---J./IND P'I) D C I C c (a) (b) (c) (d) Fig. 2.189 Solution. As shown in Fig. 2.189 (b), delta DAC has been reduced to its equivalent star. 2xl 1 2 RD = ,', 1 =0.5Q, RA=4"=0.25Q, Rc =4"=O.5Q 132 10 10 NI'- A I }I!. r .!\'. I 1U B A I 10 CD A 10 10 vv-----' 10 i (b) B (a) Fig. 2.187 6A I 1180V J 4 A 34 I 10 2A B D 4 I 10 (a) I 4A 10 C . WI' Fig. 2.188 I E (b) DC Network Theorems 133 Hence, the original network of Fig. 2.189 (a) is reduced to the one shown in Fig. 2.189 (d). As seen, there are two parallel paths between points N and B, one of resistance 1.25 Q and the other of resistance 1.5 Q. Their combined resistance is = 1.25x 1.5 =.!i Q 1.25+ 1.5 22 Total resistance of the network between points D and B is 15 13 = 0.5 + 22 =11n Example 2.92. A network of resistances isformed asfollows as in Fig. 2.190 (a) AB = 9 Q; BC = 1 Q; CA = 1.5 Qforming a delta and AD = 6 Q ; BD = 4 Q and CD =3 Q forming a star. Compute the network resitance measured between (i) A and B (ii) Band C and (iii) C and A. (Bask electricity, Bombay Univ. 1980) A A A 9 10 (c) c 10 (b) Fig. 2.190 Solution. The star of Fig. 2.190 (a) may be converted into the equivalent delta and combined in parallel with the given delta ABC. Using the rule given in Art. 2.22, the three equivalent delta resistance of the given star become as shown in Fig. 2.190 (b). When combined together, the final circuit is as shown in Fig. 2.190 (c). (i) As seen, there are two parallel paths across points A and B. (a) one directly from A to B having a resistance of 6 Q and (b) the other via C having a total resistance ( 27 9 ) = - + - =2 25 Q 20 10 . ...2...X ( 6+.zz.) 441 R 10 20 -- n ( ''' BC = ) ( 2... 22 ) 550 III 10 +6+ 20 (a) .. 6 x 2.25 _ 18 n RAB =a. . " "5) - 11 1tx (6+-ib-)_ 621 n RCA=( 2...+ 6 +.zz.) 550 10 20 . (ii) Example 2.93. State Norton's theorem and find current using Norton's theorem through a load of8 Q in the circuit shown in Fig. 2.191(a).(Circuit and Field Theory, A.M.I.E. Sec. B, 1993) Solution. In Fig. 2.191 (b), load impedance has replaced by a short-circuit. Isc =IN =200/2 = 100 A. 2 2 I I 6 r16 I tIN 6 20 r4 AB 200V 4 10 10 81 f200v lOW (a) (b) Fig. 2.191 . 134 Electrical Technology Norton's resistance RNcan be found by looking into the open terminals of Fig. 2.191 (a). For this purpose !1ABC has been replaced by its equivalent Star. As seen, RNis equal to 8n Q. Hence, Norton's equivalent circuit consists of a 100 A source having a parallel resistance of 8no. as shown in Fig. 2.192 (c). The load current ILcan be found by using the Current Divider rule. (817) IL = l00x 8 + (817) =12.5 A Example 2.94. Use delta-star conversion to find resistance between terminals 'AB' of the circuit shown in Fig. 2.193 (a). ALLresistances are in ohms. [Nagpur University April 1999] 20 40 r'Qo 20 :s,~ ~20 ~ YoQ ~ D 20 Fig. 2.193 (a) Solution. First apply delta-star conversion to CGD and EDF, so as to redraw the part of the circuit with new configuration, as in Fig. 2.193 (b). A 50 B A A 50 B D B - (). D D Fig. 2.193 (d) Fig. 2.193 (e) Simplity to reduce the circuit to its equivalents as in Fig. 2.193 (c) and later as in Fig. 2.193 (d). Convert CHJ to its equivalent star as in Fig. 2.193 (e). With the help of series-parallel combinations, calculate RADas . 2 A 6 2 4 I I h 45 '1ot... ..Jtr I \.!) RN 8 2.5 B (a) (b) (e) Fig. 2.192 Fig. 2.193 (b) A 50 HJ 0.80 B C 20 AE sa 0.8 0 0.8 0 HJ 3.20 0.80 20 B F E Fig. 2.193 (c) C r-N .. 50 H l 2.80 20 E NoN L{ H G J 1.60 1.60 20 F DC Network Theorems 135 RAB =5.33 + (1.176 x 4.12/5.296) =6.245 ohms Note: Alternatively, after simplification as in Fig. (d). "CDJ - H" star-configuration can be transformed into delta. Node H then will not exist. The circuit has the parameters as shown in Fig. 2.193 (j). Now the resistance between C and J (and also between D and 1) is a parallel combination of 7.2 and 2.8 ohms, which 2.016 ohms. Along CJD, the resistance between terminals AB then obtained as : RAB = 5.0 + (1.8 x 4.032/5.832) = 5.0 + 1.244 =6.244 ohms Example 2.94 (a). Find the resistance at the A-B terminals in the electric circuit of Fig. 2.193 (g) using Ll-Ytransformation. [U.P. Technical University, 2001] A' 50 2.80 B Fig. 2.193 if) Fig. 2.193 (g) Solution. Convert delta to star for nodes C, E, F. New node N is created. Using the formulae for this conversion, the resistances are evaluated as marked in Fig. 2.193 (h). After handing series parallel combinations for further simplifications. RAB = 36 ohms. A E A J F + :=:180v 600 200 (Load) B B Fig. 2.193 (h) Fig. 2.193 (i) Example 2.94 (b). Consider the electric circuit shown in Fig. 2.193 (i) Determine: (i) the value of R so that load of20 ohm should draw the maximum power, (ii) the value of the maximum power drawn by the load. [U.P. Technical University, 2001] Solution. Maximum power transfer takes place when load res. = Thevenin's Resistance =20 ohms, here R/60 VTH CUITentthroughload Maximum Power Load = 20 ohms, giving R =30 ohms = 180 x (60/90) =120 volts = 120/40 =3 amps = 180 watts A C E F B n 136 Electrical Technology Tutorial Problems No. 2.6 Delta/Star Conversion 1. Fmd the current in the 17o resistor in the network shown in Fig. 2.194 (a) by using (a)star/deitaconversioo and (b) Thevenin's theorem. The numbers indicate the resistanceof each member in ohms. [10/3A] 2. Convert the star circuit of Fig. 2.194 (b) into its equivalent delta circuit. Values shown are in ohms. Derive the formula used. (Elect. Techrwlogy, Indor Un;v. 1980) 10 20 90 Fig. 2.194 (a) Fig. 2.194 (b) Fig. 2.195 3. Determine the resistance between points A and B in the network of Fig. 2.195. [4.23 0] (Elect. Techrwlogy, Indor Univ. 1977) 4. Three resistances of 20 0 each are connected in star. Find the equivalent delta resistance. If the source of e.m.f. of 120 V is connected across any two terminals of the equivalent delta-connected resistances, fmd the current supplied by the source. [60 0, 3A] (Elect. Engg. Calcutta Un;v. 1980) B R D 8V 3 II 8V(a) (b) Fig. 2.196 Fig. 2.191 5. Using delta/star transformation determine the current through the galvanometer i~ the Wheatstone bridge of Fig. 2.196. [0.025 A] 6. With the aid of the delta star transformation reduce the network given in Fig. 2.197 (a) to the equivalent curcuit shown at (b) [R =5.38 0] 7. Find the equivalent resistance between points A and B of the circuit shown in Fig. 2.198. [1.4 R] 8. By first using a delta-star transformation on the mesh ABCD of the circuit shown in Fig. 2.199, prove that the current supplied by the battery is 90/83 A. 2R 40 20 ili 5 2R Fig. 2.198 Fig. 2.199 A w ':1 4 B 4 41 15 B 6 15 4 A 6 17 11 >ll5V DC Network Theorems 137 2.24. Compensation Theorem This theorem is particularly usful for the following two purposes : (a) For analysing those networks where the values of the branch elements are varied and for studying the effect of tolerance on such values. (b) For calculating the sensitivity of bridge network. As applied to d.c. circuits, it may be stated in the following to ways : (i) In its simplest form. this theorem asserts that any resistance R in a branch of a network in which a current I isflowing can be replaced,for the purposes of calculations, by a voltage equal to - IR. OR (ii) If the resistance of any branch of network is changedfrom R to (R + M) where the current flowing originally is I, the change of current at any other place in the network may be calculated by assuming that an e.m.f. - I. M has beeninjectedintothe modifiedbranch while all other sources have their e.m.f.s. suppressed and are represented by their internal resistances only. Exmaple 2.95. Calculate the values of new currents in the network illustrated in Fig. 2.200 when the resistor RJ is increases by 30 %. ~ =SA Solution. In the given circuit, the values of various branch currents are I llt =S 0 I, = 75/(5+ 10)= 5 A 12 = 13= 2.5 A Now, value of R3 = 20 + (0.3 x 20) = 26 0 AR = 60 V = - 13AR = 2.5 x 6 = - 15 V The compensating currents produced by this voltage are as shown in Fig. 2.201 (a). When these currents are added to the original currents in their respective branches the new current distribution becomes as shown in Fig. 2.201 (b) O.4A . O.SA 13=2.SA -=-7SV ~=20 0 .. Fig. 2.200 4.6A SO SO O.lA 260 2.6A 2A 200 -=-7SV 200 260 IS V (a) (b) Fig. 2.201 2.25 ~ortoo's Th1:9FeJr.. This theorem is an alternative to the Thevenin's theorem. In fact, it is the dual of Thevenin's theorem. Whereas Thevenin's theorem reduces a two-terminal active network of linear resistances and generators to an equivalent constant-voltage source and series resistance, Norton's theorem replaces the network by an equivalent constant-current source and a parallel resistance. 138 Electrical Technology This theorem may be stated as follows : (i) Any two-terminal active network containinll voltalle sources and resistance when viewed from its oUtpUlIe.,IfUtUlI:i, I,)equIvalent to a consrant-.currentsource and a Daralleiresistance. The - --~... constant current is equaf to the current which would flow m a short-circuit placed across the terminals and parallel resistance is the resistance of the network when viewed from these open- , circuited terminals after all voltage and current sources have been removed and replaced by their internal resistances. Network (with sources) Network (no sources) .4 (a) (b) Explanation As seen from Fig. 2.202 (a), a short is placed across the terminals A and B of the network with all its energy sources present. The short-circuit current Isc gives the value of constant-current source. For finding Rj' all sources have been removed as shown in Fig. 2.202 (b). The resistance of the network when looked into from terminals A and B gives Rj' The Norton's equivalent circuit is shown in Fig. 2.202 (c). It consists of an ideal constantcurrent source of infinite internal resistance (Art. 2.16) having a resistance of Rj connected in parallel with it. Solved Examples 2.96, 2.97 and 2.98 etc. illustrate this procedure. (ii) Another useful generalized form of this theorem is as follows: The voltage between any two points in a network is equal to ISc. Rj where Iscis the short-circuit current between the two points and Rj is the resistance of the network as vil;!wedfrom these points with all voltage sources being replaced by their internal resistances (if any) and current sources replaced by open-circuits. Suppose, it is required to find the voltage across resistance R3 and hence current through if [Fig. 2.202 (d)]. If short-circuit is placed between A and B, then current in it due to battery of e.m.f. EI is E/RI and due to the o~er battery is E.jR2' · EI E2_ Isc = -+--E1 G1+E2G2 Rl ~ where G1and G2 are branch conductances. Now, the internal resistance of the network as viewed from A and B simply consists of three resistances Rl' R2and R3connected in parallel between A and B. Please note that here load resistance R3 has not been removed. In the first method given above, it has to be removed. 1 1 1 1 - = -+-+-=G +G +G R; R1R2~ 123 1 Rj = G1+G2 +G3 . .. .. .. .. * After E.L. Norton, fonnerely an engineer at Bell Telephone Laboratory, U.S.A. Constant Current RJ R,. Source A \ I I 0 A ...6 E2 Isc <:R; 'j=o RJ '2= 0 B 0 Internal B Resistance Infinite (c) (d) Fig. 2.202 DC Network Theorems 139 Current through R2 is 13 = VAJ!R3' Solved example No. 2.% illustrates this approach. 2.26. How To Nortonize a Given Circuit? This procedure is based on the first statement of the theorem given above. 1. Remove the resistance (if any) across the two given tenninals and put a short-circuit across them. 2. Compute the short-circuit current Isc3. Remove all voltage sources but retain their internal resistances, if any. Similarly, remove all current sources and replace ..'1~n> Jpen-circuitsi.e. by infinite resistance. 4. Next, find the resistance RI (also called RN)of the network as looked into from the given terminals. It is exactly the same as R'h (Art. 2.16). 5. The current source (Isc)joined in parallel acrossRjbetween the two terminals gives Norton's equivalent circuit. . As an example of the above procedure, please refer to Solved Example No. 2.87, 88, 90 and 91 given below. Example 2.96. Detennine the Thevenin and Norton equivalent circuits between tenninals A and Bfor the voltage divider circuit of Fig. 2.203 (a). Solution. (a) Thevenin Equivalent Circuit Obviosuly, Vth=drop across R2= E !!l R1+~ When battery is replaced by a short-circuit. A A A -=-E B B B ~ ~ ~ Fig. 2.203. Rj = RI IIR2 =Rl R.J(RI + R2) Hence, Thevenin equivalent circuit is as shown in Fig. 2.203 (b). (b) Norton Equivalent Circuit A short placed across tenninals A and B will short out R2 as well. Hence, Isc =EIR1. The Norton equivalent resistance is exactly the same as Thevenin resistance except that it is connected in parallel with the current source as shown in Fig. 2.203 (c) Example 2.97. Using Norton's theorem, find the constant-current equivalent of the circuit shown in Fig. 2.204 (a). Solution. When tenninals A and B are short-circuited as shown in Fig. 2.204 (b), total resistance of the circuit, as seen by the battery, consists of a 10 Q resistance in series with a parallel combination of 10 Q and 15 Q resistances. . 15x 10 . . total resIstance = 10 + - = 16 Q 15 + 10 .. batterycurrent I = 100/16= 6.25 A D (a) (b) (c) Fig. 2.204 This current is divided into two parts at point C of Fig. 2.204 (b). Current through A B is Isc =6.25 x 10/25=2.5 A Since the battery has no internal resistance, the input resistance of the network when viewed fromA and B consists of a 15.0 resistance in series with the parallel combination of 10.0 and 10.0. Hence, RI =15 + (10/2) =20 .a Hence, the equivalent constant-current source is as shown in Fig. 2.204 (c). Example 2.98. Apply Norton's theorem to calculate currentflowing through 5 - Q resistor of Fig. 2.05 (a). Solution. (i) Remove 5 - .a resistor and put a short across terminals A and B as shown in Fig. 2.205 (b). As seen, 10 -.a resistor also becomes short-circuited. (ii) Let us now find Isc' The battery sees a parallel combination of 4.0 and 8.0 in series with a 4 .a resistance. Total resistance seen by the battery =4 + 4 II8 =20/3.0. Hence, 1= 20 + 20/3 = 3 A. This current divides at point C of Fig. 2.205 (b). Current going along path CAB gives Isc- Its value =3 x 4/12 = 1 A. A ~ISC B A 5 B ~ ~ ~ Fig. 2.205 (iii) In Fig. 2.205 (c), battery has been removed leaving behind its internal resistance which, in this case, is zero. Resistance of the network looking into the terminals A and B in Fig. 2.205 (d) is Rj = 10 1110=5n (iv) Hence, fig. 2.205 (e), gives the Norton's equivalent circuit. 140 Electrical Technology IOn c Isn I Ion c Isn a&.a. A........ -A ... A ....J..OOV Ion -T-IOOv Ion r=O l lsc B I B 4 8 I 4 C 8 A 20V Cl 41 IOj 'i 10 B (a) (b) 4 8 8 rrTA 2r1:A (+) 5 I B. LL. DC Network Theorems 141 (v) Now, join the 5 - 0 resistance back across terminals A and B. The current flowing through it, obviously, is lAB =1 x 5/10 =0.5 A. Example 2.99. Find the voltage across points A and B in the network shown in Fig. 2.206 (a) by using Norton' s theorem. Solution. The voltage between points A and B is VAB=lsc Rj where lsc = short-circuitcurrentbetweenA andB Rj = Internal resistance of the network as viewed from points A and B. Whenshort-circuitis placedbetweenA andB, the currentflowingin it due to 50-Vbatteryis = 50/50 =1 A - from A to B = 100120=5 A -fromB to A = 1-5=-4A -fromBtoA Current due to 100 V battery is lsc i : 50n ! I I a a$. i Ja Ja I 0 '"'~, ~~ =<9 !I I i I 10' L ', Zero Resistance o Fig. 2.206(a) Fig. 2.206(b) Now, suppose that the two batteries are removed so that the circuit becomes as shown in Fig. 2.206 (b). The resistance of the network as viewed from pointsA and B consists of three resistances of 10 0, 200 and 50 0 ohm connected in parallel (as per second) statement of Norton's theorem). 1 _ -1. +..!... -1.. _ 100 .. R.I - 10 20 + 50' hence R" - 17 Q .. VAB = - 4 x 100/17=- 23.5 V The negative sign merely indicates that point B is at a higher potential with respect to the point A. Example 2.100. Using Norton's theorem. calculate the current flowing through the 15 Q load resistor in the circuit of Fig. 2.207 (a). All resistance values are in ohm. Solution. (a) Short-Circuit Current lsc As shown in Fig. 2.207 (b), terminals A and B have been shorted after removing 15 0 resistor. We will use Superposition theorem to find lsc. (i) When Only Current Source is Present In this case, 30-V battery is replaced by a short-circuit. The 4 A current divides at point D between parallel combination of 4 0 and 6 O. Current through 6 0 resistor is lsc' = 4 x 4/(4 + 6) = 1.6 A (ii) When Only Battery is Present In this case, current source is replaced by an open-circuit so that no current flows in the branch CD. The current supplied by the battery constitutes the short-circuit current .. Isc" = 30/(4 + 6) = 3 A .. Isc = Is/' - Isc'= 3 - 1.6= 1.4A - from B to A -from A toB - from A to B 500 0 L_ so V 400 1200 1O 0 -=-100 V T I (j (b) Norton's Parallel Resistance As seen from Fig. 2.207 (c) RI =4 + 6 =10 Q. The 8,0. resistance does not come into the picture because of an open in the branch CD. Fig. 2.207 (d) shows the Norton's equivalent circuit along with the load resistor. IL = 1.4 x 10 (10 + 15)=0.56 A Example 2.101. Using Norton's current-source equivalent circuit of the network shown in Fig, 2,208 (a), find the current that wouldflow through the resistor R2when it takes the values of 12, 24 and 360. respectivley, [Elect. Circuits, South Gujarat Univ. 1987] Solution. In Fig. 2.208 (b), terminals A and B have been short-circuited. Current in the shorted path due to EI is =120/40 =3 A from A to B. Current due to E2 is 180/60 =3 A from A to B. Hence Isc =6A. With batteries removed, the resistance of the network when viewed from open-circuited terminals is =40 II60 =24 Q. (i) When RL = 120. (ii) When RL = 24 0. (iii) When RL = 360. ~ A ~ IL = 6x24(24+12)=4A IL = 6/2= 3A, IL = 6 x 24/(24 + 36) = 2.4 A A ~ Example 2.102. Using Norton's theorem, calculate the current in the 6-Q resistor in the network of Fig, 2,209 (a), All resistance are in ohms, 12A i A 4 10 4 10 A c c B (b) D 8 6 2 12A i 8 ! 2 Isc B D (a) 142 Electrical Technology 4 C 6 A 4 C 6 A 4 C 6 A A T-"/'N'--- E2. [ . EIR2 + E2RI ' .. EI E2 ] (,) Vth = ; Rth =RIll R2 (u) Isc =- R + - R ; Rp =RIll R2 RI + R2 I 2 6. Obtain (i) Thevenin and (ii) Norton equivalent circuit with respect to the terminals A and B of the net.Workof Fig. 2.244. Numbers represent resistances in ohm. [(i) Vth =4 V; Rth =14/90 (ii) Isc = 2.25 A; Rp = 14/90] 7. The Network equivalent of the network shown in Fig. 2.245 between terminals A and B is a parallel resistance of 10 O. What is the value of the unknown resistance R ? [60 0] A 4 4 A 1" IE, 0 -6V -18V T T T <. 63 R2 2 6 <:" R) 3.6 I 1 B I I L I B <> Fig. 2.242 Fig. 2.243 .Fig. 2.244 Electrical Technology 2 A 6V 4~ 2 (b) B Fig. 2.245 Fig.2.246 8. The circuit shown in Fig. 2.246 (a) is the Norton equivalent of the circuit to the left ofAB in Fig. 2.246 (b). What current will flow if a short is placed across AB ? [1 A] 9. The Norton equivalent circuit of two identical batteries connected in parallel consists of a 2-A source in parallel with a 4-n resistor. Find the value of the resistive load to which a single banery will deliver maximum power. Also calculate the value of this maximum power. [8 0; 2 W] 10. The Thevenin equivalent circuit of a certain consists of a 6-V d.c. source in series with a resistance of 3 n. The Norton equivalent of another circuit is a 3-A current source in parallel with a resistance of 6 n. The two circuits are connected in parallel like polarity to like. For this combination, determine (0 Norton equivalent (ii) Thevenin equivalent (iii) maximum power it can deliver and (iv) value of load resistance from maximum power. [(i) 5 A, 2 0 (it')10 V, 2 0 (iii) 12.5 W (iv) 20] 11. For the ladder network shown in Fig. 2.247, find the value of RLfor maximum power transfer. What is thevalueof thisPmax [20,9/16 W] 6 3 3 A 3 6 A 3 6 -=-18V 6 6 3V 6V 3V 6V B B Fig. 2.247 Fig. 2.248 Fig. 2.249 12. Calculate the value of RLwhich will draw maximum power from the circuit qf Fig. 2.248. Also, find the value of this maximum power. [60; 1.5 W] 13. Find Norton's equivalent circuit for the network shown in Fig. 2.249. Verify it through its Thevenin's equivalent circuit. [1 A, Parallel resistance = 6 0] 14. State the Tellegen's theorem and verifyit by an illustration. Comment on the applicability ofTellegen's theorem on the types of networks. (Circuit and Field Theory, A.M.I.E. Sec. B, 1993) Solution. Tellegen's Theorem can be stated as under: For a network consisting of n elements if ii' i2, in are the currents flowing through the elements satisfying Kirchhoffs current law and vI' v2 vn are the voltages across these elements satisfying Kirchhoffs law, then !Vkik = 0 k=1 where vkis the voltage across and ikis the current through the kthelement. In other words, according to Tellegen' s Theorem, the sum of instantaneous powers for the n branches in a network is always zero. 164 t L A 30 1- 1- t )IA 2.4 ,6V 16V 16V '1 oB (a) This theorem has wide applications. It is valid for any lumped network that contains any elements linear or non-linear, passive or active, time-varient or time-invariant. Explanation: This theorem will be explained with the help of the simple circuit shown in Fig. 2.250. The total resistance seen by the battery is =8+4114=100. Battery current 1= 100/10= 10A. This current divides equally at point B, Drop over 8 0 resistor =8 x 10 =80 V Drop over 4 0 resistor =4 x 5 =20 V Drop over 1 0 resistor = 1 x 5 =5 V Drop over 3 0 resistor =3 x 5 = 15 V Accordingto Tellegen's Theorem, = 100x 10- 80x 10- 20x 5- 5x 5- 15x 5=0 DC Network Theorems 165 8 00 4 3 Fig. 2.250 (b) Milliman's Theorem 15. Use Millman's theorem, to find the potential of point A with respect to the ground in Fig. 2.251. [VA=8.18 V] 16. Using Millman's theorem, find the value of output voltage Voin the circuit of Fig. 2.252. All resistances are in ohms. [4 V] 3 A 6 8 v t )6A 12 V + 9 12V I )8V B Fig, 2.251 Fig. 2.252 Va [ --1 ' 0 I I 112< R! I I I I L_ __.J Load 2V Fig. 2.253 (b) MPT Theorem 17. In Fig. 2.253 what value of R will allow maximum power transfer to the load? maximum total load power. All resistances are in ohms. Also calculate the [40 ; 48 W] OBJECTIVE TESTS - 2 1. Kirchhoffs current law is applicable to only (a) closed loops in a network (b) electronic circuits (c) junctions in a network (d) electric circuits. 2. Kirchhoff's voltage law is concerned with (a) IR drops (b) battery e.m.fs. (c) junction voltages (d) both (a) and (b) 3. According to KVL, the algebraic sum of all IR drops and e.m.f.s in any closed loop of a network is always (a) zero (b) positive (c) negative (d) determined by battery e.m.fs. 4. The algebraic sign ~f an IR drop is primarily dependent upon the (a) amount of current flowing through it (b) value of R (c) direction of current flow (d) battery connection. 5. Maxwell's loop current method of solving electrical networks (a) uses branch currents (b) utilizes Kirchhoff's voltage law (c) is confined to single-loop circuits (d) is a network reduction method. 6. Point out of the WRONG statement. In the 166 node-voltage technique of solving networks, choice of a reference node does not (a) affect the operation of the circuit (b) change the voltage across any element (c) alter the p.d. between any pair of nodes (d) affect the voltages of various nodes. .7. The nodal analysis is primarily based on the application of (a) KVL (b) KCL (c) Ohm's law (d) both (b) and (c) (e) both (a) and (b). 8. Superposition theorem can be applied only to circuits having-elements. (a) non-linear (b) passive (c) linear bilateral (d) resistive 9. The Superposition theorem is essentially based on the concept of (a) duality (b) linearity (c) reciprocity (d) non-linearity 10. While Thevenining a circuit between two terminals, Vthequals (a) short-circuit terminal voltage (b) open-circuit terminal voltage (c) EMF of the battery nearest to the termnals (d) net voltage available in the circuit. 11. Thevenin resistance Rthis found (a) between any two 'open' terminals (b) by short-circuiting the given two terminals (c) by removing voltage sources along with their internal resistances (d) between same open terminal as for Vth. 12. While calculating Rth' constant-current sources in the circuit are : (a) replaced by 'opens' (b) replaced by 'shorts' (c) treated in parallel with other voltage sources (d) converted into equivalent voltage sources. 13. Thevenin resistance of the circuit of Fig. Electrical Technology A 14. 2.254 across its terminals A and B is-- ohm. (a) 6 (b) 3 (c) 9 (d) 2 The load resistance needed to extract maximum power from the circuit of Fig. ~ 6 2.255 is--ohm (a) 2 (b) 9 (c) 6 (d) 18 The Norton equivalent circuit for the network of Fig. 2.255 between A and B is-current source with parallel resistance of3 3V B Fig. 2.254 A 3 18 V B 15. Fig. 2.255 (a) 2A, 6 0 (c) 2 A, 3 0 (b) 3 A, 2 0 (d) 3 A, 9 0 16. The Norton equivalent of a circuit consists of a 2 A current sources in parallel with a 4 0 resistor. Thevenin equivalent of this circuit is a-volt source in series with a 4 0 resistor. (a) 2 (b) 0.5 (c) 6 (d) 8 17. If two identical 3 A, 4 0 Norton equivalent circuits are connected in parallel with like polarity to like the combined Norton equivalent circuit is (a) 60,40 (b) 6A,20 (c) 3 A, 2 0 (d) 6 A, 8 0 18. Two 6 V, 2 0 batteries areconnected in series siding. This combination can be replaced by a singleequivalentcurrentgenemtorof-with a parallel resistance of-ohm (a) 3 A, 4 0 (b) 3 A, 2 0 (c) 3 A, 1 0 (d) 6 A, 2 0 19. Two identical 3-A, I 0 batte~ies are connected in parallel with like polarity to like. The Norton equivalent circuit of this combination is (a) 3 A, 0.5 0 .(b) 6A,l 0 DC Network Theorems (c) 3 A, 1 Q (d) 6 A, 0.5 Q 20. Thevenin equivalent circuit of the network 5 a + 2i ~10 b Fig. 2.256 shown in Fig. 2.256 is required. The value of open-circuit voltage across terminals a and b of this circuit is-volt. (a) zero (c) 2 i/5 (b) 2 i/1O (d) 2 i/5 167 21. For a Hnear network containing generators and impedances, the ratio of the voltage to the current produced in other loop is the same as the ratio of voltage and current obtained obtained when the positions of the voltage source and the ammeter measureing the current are interchanged. This network theorem is known as-theorem. (a) Millman's (b) Norton's (c) Tellegen's (d) Reciprocity (Circuits and Field Theory, A.M.I.E. Sec. B., 1993) 22. A 12 volt source will an internal resistance of 1.2ohms is connected across a wire-would resistor. Maximum power will be dissipated in the resistor when its resistance is equal to (a) zero (b) 1.2 ohm (c) 12 ohm (d) infinity (Grad. l.E.T.E. Dec. 1985) q'ZZ P'IZ v'OZ P'61 q '81 P'LI P'91 q'SI .J'N v'£I v'U p'U q '01 q'6 J '8 P'L P'9 q'S J'" v'f p'Z J'I S1I3:MSNV 3 WORK, POWER AND ENERGY 3.1. Effect of Electric Current It is a matter of common experience that a conductor, when carrying current, becomes hot after some time. As explained earlier, an electric current is just a directed flow or drift of electrons through a substance. The moving electrons as the pass through molecules of atoms of that substance, collide with other electrons. This electronic collision results in the production of heat. This explains why passage of current is always accompanied by generation of heat. 3.2 Joule's Law of Electric Heating The amount of work required to maintain a current of I amperes through a resistance of R ohm for t second is W.D. = PRt joules = VIt joules (": R = VI!) = Wt joules (": W = VI) = VtlR joules (": I = VIR) This work is converted into heat and is dissipated away. The amount of heat produced is H = work done = W.D. mechanical equivalent of heat J J = 4,186 joules/kcal = 4,200joulesI kcal(approx) 2 H = I RtI4,200 kcal = VltI4,200kcal = WtI4,200 kcal = VtI4,200 R kcal where .. 3.3 Thermal Efficiency It is defined as the ratio of the heat actualy utilized to the total heatJ>roducedelectrically. Con- sider the case of the electric kettle used for boiling water. Out of the total heat produced (i) some goes to heat the apparatus itself i.e. kettle (ii) some is lost by radiation and convection ect. and (iii) the rest is utilized for heating the water. Out of these, the heat utilized for useful purpose is that in (iii). Hence, thermal efficiency of this electric apparatus is the ratio of the heat utilized for heating the waterto the total heatproduced. · Hence. the relation between heat produced electrically and heat absorbed usefully becomes Vlt -xll = ms(82-81) J Example 3.1. The heater element of an electric kettle has a constant resistance of 100 Q and the applied voltage is 250 V. Calculate the time taken to raise the temperature of one litre of water from J5°C to 90°C assuming that 85% of the power input to the kettle is usefully employed. If the water equivalent of the kettle is 100 g. find how long will it take to raise a second litre of water through the same temperature range immediately after thefirst. (Electrical Engineering, Calcutta Univ. 1980) Solution. Mass of water = lOOOg = I kg (": I cm3 weight I gram) 168 Work, Power and Energy 169 Heat taken by water = 1x (90- 15) =75 kcal Heat taken by the kettale = 0.1 x (90- 15) =7.5 kcal Total heat taken = 75 + 7.5 =82.5 kcal Heat produced electrically H = PRt/J kcal Now,l =250/100 =2/5 A, J =4,200 J/kcal; H =2.52 X 100 x t/42oo kcal Heat actually utilized for heating one litre of water and kettle = 0.85 x 2.52 x 100 x t/4,2oo kcal 0.85x6.25x1OOxt . .. 4,200 = 82.5 :. t =10 nun 52 second In the second case, heat would be required only for heating the water because kettle would be already hot. .. 75 = 0.85x6.25x100xt 4,200 .. t = 9 min 53 second Example 3.2. Two heater A and B are in parallel across supply voltage Y. Heater A produces 500 kcal in 200 min. and B produces 1000kcal in 10 min. The resistance of A is 10 ohm. What is the resistanceof B ? If the sameheatersare connectedin seriesacrossthe voltageV,howmuchheat will be prduced in kcal in 5 min ? (Elect. Science - II, Allahabad Univ. 1992) V2t = - kcal JR y2x (20 x 60) IOxJ 2 1000 = Y x (lOx60) RxJ From Eq. (i) and (ii), we get, R = 2.5n. Whenthe two heatersare connectedin series,let H be the amountof heat producedin kcal. Sincecombinedresistancis (10+ 2.5)=12.5n, hence H = y2x(5x60) 12.5x J Dividing Eq. (iii) by Eq. (i), we have H =100 keal. Example 3.3 An electric kettle needs six minutes to boU2 kg of waterfrom the initial temperature of 20°e. The cost of electrical energy requiredfor this operation is 12 paise. the rate being 40 paise per kWh. Find the kW-rating and the overall efficiency of the kettle. (F.Y. Engg. Pdne Univ. Nov. 1989) Solution. Heat produced For heater A, 500 = ..,(i) For heater B, ...(ii) ...(iii) . 12paise Solution. Input energy to the kettle = =0.3 kWh 40 paise/k'-- energy in kWh 0.3 Input power = m:__:_L_u_ = (6/60) =3 kW Hence, the power rating of the electric kettle is 3 kW Energy uti1isedin heating the water = mst = 2 x 1 x (100 - 20) = 160 kcal = 160/860 kWh =0.186 kWh. Efficiency =output/input =0.186/0.3 =0.62 =62%. 3.4. S.I. Units 1. Mass. It is quantity of matter contained in a body. Unit of mass is kilogram (kg). Other multiples commonly used are : 1 quintal = 100kg, 1tonne= 10quintals= 1000kg 170 2. Force. Unit of force is newton (N). Law of Motion i.e. F =ma.2 If m = 1 kg; a = Im1s , then F = 1 newton. Hence, one newton is that force which can give an acceleration of I mls2 to a mass of I kg. Gravitational unit of force is kilogram-weight (kg-wt). It may be defined as follows: or It is the force which can impart an acceleration of 9.8 mls2 to a mass of 1 kg. It is the force which can impart an acceleration of 1 mls2 to a mass of 9.8 kg. Obviously, 1 kg-wt. = 9.8 N 3. Weight. It is the force with which earth pulls a body downwards. Obviously, its units are the same as for force. (a) Unit of weight is newton (N) (b) Gravitational unit of weight is kg-wt.* Note. If a body has a mass of m kg, then its weight, W =mg newtons =9.8 newtons. 4. Work, If a force of F moves a body through a distance S in its direction of application, then Work done W = FxS (a) Unit of work is joule (1). If.. in the above equation, F =IN: S =1 m ; then work done =1 m.N or joule. Hence, one joule is the work done when a force of 1 N moves a body through a distance of 1 m in the direction of its application. (b) Gravitational unit of work is m-kg. wt or m-kg**. If F = I kg-wt; S = 1 m; then W.D. = 1 m-kg. Wt =1 m-kg. Hence, one m-kg is the work done by a force of one kg-wt when applied over a distance of one metre. Obviously, 1m-kg =9.8 m-Nor J. 5. Power. It is the rate of doing work. Its units is watt (W) which represents I joule per second. I W = 1 Jls . If a force of F newton moves a body with a velocity of v m.ls then power = F x Vwatt If the velocity v is in kmls, then power = F x v kilowatt If the velocity v is in kmls, then power = F x v kilowatt 6. Kilowatt-hour (kWh) and kilocalorie(kcal) . 1 kWh = 1000 x 11. x 3600 s =36 X 105 J . s I keal =4,186 J :. 1 kWh =36 x 105/4,186=860 keal 7. Miscellaneous Units . J (l) I watt hour (Wh) =1- x 3600s =3600 J s (if) 1 horse power (metric) =75 m-kg/s =75 x 9.8=735.5 Jls or watt (iif) 1kilowatt(kW)=1000W and 1megawatt(MW)=106W 3.5. Calculation of Kilo-watt Power of a Hydroelectric Station Let Q =water discharge rate in cubic metres/second (m3/s), H = net water headin metre (m). g =9.81 11; overall efficiency of the hydroelectric station expressed as a fraction. Since I m3of water weighs 1000 kg., discharge rate is 1000 Q kg/so When this amount of water falls through a height of H metre, then energy or work available per second or available power is Electrical Technology Its definition may be obtained from NeWton's Second = 1000 QgH Jls or W =QgH kW * Often it is referred to as a force of I kg, the word 'wl' being omitted. To avoid confusion with mass of 1 kg, the force of 1 kg is written in engineering literature as kgf instead of kg. wt. ** Generally the work 'wl' is omitted and the unit is simply written as m-kg. Work, Power and Energy 171 Since the overall station efficiency is 11,power actually available is =9.8111 QHkW. Example 3.4. A de-icing equipmentfitted to a radio aerial consists of a length of a resistance wireso arrangedthat when a currentispassedthroughit,parts of the aerialbecomewarm. The resistance wire dissipates 1250 W when 50 V is maintained across its ends. It is connected to a d.c. supply by 100 metres of this copper wire, each conductor of which has resistance of 0.006 QIm. Calculate (a) the current in the resistance wire (b) the power lost in the copper connecting wire (c) the supply voltage required to maintain 50 V across the heater itself. Solution. (a)Current=wattage/voltage = 1250/50=25 A (b) Resistance of one copper conductor = 0.006x 100=0.6 {} Resistanceof both copperconductors = 0.6x 2=1.2 {} Power loss = PRwatts =252x 1.2=750 W (c) Voltage drop over connecting copper wire = IR volt =25 x 1.2=30 V :. Supply voltage required = 50 + 30 =80 V Example 3.5 A factory has a 240-V supply from which thefollowing loads are taken: Lighting : Three hundred 150-W,four hundred 100 Wand five hundred 60-W lamps Heating : 100 kW Motors : A total of 44.76 kW (60 b.h.p.) with an average efficiency of 75 percent Misc. : Various load taking a current of 40 A. Assuming that the lighting load is onfor a period of 4 hours/day, the heatingfor 10 hours per day and the remainderfor 2 hours/day, calculate the weekly consumption of thefactory in kWh when working on a 5-day week. What current is taken when the lighting load only is switched on ? Solution. The power consumed by each load can be tabulated as given below : Power consumed Lighting 300 x 150 400 x 100 500 x 60 = 45,000 = 45 kW = 40,000 = 40 kW = 30,000 = 30 kW Total = 115kW Heating = 100kW Motors = 44.76/0.75= 59.7 kW Misc. = 240x 40/1000= 9.6 kW Similarly,the energyconsumed/daycan be tabulatedas follows: Energy consumed/ day Lighting = 115kW x 4 hr = 460 kWh Heating = 100kW x 10hr = 1,000kWh Motors = 59.7kW x 2 hr = 119.4kWh Misc. = 9.6 kW x 2 hr = 19.2kWh Totaldailyconsumption = 1,598.6kWh Weeklyconsumption = 1,598.6x 5 =7,993kWh Current taken by the lighting load alone = 115x 1000/240=479 A Example 3.6. A Disel-electric generating set supplies an output of 25 kW. The calorific value of thefuel oil used is 12,500 kcal/kg. If the overall efficiency of the unit is 35% (a) calculate the mass of oil required per hour (b) the electric energy generated per tonne of the fuel. Solution. Output =25 kW, Overall 11= 0.35,Input= 25/0.35= 71.4kW :. inputper hour = 71.4 kWh =71.4 x 860 =61,400 kcal Since 1kg of fuel-oilproduces12,500kcal (a) :. massof oil required (b) 1tonneof fuel = 61,400/12,500 = 4.91 kg = 1000kg 172 ElectricalTechnology Heatcontent = lOOOx 126'500=12.5x lO6kcal = 12.5 x 10/860 =14,530 kWh Overall 11=0.35% :. energy output = 14,530 x 0.35 = 5,088kWh Example 3.7. The effective water head for a 100 MW station is 220 metres. The station supplies full loadfor 12 hours a day. If the overall efficiency (~fthe station is 86.4%,find the volume of water used. SolutioIL Energy supplied in 12 hours =100 x 12=1200 MWh = 12 X lO5 kWh =12 X lO5 x 35 X lO5 J =43.2 X lOll J . II 12 Overall 11=86.4% =0.864 :. Energymput=43.2 x lO /0.864 =5 x 10 J Suppose m kg is the mass of water used in 12 hours, then m x 9.81 x 220 =5 x lOI2 12 8 .. m = 5 x 10 /9.81x 220=23.17 x lO kg Volumeof water = 23.17x lO81103=23.17x 105m3 (": 1 m3 of water weighs lO3kg) Example 3.8. Calculate the current required by a 1,500 volts d.c. locomotive when drawing 100 tonne load at 45 km.p.h. with a tractive resistance of 5 kg/tonne along (a) level track (b) a gradient of I in 50. A sume a motor efficiency of 90 percent. Solution. As shown in Fig. 3.1 (a), in this case, force required is equal to the tractive resistance only. (a) Force required at the rate of 5 kg-wtltonne = 100 x 5 kg-wt. =500 x 9.81 =4905 N Distance travelled/second = 45 x 1000/3600=12.5 mls Power output of the locomotive = 4905 x 12.5 J/s or watt =61,312 W 11= 0.9 :. Power input = 61,312/0.9 = 68,125 W :. Curmet drawn = 68,125/1500 = 45.41 A (a) (b) W Fig. 3.1 (b) When the load is drawn along the gradient [Fig. 3:1 (b)], component of the weight acting downwards=100x 1150=2 tonne-wt= 2000kg-wt= 2000x 9.81 = 19,620N Total force required = 19,620 + 4,905 = 24,525N Poweroutput = forcex velocity= 24,525x 12.5watt Powerinput=24,525x 12.5/0.9W ; ~urrentdrawn= 2~~25_x_~:.5 =227A Example 3.9. A room measures 4 m x 7 m x 5 m and the air in it has to be always kept 15°C higher than that of the incoming air. The air inside has to be renewed every 35 minutes. Neglecting radiation loss. calculate the ratin} of the heater suitablefor this purpose. Take specific heat of air as 0.24 and density as 1.27 kg/m . Solution. Volume of air to be changed per second =4 x 7 x 5/35 =60 = 1/15 m3 Work, Power and Energy Mass of air to be changed/second =(1115)x 1.27 kg Heat required/second = mass/second x sp. heat x rise in temp. = (1.27/15) x 0.24 x 15 kcaVs=0.305 kcaVs = 0.305 x 4186 J/s =1277 watt. Example 3.10. A motor is being self-started against a resisting torque of 60 N-m and at each start. the engine is cranked at 75 r.p.m.for 8 seconds. For each start, energy is drawnfrom a leadacid battery. If the battery has the capacity of /00 Wh, calculate the number of starts that can be made with such a battery. Assume an overall efficiency of the motor and gears as 25%. (Principles of Elect. Engg.-I, Jadavpur Univ. 1987) Solution. Angular speed 00 = 2n N/60 rad/s =2n x 75/60 =7.85 rad/s Power required for rotating the engine at this angular speed is P =torque x angular speed =ooT watt =60 x 7.85 =471 W Energy required per start is = power x time per start =471x 8 = 3,768watt-s =3,768 J = 3,768/3600 = 1.047 Wh Energy drawn from the battery taking into consideration the efficiency of the motor and gearing = 1.047/0.25=4.188 Wh No. of start possible with a fully-charged battery =100/4.188=24 (approx.) Example 3.11. Find the amount of electrical energy exdpended in raising the temperature of 45 litres of water by 75°C. To what height could a weight of 5 tonnes be raised with the expenditure of the same energy? Assume efficiencies of the heating equipment and lifting equipment to be 90% and 70% respectively. . (Elect.Engg. A.M.Ae. S.I. Dec.1991) Solution. Massof waterheated=45 kg. Heatrequired=45x 75= 3,375kcal Heatproducedelectrically=3,375/0.9=3,750kcal. Now, 1kcal=4,186J :. electrical energy expended =3,750 x 4,186 J Energy available for lifting the load is =0.7 x 3,750 x 4,186 J If h metre is the height through which the load of 5 tonnes can be lifted, then potential energy of the load=mgh joules =5 x 1000 x 9.81 h joules .. 5000 x 9.81 x h =0.7 x 3,750 x 4,186 :. h =224 metres 173 Example 3.12. An hydro-electric station has a turbine of efficiency 86% and a generator of efficiency 92%. The effecitive head of water is 150 m. Calculate the vol",me of water used when delivering a load of 40 MWfor 6 hours. Water weighs /000 kglm3. Solution. Energy output = 40 x 6 =240 MWh = 240 x 103x 36 X 105=864 X 109 J . 864 X 109 11 Overall 11=0.86x 0.92 :. Energymput= =10.92 x 1.0 J 0.86 x 0.92 Since the head is 150 m and I m3of water weighs 1000 kg, energy contributed by each m3 of 4 water =150 x 1000 m-kg (wt) 150 x 1000 x 9.8] J =147.2 x 10 J ]092 X 1011 .. Volumeof waterfor the requiredenergy=' .A =74.18X 104m3 147.2xlO Example3.13. An hydroelectricgeneratingstationissuppliedforma reservoiorof capacity6 millionm3 at a headof 170m (i) What is the available energy in kWh if the hydraulic efficiency be 0.8 and the electrical efficiency 0.9 ? (ii) Find thefall in reservoir level ajier a load of 12,000 kW has been suppliedfor 3 hours, the area of the reservoir is 2.5 km2. (iii) If the reservoir is supplied by a river at the rate of 1.2 m3Is, what does thisflow represent in kW and kWh/day? Assume constant head and efficiency. Water weighs 1 tonne/m3. (Elect. Engineering-I, Osmania Univ. 1987) 174 Electrical Technology Solution. (i) Wt. of water W = 6 X 106 X 1000 kg wt =6 x 109x 9.81 N Water head = 170m Potential energy stored in this much water = Wh=6x 109x9.81 x 170J= 1012J Overall efficiency of the station = 0.8 x 0.9=0.71 .. energyavailable = 0.72 x 1013J =72 x 1011/36X 105 = 2 X 106 kWh (ii) Energy supplied = 12,000 x 3 =36,000 kWh Energy drawn from the reservoir after taking into consideration the overall efficiency of the station = 36,000/0.72 =5 x 104kWh . = 5 X 104 x 36 X 105=18 X 1010 J If m kg is the mass of water used in two hours, then, since water head is 170 m 10 mgh = 18x 10 or m X 9.81 x 170 = 18 x 1010 :. m =1.08 X 108 kg If h metre is the fall in water level, then h x areax density = mass of water .. h x (2.5x 1O~x 1000 = 1.08x 108 :. h =0.0432 m =4.32 en (iii) Mass of water stored per second = 1.2 x 1000 = 1200 kg Wt. of water stored per second = 1200 x 9.81N Power stored = 1200 x 9.81 x 170 J/s = 2,0'00kW Poweractuallyavailable = 2,000x 0.72=1440 kW Energy delivered Iday = 1440 x 24 = 34,560kWh Example 3.14. The reservoirfor a hydro-electric station is 230 mabove the turbine house. The annual replenishment of the reservoir is 45 X1010kg. What is the energy available at the generating station bus-bars ifthe loss of headin the hydraulicsystemis 30 mand the overallefficiencyof the station is 85%. Also, calculate the diameter of the steel pipes needed if a maximum demand of 45 MW is to be supplied using two pipes. (Power System, Allahabad Univ. 1991) Soltion. Actual head available =230 - 30 =200 m Energy available at the turbine house = mgh = 45 X 1010x 9.81 x 200 =88.29 X 1013J 13 = 88.29 x 10 =24.52 X 107 kWh 36 x 105 11 = 0.85 Enery output = 24.52 X 107x 0.85 =20.84 x 107kWh , The kinetic energy of water is just equal to its loss of potential energy. t mi = mgh :. v = .J2 gh = .J2 x 9.81x 200 = 62.65 mls Power available from a mass of m kg when it flows with a velocity of v mls is - 1 2 1 2 P = 2" mv =2" x m X62.65 J/s or W Equating this to the maximum demand on the station, we get 126 2" m 62.65 = 45 x 10 :. m=22,930 kg!s If A is the total area of the pipes in m2, then the flow of water is Av m3/s. Mass of water flowing! second = Av x 103 kg (:. 1 m3 of water = 1000 kg) 3' 22,930 2 .. A x v x 10 = 22,930 orA = 3 = 0.366m 62.65x 10 If 'd' is the diameter of each pipe, then rcJ114=0.183 :. d = 0.4826m Overall .. Work, Power and Energy 175 Example 3.15. A large hydel power station has a head of 324 m and an average flow of ] 370 cubic metres/sec. The reservoir is a lake covering an area of6400 sq. km, Assuming an efficiency of 90% for the turbine and 95% for the generator, calculate (i) the available electric power.. . (ii) the number of days this power could be supplied for a drop in water level by ] metre. (AMIE Sec. B Power System I (E-6) Winter 1991) Solution. (I) Available power =9.8111 QH kW =(0.9 x 0.95) x 1370 x 324 =379, 524 kW = 379.52 MW. (ii) If A is the lake area in m2and h metre is the fall in water level, the volume of water used is =A x h = m3. The time ~uired to discharge this water is Ah I Q second. 6 2 3 Now, A = 6400 x 10 m ; h = 1 m; Q =1370 m Is. ... t =6400 X 106x 1/1370=4.67 x 106second = 540686 days Example 3.16. The reservoir area of a hydro-electric generating plant is spread over an area of 4 sq km with a storage capacity of 8 million cubic-metres. The net head of water available to the turbine is 70 metres. Assuming an efficiency of 0.87 and 0.93 for water turbine and generator respectively, calculate the electrical energy generated by the plant. Estimate the difference in water level if a load of 30 MWis continuously supplied by the generatorfor 6 hours. (Power System I-AMIE Sec. B, Summer 1990) Solution. Since 1 cubic metre of water weighs 1000 kg., the reservoir capacity =8 x 106 3 6 9 m =8 x 10 x 1000kg. =8 x 10 kg. . 999 Wt. of water, W=8 x 10 kg. Wt. 8 x 10 x 9.81=78.48 x 10 N. Net water head =70 m. Potential energy stored in this much water =Wh =78.48x 109x 70=549.36 X IOIOJ Overall efficiency of the generating plant =0.87 x 0.93 =0.809 .1010 Energy available =0.809x 549.36x 10 J =444.4 x 10 J =444.4x 1010/36x 105=12.34 X 105kWh Energy supplied in 6 hours =30 MW x 6 h =180 MWh = 180,000 kWh Energy drawn from the reservoir after taking into consideration, the overal efficiency of the station =180,000/0.809 =224,500 kWh =224,500x 36 x 105 = 80.8X 1010J If m kg. is the mass of water used in 6 how'S,then since water head is 70 m" 8 010 8 10 . , mgh = 0.8 x 1 or m x 9. 1x 70=80.8x 10 ,. m=1.176 x 10 kg. If h is the fall in water level. then h x area x density =mass of water ... h x (4x 1O~x 1000=1.176x 109 ... h=0.294m =29.4em. Example 3.17. A proposed hydro-electric station has an available head of 30 m, catchment area of 50 x 106sq.m, the rainfallfor which is 120 em per qnnum. .If 70% of the total rainfall can be collected, calculate the power that could be generated. Assume thefollowing efficiencies: Penstock 95%, Turbine 80% and Generator 85. (Elect. Engg. AMIETE See. A Part II Dec. 1991) Solution. Volume of water available =0.7(50 x 106x 1.2)=4.2 x 107m3 Mass of water available =4.2 x 107x 1000=4.2 x IOlOkg This quantity of water is available for a period of one year. Hence, quantity available per second =4.2 x 1010/365x 24 x 3600 =1.33 x 103. Available head =30 m Potential energy available =mgh =1.33 x 103x 9.8 x 30 =391 X 103 J Since this energy is available per second, hence power availabel is =391 x HYJ/s=391x I r as shown in Fig. 4.18 though its starting points is coincident with that of r. The variations of the potential and electric intensity with distance for a charged sphere are shown in Fig. 4.18. 4.15. Equipotential Surfaces An equipotential surface is a surface in an electric field such that all points on it are at the same potential. For example, different spherical surfaces around a charged sphere are equipotential surfaces. + + One important property of an equipotential surface is that the ca Q+ direction of the electric field strength and flux density is always at right angleSto the surface. Also, electric flux emerges out nonnal + r2:-- + to such a surface. If, it is not so, then there would be some + I I component of E along the surface resulting in potential difference + ~ '1-1 between various points lying on it which is contrary to the + I definition of an equipotential surface. --L:V 47tEor Electrostatics 193 EQUIPOTENTIAL SURFACE Fig. 4.17 I I I I 1 I I E I L__~ ~I ~I ~ I ~- ~ 47tEOd2 I U ~ 15 I I ~_ I -DISTANCE Fig. 4.18 Fig. 4.19 4.16. Potential and Electric Intensity Inside a Conducting Sphere It has been experimentally found that when charge is given to a conducting body say, a sphere then it resides entirely on its outer surface i.e., within a conducting body whether hollow or solid, the charge is zero. Hence, (I) flux is zero (ii) field intensity is zero (iil) all points within the conductor are at the same potential as at its surface (Fig. 4.19). Example 4.9. Three concentric spheres of radii 4, 6 and 8 cm have charges of + 8, - 6 and + 4 JlJlCrespectively. What are the potentials and field strengths at points, 2, 5, 7 and 10 cmfrom the centre. Solution. As shown in Fig. 4.20, let the three spheres be marked A, B and C. It should be remembered that (I) the field intensity outside a sphere is the same as that obtained by considering the charge at its centre (ii) insidce the sphere, the field strength is zero (iii) potential anywhere inside a sphere is the same as at its surface. (;) Consider point 'a' at a distance of 2 cm from the centre O. Since it is inside all the spheres, field strength at this point is zero. Potential at 'a' = ~ ~=9XI09 ~ Q ~4REod ~d (8 0-12 6 0-12 4 0-1' ) = 9 X 109 ~~ _ x 1 + x 1 - =1.35V 0.04 0.06 0.08 (ii) Since point 'b' is outside sphere A but inside B and C. . . Electrical field = Q? =9 x 109 Q N/C 4REd- d o 8 X 10-12 = 9 x 109X- -- =28.8 N/C 0.052 'h' = 9 x 109x ( _8~ IO-I~ _ 6 X 10-12 + 4 x 10-12 ) =0.99 V 0.05 0.06 0.08 (iii) The field strength at point 'c' distant 7 cm from centre 0 9 [ 8 X 10-12 6 X 10-12 ] = 9 x 10 x - =3.67 N/C 0.072 0.072 . , . 9 [ 8 X 10-12 6 X 10-12 4 X 10-12 ] PotentIal at c = 9 x 10 x -- -- - - - + - -.- =0.71 V 0.07 0.07 0.08 (iv) Field strength at 'd' distant 10 cm from point 0 is [8 0-1' 6 0-12 4 0 -12 ] = 9xlO9x xl - _ x~+ _xl__ =5.4N/C O.e 0.12 0.12 [ 8 X 10-12 6 X 10-12 4 X 10-12 ] Potential at 'd' = 9 x 109X -- __on --- + =0.54 V QI QI QI Example 4.10. Two positive point charges of 12 x 10-10C and and 8 x 10-10C are placed 10 cm apart. Find the work done in bringing the two charges 4 cm closer. Solution. Suppose the 12 x 10-10C charge to be fixed. Now, the potential of a point 10 em 12 x 10-10 from this charge = 9 x 109 0.1 =108 V The potential of a point distant 6 em from it 2 0-10 = 9 X 109 X I x I - =180 V 0.06 .. potentialdifference = 180 - 108=72 V Work done = chargex p.d.=8 x 10-10X72 =5.76 X 10-8 joule Example 4.11. A point charge of 10-9C is placed at a point A infree space. Calculate.- (i) the intensity of electrostaticfield on the suiface of sphere of radius 5 cm and centre A. (ii) the difference of potential between two points 20 cm and 10 cm awayfrom the charge at A. (Elements of Elect.-I, Banglore Univ. 1987) E = Q/4REo / = 1O-9/47t x 8.854 X 10-12 x (5 X 10-2)2 =3,595 VIm = Q/4REod = 1O-9/47t x 8.854 X 10-12 x 0.2 =45 V = 1O-9/47tx 8.854 X 10-12x 0.1 =90 V = 90 - 45 =45 V 194 Potential at Solution. (i) (ii) Potential of first point Potential of second point :. p.d. between two points Electrical Technology . d Fig.4.20 Electrostatics 195 4.17. Potential Gradient It is defined as the rate of change of potential with distance in the direction of electric force dV dx Its unit is volt/metre although volt/cm is generally used in practice. Suppose in an electric field of strength E, there are two points dx metre apart. The p.d. between them is dV = E. (- dx) = - E . dx :. E = - ~~ ...(i) The -ve sign indicates that the electricfield is directed outward, while the potential increases inward. Hence, it means that electric intensity at a point is equal to the negative potential gradient at that point. i.e. 4.18. Breakdown Voltage and Dielectric Strength An insulator or dielectric is a substance within which there are no mobile electrons necessary for electric conduction. However, when the voltage applied to such an insulator exceeds a certain value, then it breaks down and allows a heavy electric current (much larger than the usual leakage current) to flow through it. If the insulator is a solid medium, it gets punctured or cracked. The disruptive or breakdown voltage of an insulator is the minimum voltage required to break it down.* Dielectric strength of an insulator or dielectric medium is given by the maximum potential difference which a unit thickness of the medium can withstand without breaking down. In other words, the dielectric strength is given by the potential gradient necessary to cause breakdown of an insulator. Its unit is volt/metre (Vim) although it is usually expressed in kV/mm. For example, when we say that the dielectric strength of air is 3 kVlmm, then it means that the maximum p.d. which one mm thickness of air can withstand across it without breaking down is 3 kV or 3000 volts. If the p.d. exceeds this value, then air insulation breaks down allowing large ~le~tric current to pass through. Dielectric strength of various insulating materials is very important factor in the design of highvoltage generators, motors and transfonners. Its value depends on the thickness of the insulator, temperature, moisture, content, shape and several other factors. For exanlple doubling the thickness of insulation does not double the safe working voltage in a machine.** Note. It is obvious that the electric intensity E, potential gradient and dielectric strength are dimensionally equal. 4.19. Safety Factor of a Dielectric It is given by the ratio of the dielectric strength of the insulator and the electric field intensity established in it. If we represent the dielectric strength by Ebdand the actual field intensity by E, then safety factor k = EbdIE For example, for air Ebd=3 X 106Vim. If we establisha field intemityof 3 x 105Vimin it, then, k =3 x 106/3 X 105= 10. * Flashover is the disruptive discharge which taken places over the surface of an insulator and occurs when the air surrounding it breaks down. Disruptive conduction is luminous. ** The relation between the breakdown voltage V and the thickness of the dielectric is given approximately by the relation V =At~1l where A is a constant depending on the nature of the medium and also on the thickness t. The above statement is known as Baur's law. 196 Electrical Technology 4.20. Boundary Conditions There are discontinuities in electric fields at the boundaries between conductors and dielectrics of different perrnittivities. The relationships existing between the electric field strengths and flux densities at the boundary are called the boundary conditions. With reference to Fig. 4.21, first boundary conditions is that the nonnal component of flux density is continuous across a surface. As shown, the electric flux approaches the boundary BB at an angle 91 and leaves it at 92, DIn and D2n are the nonnal components of DI and D2. According to first boundary condition, DIn = D2n ...(i) The second boundary condition is that the tangential field strength is continuous across the boundary I .. Ell = E21 ...(ii) B Q -Dtn Since and .. This gives the law of electric flux refraction at a boundary. It is seen that if EI> £Z, 91> 92, TABLE NO. 4.1 Dielectric Constant and Strength (*indicates average value) In Fig. 4.21, we see that DIn = DI cos 91 and D2n = D2cos 92 Also EI = D/EI and Ell = DI sin 9/EI IB Similarly, E2 = Dz/E2 and E21 = D2 sin 91/£Z Fig. 4.21 qn = D2n E2 .. and-= - Ell tan 91 E21 tan 92 Insulating material Dielectric constant or relative Dielectric Strength in permittivity (e,) kV/mm Air 1.0006 3.2 Asbestos* 2 2 Bakelite' 5 15 Epoxy 3.3 20 Glass 5-12 12-100 Marble* 7 2 . Mica 4-8 20-200 Micanite 4-5-6 25-35 Mineral Oil 2.2 10 Mylar 3 400 Nylon 4.1 16 Paper 1.8-2.6 18 Paraffin wax 1.7-2.3 30 Polyethylene 2.3 40 Polyurethane 3.6 35 Porcelain 5-6.7 15 PVC 3.7 50 Quartz 4.5-4.7 8 Rubber 2.5-4 12-20 Teflon 2 20 Vacuum 1 infinity Wood 2.5-7 --- Electrostatics 197 Example 4.12. Find the radius of an isolated sphere capable of being charged to 1 million volt potential before sparking into the air. given that breakdown voltage of air is 30.000 V/cm. Solution. Let r metres be the radius of the spheres, then V = ~ =106 V 41tEor Breakdown voltage . = 30,000 Vlcm =3 x 106VIm Since electric intensity equals breakdown voltage Q 6 .. E = ~=3xlO VIm 41tEor Dividing (i) by (ii), we get r = 1/3= 0.33 metre Example 4.13. A parallel plate capacitor having waxes paper as the insulator has a capacitance of 3800 pF. operating voltage of 600 V and safety factor of 2.5. The waxed paper has a relative permittivity of 4.3 and breakdown voltage of 15 x ](/ Vim. Find the spacing d between the two plates of the capacitor and the plate area. Solution. Breakdown voltage Vbd=operating voltage x safety factor =600 x 2.5 =1500 V 6 -4 Vbd=dxEbd or d = 1500/15x 10 = 10 m=O.1 mm C= EoErAid or . A = CdIEoEr=3800X10-9x 10-4/8.854X 10-12x 4.3 :: 0.01 m2 Example 4.14. Two brass plates are arranged horizontally. one 2 cm above the other and the lower plate is earthed. The plates are charged to a difference of potential of 6,000 volts. A drop of oil with an electric charge of 1.6 x 10,19C is in equilibrium between theplates so that it neither rises norfalls. What is the mass of the drop ? Solution. The electric intensity is equal to the potential gradient between the plates. 5 g = 6,000/2 =3,000 volt/em =3 x 10 VIm E = 3 X 105VIm or NIC force on drop = Ex Q =3 X 105 x 1.6 X 10-19=4.8 X 10-14 N Wt. of drop = mg newton m x 9.81 = 4.8 x 10-14 :. m=4.89x 10-15kg ...(i) ...(ii) .. .. .. Example 4.15. A parallel-plate capacitor has plates 0.15 mm apart and dielectric with relative permittivity of 3. Find the electricfield intensity and the voltag~betw.eenplates if the surface charge is 5 x 1O-4IlC/cm2. (Electrical Engineering, Calcutta Univ. 1988) Solution. The electric intensity between the plates is D -4 2 -6 2 E = - volt/metre; Now, C1=5 x 10 1.1Clem =5 x 10 Clm EOEr . Since,chargedensityequalsflux density D 5 x 10-6 E = - = 12 =188,000 VIm =188 kV/m EOEr 8.854x 10- x 3 Now potential difference V = Ex dx =188,000 x (0.15 x 10-3)=2.82 V Example 4.16. A parallel-plate capacitor consists of two square metal plates 500 mm on a side separated by 10 mm. A slab of Teflon (Er =2.0) 6 mm thick is placed on the lower plate leaving an air gap 4 mm thick between it and the upper plate. If 100 V is applied across the capacitor, find the electricfield (Eo) in the air. electricfield Erin Teflon.flux density Da in air.flux density Drin Teflon and potential difference Vracross Teflon slab. (Circuit and Field Theory, A.M.I.E. See B, 1995) S . C EoA. 8.854 x 10-12x (0.5)2 3 16 10-10 oluhon. = = -. x F (dll Eel + d21 Er2) (6 X 10-312) + (4 x 10-3/1) .. Electrical Technology Q = CV =3.16 X 10-10x 100=31.6 X 10-9 C D = Q/A =31.6 x 10-9/(0.5)2 = 1.265 X 10-7 C/m2 The charge or flux density will be the same in both media i.e. Da=DI =D In air, Eo = D/Eo=1.265x 10-7/8.854x 10-12=14,280Vim In Teflon, EI = D/EoEr= 14,280/2 = 7,140 Vim VI = Elxdl=7,I40x6xlO-3=42.8V Example 4.17. Calculate the dielectric flux in micro-coulombs between two parallel plates each 35 em square with an air gap of 1.5 mmm between them. the p.d. being 3.000 V. A sheet of insulating material 1 mm thick is inserted between the plates. the permittivity of the insulating material being 6. Find out the potential gradient in the insulating material and also in air if the voltage across the plates is raised to 7.500 V. (Elect. Engg.-I, Nagpur Univ. 1993) Solution. The capacitance of the two parallel plates is C = EoErAid Now, Er= 1 -for air A = 35 x 35 x 10-4= 1,225 X 10-4m2;d = 1.5 X 10-2m -12 0 -4 .. C = 8.854x 10 x 1,~25x 1 F = 7.22 X10-16F 1.5x 10-' Charge Q = CV =7.22 X 10-10 x 3,000 coulomb Dielectric flux =7.22 x 3,000 x 10-10C -6 = 2.166 x 10 C =2.166 JlC With reference to Fig. 4.23, we have VI = EI XI= 0.5 X 10-3EI ; V2= 10-3E2 Now V = VI+ V2 .. 7,500 = 0.5 x 10-3EI + 10-3E2 or EI+2E2= 15xlO6 ...(i) Also D = EoErlEI = EoEr2E2 :. EI = 6 E2 ...(ii) From (i) and (ii), we obtain EI =11.25X 106Vim; E2=1.875X 106Vim Example 4.18. An electric field in a medium with relative permittivity of 7 passes into a medium of relative permittivity 2. If E makes an angle of 60° with the normal to the boundary in thefirst dielectric. what angle does the field make with the normal in the second dielectric? (Elect. Engg. Nagpur Univ. 1991) 198 0.5 ~~lmm --j AIR 7,500 V Fig. 4.22 Solution. As seen from Art. 4.19. tan a E tan 60° 7 r;; - a l = _L or - _a - = u :. tan a2= ,,3 x 2/7 = 4.95 or a2= 26°20' tan 2 E2 tan 2 2 Example 4.19. Two parallel sheets of glass having a uniform air gap between their inner surfaces are sealed around their edges (Fig. 4.23). They are immersed in oil having a relative permittivity of6 and are mounted vertically. The glass has a relative permittivity of 3. Calculate the values of electricfield strength in the glass and the air when that in the oil is 1.2 kV/m. Thefield enters the glass at 60° to the horizontal. Solution. Using the law of electric flux refraction, we get (Fig. 4.23). tan aitan al = EiE) =Eo EdEo Er! = (EriEr!) Electrostatics .. (6/3) tan 60° = 2 x 1.732 = 3.464; 73.9° 82 = Similarly tan 83 = (£r/£r2)tan 82= (1/6)tan 73.9° = 0.577;:. 83= 30° As shown in Art. 4.20. D'n = D211or D, cos 81= D2cos 82 .. D2 = D] x cos 8/cos 82 or £0£r2E2 = £0£r' E, x cos 8/cos 82 :. 6 E2 = 3 x 1.2X103x cos600/cos73.9° .. E2 = 1082VIm Now, £0 £r3E3 cos 83 = £0 £r2E2 cos 82 .. E3 = E2 (£rl£r3) X (cos 8/cos 83) = 1082 (611) (cos 73.9°/cos 30°) = 1J179VIm Tutorial Problems No. 4.1 199 OIL I I AIR I I OIL <2> CD 101 "I I CD Fig. 4.23 1. Two parallel metal plates of large area are spaced at a distance of 1 em from cal:h other in air and a p.d. of 5,000 V is maintained between them. If a sheet of glass 0.5 em thick ~d having a relative pennittivity of 6 is introduced between the plates, what will be the maximum electric stress and where will it occur? [8.57 kV/cm; in air] 2. A capacitor, fonned by two parallel plates of large area, spaced 2 em apart in air, is connected to a 10,000 V d.c. supply. Calculate the electric stress in the air when a flat sheet of glass of thickness 1.5 em and relative permittivity 7 is introduced between the plates. [1.4 x 108VIm) 3. A capacitor is made up of two parallel circular metal discs separated by three layers of dielectric of equal thickness but having relative pennittivities of 3, 4 and 5 respectively. The diameter of each disc is 25.4 em and the distance between them is 6 em. Calculate the potential gradient in eal:h dielectric '.vhen a p.d. of 1,500 V is.applied between the discs. [319.2; 239.4; 191.5 kV/m] 4. A capacitor, fonned by two parallel plates of large area, spaced 2 em apart in air, is connected to a 10,000 V d.c. supply. Calculate the electric stress in the air when a flat sheet of glass of thickness n.5 em and relative pennittivity 5 is introduced between the plates. [0.625 x 104VIm] 5. The capacitance of a capacitor fonned by two parallel metal plates, each having an effective surface area of 50 cm2and separated by a dielectric I mm thick, is 0.0001 J.1F'.The plates are charged to a pd. of 200 V. Calculate (u) the charge stored (b) the electric flux density (c) the.relative permittivity of the dielectric. [(a) 0.02 pC (b) 4 pC/m2 (c) 2.26] 6. A capacitor is constructed from two parallel metallic circular plates separated by three layers of dielectric each 0.5 em thick and having relative permittivity of 4, 6 and 8 respectively. Ii the metal discs are 15.25 em in diameter, calculate the potential gradient in each dielectric when the applied voltage is 1,000 volts. (Elect. Engg.-I Delhi Univ. 1978) 7. A point electric charge of 8 /-ICis kept at a distance of I metre from another point charge of - 4 /-IC in free space. Oetennine the location of a point along the line joining two charges where in the electric field intensity is zero. (Elect. Engineering, Kerala Univ. 1981) OBJECTIVE TESTS - 4 1. Theunitof absolutepelmittivityof a medium is (0) joule/coulomb (b) newton-metre (c) farad/metere (d) farad/coulomb 2. If relative pennittivity of mica is 5, its absolute pennittivity is (a) 5 ~ (c) ~5 (b) 5/£0 (d) 8.854 X 10-12 p'zy :J'n v'OI P'6 P'8 q'L v'9 P'S p't q'£ v"Z; :J'1 H3MSNV 200 Electrical Technology 3, Two similar electric charges of I C each are (c) charge placed I m part in air. Force of repulsion (d) potential between them would be nearly newton 9. The 51 unit of electric intensity is (a) I (b) 9 x 109 (a) N/m (b) VIm (c) 4n (d) 8.854 x 10-12 (c) N/C (d) either(b)or(c) 4, Electric flux emanating from an electric 10, According to Gauss's theorem, the surface charge of + Q coulomb is integral of the normal component of elec- (a) QI£o (b) QI£r tric flux density D over a closed surface con- (c) QIEotr (d) Q taining charge Q is 5, The unit of electric intensity is (a) Q (b) Q/£o (a) joule/coulomb (c) £oQ (d) Q2/£o (b) newton/coulomb 11, Which of the following .is zero inside a (c) volt/metre charged conducting sphere ? (d) both (b) and (c) (a) potential 6, If D is the electric flux density, then value of (b) electric intensity electric intensity in air is (c) both (a) and (b) (a) DI£o (b) DI£oEr (d) both (b) and (c) (c) dV/dt (d) Q/EA 12, In practice, earth is chosen as a place of zero 7, For any medium, electric flux density D is electric potential because it related to electric intensity E by the equation (a) is non-conducting (a) D=£oE (b) D = EotrE (b) is easily available (c) D = EI£o£r (d) D = EoEI£r (c) keepslossing and gainingelectric charge 8, Inside a conducting sphere,...remainsconstant every day (a) electric flux (d) has almost constant potential. (b) electric intensity 5 CAPACITANCE 5.1. Capacitor A capacitor essentially consists of two conducting surfaces separatecl,vy a layer of an insulating medium calleddielectric. The conducting surfaces may be in the formof eithercircular (or rectangular) plates or be of spherical or cylindrical shape. The purpose of a capacitor . is. to store electrical energy by electrostatic stress in the dielectric (the word 'condenser' is a misnomer since a capacitor does not 'condense' electricity as such, it merely stores it). A parallel-plate capacitor is shown in Fig. 5.1. One plate isjoined to the positive end of the supply and the other to the negative end or is earthed. It is experimentally found that in the presence of an earthed plate B, plate A is capable of withholding more charge than when B is not there. When such a capacitor is put across a battery, ~ere is a momentary flow of electrons from A to B. As negatively-charged electrons are withdrawn from A, it becomes positive and as these electrons collect on B, it becomes negative. Hence, a p.d. is established between plates A and B. The transient flow of elec~ons gives rise to charging current. The strength of the charging current is maximum v,~pn the two plates are uncharged but it then decreases and finally ceases when p.d. across the plates becomes slowly and slowly equal and opposite to the battery e.m.f. B + + + Fig. 5.1 5.2. Capacitance The property of a capacitor to 'store electricity' may be called its capacitance. As we may measure the capacity of a tank, not by the total mass or volume of water it can hold, but by the mass in kg of water required to raise its level by one metre, similarly, the capacitance of a capacitor is defined as "the amount of charge required to create a unit p.d. between its plates." Suppose we give Q coulomb of charge to one of the two plate of capacitor and if a p.d. of V volts is established between the two, then its capacitance is C = Q = charge V potential diffemce Hence, capacitance is the charge required per unit potential difference. By definition, the unit of capacitance is coulomb/volt whi~h is also calledfarad (in honour of Michael Faraday) .. 1 farad = I coulomb/volt One farad is defined as the capacitance of a capacitor which requires a charge of one coulomb to establish a p.d. of one volt between its plates. One farad is actually too large for practical purposes. Hence, much smaller units like microfarad (J.1F),nanofarad (oF) and micro-microfarad (J..I.I.lF)or picofarad (pF) are generally employed. 1 J.IF = 10-9 F; I of = 10-9 F ; J..I.I.lF or pF = 1O-12F Incidentally, capacitance is that property of a capacitor which delays and change of voltage across it. 201 202 Electrical Technology 5.3. Capacitance of an Isolated Sphere Consider a charged sphere of radius r metres having a charge of Q coulomb placed in a medium of relative permittivity Eras shown in Fig. 5.2. It has been proved in Art 4.13 that the free surface potential V of such a sphere with respect to infi~ty (in practice,earth)is given byA~ V = :. 41tEoErr QV =41tEoE r r 1~ By definition, QIV = capacitanceC :. = 4 1t EoEr r F - in a medium = 4 1tEor F - in air Fig. 5.2 Note: it issometimesfeltswprisingthatanisolatedspherecanactas a capacitorbecause,at firstsight,it appearsto have one plateonly. The questionarisesas to whichis the second surface. But if we remember that the surface potential V is with reference to infinity (actually earth) then it is obvious that the other surface is earth. The capacitance 4 1tEor exists between the surface of the sphere and earth. 5.4. Spherical Capacitor (a) When outer sphere is earthed Consider a spherical capacitor consisting of two concentric spheres of radii 'a' and 'b' metres as shown in Fig. 5.3. Suppose, the inner sphere is given a charge of + Q coulombs. It will induce a charge of - Q coulombs on the inner surfaces which will go to earth. If the dielectric medium between the two spheres has a relative permittivity of Er'then the free surface potential of the inner sphere due to its own charge Q/4 1tEoEra volts. The potential of the inner sphere due to - Q charge on the outer sphere is - Q/4 1t Eo Er b -= (remembering that potential anywhere inside a sphere is the same as that its surface). Fig. 5.3 :. Total potential difference between two surfaces is V = Q Q ~+ 41tEo Er a 41tEo Er b . fiQ, B = 41t ~o ~ (~ -i) = 41t :~~ (b~a) Q _ 41t EoEr ab . C _ 4 ab F - - .. - 1tEoE- V b-a rb-a (b) When inner sphere is earthed Such a capacitor is shown in Fig. 5.4. If a charge of + Q coulombs is given to the outer sphere A, it will distribute itself over both its inner and outer surfaces. Some charge Q2 coulomb will remain on the outer surface of A because it is surrounded by earth all around. Also, some charge + Q) coulombs will shift to its inner side because there is an earthed sphere B inside A. Obviously, Q = QI + Q2 The inner charge + Q) coulomb on A induces - Q. coulomb on B but the other induced charge of + QI coulomb goes to earth. Now, there are two capacitors connected in parallel : (i) One capacitor consists of the inner surface of A and the outer surface of B. Its capacitance, as found earlier, Fig. 5.4 . C 4 ab IS . = 1tEoEr-b-a (ii) The second capacitor consists of outer surfaces of B and earth. Its capacitance is C2=41t Eob - if surrounding medium is air. Total capacitance C = C. + C2. 5.5. Parallel-plate Capacitor (i) Uniform Dielectric-Medium A parallel-plate capacitor consisting of two plates M and N each of area A mZseparated by a thickness d metres of a medium of relative permittivity E, is shown in Fig. 5.5. If a charge of + Q coulomb is given to plate M, then flux passing through the medium is 'II=Q coulomb. Flux density in the medium is D = ",=Q A A Electric intensity E = Vld and D =E E Q = EV . Q _ EA Ad" -y-y C = EoE, A farad d E A = -!!..- farad d (ii) Medium Partly Air As shown in Fig. 5.6, the medium consists partly of air and partly of parallel-sided dielectric slab of thickness t and relative permittivity E,. The electric flux density ~t ~ D =QIA is the same in both media. But electric intensities are different. D Capacitance M + N + + or A + I-- d -J .. Fig.5.5 ... in the medium ... in air p.d. between plates, EO V = E, . t + Ez (d - t) D D D ( t ) = -t+-(d-t)=- -+d-t EoE, Eo Eo E, Q = -Cd -(t-tIE)] EoA ' 203 - in a medium ...(i) - with air as medium - - I-d-J Fig. 5.6 Q = EoA or C = EoA V [d-(t-tIE,)] [d-(t-tIE,)] If the mediumweretotallyair, thencapacitancewouldhavebeen C ;", EOAid From (ii) and (iii), it is obvious that when a dielectric slab of thickness t and relative permittivity E, is introduces between the plates of an air capacitor, then its capacitance increase because as seen from (ii), the denominator decreases. The distance between the plates is effectively reduces by (t - tl£,). To bring the capacitance back to its original value, the capacitor plates will have to be further separated by that much distance in air. Hence, the new separation between the two plates would be = [d + (t - t I £,)] The expression given in (i) above can be written as C = £0 A d 1£, If the space between the plates is filled with slabs of different thickness and relative permittivities, then the above expression can be generalized into C = £0 A rod1£, The capacitance of the capacitor shown in Fig. 5.7 can be written as or ...(ii) (iii) Composite Medium The above expression may be derived independently as given under: If V is the total potential difference across the capacitor plates and Vi' V2, V3, the potential differences across the three dielectric slabs, then V = V) + V2 + V3=E)l) + E212 + E313 D D D = -.1)+-.12+-.13 Eo Er) Eo Er2 EoEr3 D ( I) 12 (3 ) Q ( 1) 12 (3 = E;;" ~+ Er2 + Er3 = EOA ~+ Er~+ Er3) C = Q = EOA V ( 1) 12 (3 ) -+-+- Er) Er2 Er3 5.6. Special Cases of Parallel-plate Capacitor Consider the cases illustrated in Fig. 5.8. (i) As shown in Fig. 5.8 (a), the dielectric is of thickness d but occupies only a part of the area. This + arrangement is equal to two capacitors in parallel. Their Al capacitances are EOA Eo Er~ C) = d and C2 = d Total capacitance of the parallel-plate capacitor is A2 C = C) + C2= EOdA)+ EoEr~ (ii) The arrangement shown in Fig. 5.8 (b) consists of two capacitors connected in parallel. 204 hi +12+13-1 Fig. 5.7 Electrical Technology C = .. t--- d --t + + + + (a) (b) Fig. 5.8 (a) one capacitor having plate area A) and air as dielectric. Its capacitance is C) = EodA (b) the other capacitor has dielectric partly air and partly some other medium. Its capacitance is [Art 5.5 (ii)]. C2= Eo~ . Total capacitance is C =C) + C2 . [d - (1 - 1/ Er) 5.7. Multiple and Variable Capacitors Multiple capacitors are shown in Fig. 5.9 and Fig. 5.10. The arrangement of Fig. 5.9. is equivalent to two capacitors joined in parallel. Hence, its (a) Fig. 5.9 A ------. ------ B ------. ------. ------ ------. c (b) Fig.5.10 + + Capacitance 205 capacitance is double that of a single capacitor. . times the capacitance of single capacitor. If one s~t of plates is fixed and the other is capable of rotation, then capacitance of such a multiplate capacitor can be varied. Such variablecapacitance air capacitors are widely used in radio work (Fig. 5.11). The set of fixed plates F is insulated from the other set R which can be rotated by turning the knob K. The common area between the two sets in varied by rotating K, hence the capacitance between the two is altered. Minimum capacitance is obtained whenR is completely rotated out of F and maximum when R is completely rotated in i.e. when the two sets of plates completely overlap each other. The capacitance of such a capacitor is _ (n -1) .EOEr A - d where n is the number of plates which means that (n - 1) is the number of capacitors. Example 5.1. The voltage applied across a qapacitor having a capacitance of 10 11F is varied thus: Similarly, the arrangement of Fig. 5.10 has four K Fig. 5.11 Thepod.is increased uniformlyfrom. o to 600 V in seconds. It is then maintained constant at 600 V for J second and subsequently decreased uniformly to zero in five seconds. Plot a graph showing the variation of current during these 8 seconds. Calculate (a) the charge (b) the energy stored in the capacitor when the terminal voltage is 600. (Principles of Elect. Engg.-I, Jadavpur Univ. 1987) 8 Solution. The variation of voltage across the capacitor is as . shown in Fig. 5.12 (a). The charging current is given by . dq d dv l = -=-(Cv)=C.- dt dt dt Chargingcurrentduringthe first stage =lOx 1O-{)x (600/2) =3 x 10-3A =3 mA Charging current during the second stage is zero ~cause . t dv/dt =0 as the voltage remains constant. (b) Charging current through the third stage . -{) ( 0-600 ) -3 Fig. 5.12 = 10x 10 x 5 =- 1.2 x 10 A =- 1.2mA Thewaveformof the chargingcurrentor capacitorcurrentis shownin Fig. 5.12(b). (a) Charge when a steady voltage of 600 V is applied is = 600x 10x'lO-{)= 6 x 10-3 C (b) Energy stored =! C V2 =1 X 10-5 X 6002 =1.8 J Example 5.2. A voltage of V is applied to the inner !>phereof a spherical capacitor, whereas the outer .\phere is earthed. The inner sphere has a radius 4a and the outer one of bo~fb isfixed and a may be varied. prove that the maximum stress in the dielectric cannot be reduced below a value of 4 V/b. Solution. As seen from Art. 5.4, V = . Q ( 1_b !.) ...(i) 1tEo Er a As per Art. 4.15, the value of electric in density at any radius x between the two spheres is given . by E = Q, or Q = 4 1t EoEri E . 4 1t Eo Er X . Substituting in this value in (z)above, we get2 V = ~ 1tEoEr x E ( 1._ 1. 41tEo Er a b ) 600 V 1 400 200 1234567 .t o (a) 3 1 : -1.2 2 3 8 or E= V (11a -lib) x2 206 Electrical Technology As per Art. 5.9, the maximum value of E occurs as the surface of inner sphere i.e. when x =a For E to be maximum or minimum, dEldn = O. · d ( II ) 2 d 2 .. - - - - a = 0 or - (a - a Ib) =0 da a b da or I - 2 alb = 0 or a =bl2 V V V Now, E = .., :. Emax=- --2 = - --- z- (lla -lib) a (a - a Ib) . V 4bV 4bV 4V Smce, a =bl2 :. E = -.. - - .-- = -- -- - =- = - ",ax (bI2-b2/4b) 2b2_b2 b2 b Example 5.3. A capacitor consists of two similar squWe aluminium plates, each 10 em x 10 cm mounted parallel and opposite to each other. What is their capacitance in ~~ F when distance between them is I cm and the dielectric is air? {{the capacitor is given a charge (~{500 ~~ C. what will be the difference of potential between plates? How will this be affected if the space between the plates isfilled with wax which has a relative permittivity of 4 ? Solution. C = EoAid farad Here Eo = 8.854 X 10-12 F/m; A = 10 x 10=100 cm2=10- 2 m2 d = I cm =10-2m 8.854x 10-12x 10-2 =8.854X 10-12F =8.854~~ 10-2 Q Q 500x10-12 C V :. V = C or V = -- = 56.5volts. .. C = Now C = When was is introduced, their capacitance is increased four times because C = EoErAid F = 4 x 8.854 = 35.4 ~~ F The p.d. will obviously decrease to one fourth value because charge remains constant. :. V = 56.5/4 =14.1 volts. Example 5.4. The capacitance of a capacitor formed by two parallel metal plates each 200 cm2 in area separated by a dielectric 4 mm thick is 0.0004 microfarads. A p.d. of 20,000 V is applied. Calculate (a) the total charge on the plates (b) the potential gradient in Vim (c) relative permittivity of the dielectric (d) the electricflux density. (Elect. Engg. I Osmaina Univ. 1988) Solution. C = 4 X 10-4 ~F; V= 2 X 104 V -4 4 -6 (a) :. Total charge Q = CV =4 x.lO x 2 x 10 ~C =8 ~C = 8 x 10 C 4 = dV = 2 x 10 =5 X 106Vim dx 4x 10-3 D = QIA =8 x 10-6/200 X 10-4 =4 X 10-4 Glm2 E = 5 X 106 Vim E = ~ =_ 4 X 10-4 =9 r EoX E 8.854 X 10-12x 5 X 106 Example 5.5. A parallel plate capacitor has 3 dielectrics with relative permittivities of 5.5, 2.2 md /.5 respectively. The area of each plate is 100 cm2 and thickness of each dielectric I mm. Calculate the stored charge in the capacitor when a potential difference of 5,000 V is applied across the composite capacitor so formed. Calculate the potential gradient developed in each dielectric of the capacitor. (Elect. Engg. A.M.Ae.S.I. June 1990) (b) Potential gradient (c) (d) Since D = Eo Er E .. Solution. As seen from Art. 5.5, A -12 -4 8 854 0-14 C = Eo _8.854 x 10 x (100 x 10 ) =' x I = 292 pF ( !!L+!!.L+!!l ) ( 10--' + 1r.3 + 10-3 ) 10-3x 0.303 Erl Er2 Er3 5.5 . 1.5 Capacitance Q =CV =292 X 10-12x 5000 =146 x 10-8coulomb D = QIA =146 x 10-8/(100 x 10-4) = 146 x 1O~ C/m2 ~ I 8 -12 6 gl = EI =Dl£o Erl =146x 10 8. 54 x 10 x 5.5=3 x 10 VIm 6 6 g2 = E2 = DIEo Er2 = 7.5 x 10 Vim; g3 = DIf.o Er3 = II x 10 Vim Example 5.6. An air capacitor has two parallel plates 10 cm2in area and 0.5 cm apart. When a dielectric slab of area 10 cm2 and thickness 0.4 cm was inserted between the plates, one of the plates has to be moved by 0.4 cm to restore the capacitance. What is the dielectric constant of the slab? (Elect. Technology, Hyderabad Univ. 1992 ) Solution. The capacitance in the fIrst case is O.5cmOAcm C = EoA =EoXlOxl0-4 =EO ~ a d 0.5 X 10-2 5 The capacitor, as it becomes in the second case, is shown in Fig. 5.13. The capacitance is 207 C = EoA _ EOX 10-3 =~ In 'LdlEr (0.5:rlO-3) (;r +4) Sl' nce C - C . EO- EO ." - , a - m .. - - 5 (51 4 .. ~r - 5 Er + ) Note. We may use the relation derived in Art. 5.5 (ii) Separation = (t - tIEl) :. 0.4 =(0.5 - 0.5/Er) or Er= 5 Example 5.7. A parallel plate capacitor of area, A, and plate separation, d, has a voltage, Vo' applied by a battery. The battery is then disconnected and a dielectric slab of permittivity EI and thickness, dl, (dl < d) is inserted. (a) Find the new voltage VIacross the capacitor, (b) Find the capacitance Cobefore and its value CI after the slab is introduced. (c) Find the ratio VINo and the ratio ClCo when di = d/2 and EI =4 Eo' (Electromagnetic Fields and Waves AMIETE (New Scheme) June 1990) .EoA A C - --' Co - d' I - ( (d - d ) d ) 1+-1. EO EI . _ _' _ A _ 8EoA Smcedl -dl2 and EI-4£0 .. CI - ( ) -- d d 5d -+ 2Eo 2x4EO (a) Since the capacitor charge remains the same C Co foA 5d 5Vo Q= 0 VO=CI VI:' VI = Vo-=Vox-x- =- CI d 8EoA 8 8E A d 5 (c) As seen from above, VI = Vo5/8 ; CI Co= Sd o x - =_8 EOA AIR Fig. 5.13 Solution. (b) Tutorial Problems No. 5.1 1. Two parallel plate capacitors have plates of an equal area, dielectrics of relative pennittivities Erland Er2and plate spacing of dl and d2. Find the ratio oftheir capacitances if Er/Er2= 2 and d/d2 = 0.25. [C/C2 = 8J 2. A capacitor is made of two plates with an area of 11cm2 which are separated by a mica sheet 2 mm thick. If for mica Er= 6, find its capacitance. If, now, one plate of the capacitor is moved further to give an air gap 0.5 mm wide between the plates and mica, find the change in capacitance. [29.19 pF, 11.6 pFJ 3. A parallel-plate capacitor is made of two plane circular plates separated by d cm of air. When a parallel-faced plane sheet of glass 2 mm thick is place between the plates, the capacitance of the system is increased by 50% of its initial value. What is the distiiIlcebetween the plates if the dielectric constant of the glass is 6 ? [0.5 X 10-3 m] 4. A p.d. of 10kV is applied to the terminals of a capacitor consisting of two circular plates, each having an area of 100 cm2 separated by a dielectric 1 mm thick. If the capacitance is 3 x 10-41.1F, calculate (a) the total electric flux in coulomb (b) the electric flux density and (c) the relative permittivity of the dielectric. [(a) 3 x 10-6C (b) 3 x 10-41.1C/m2 (c) 3.39] 5. Two slabs of material of dielectric strength 4 and 6 and of thickness 2 mm and 5 mm respectively are inserted between the plates of a parallel-plate capacitor. Find by how much the distance between the plates should be changed so,as to restore the potential of the capacitor to its original value. [5.67 mm] 6. The oil dielectric to be used in a parallel-plate capacitor has a relative permittivity of 2.3 and the maximum working potential gradient in the oil is not to exceed 106V/m. Calculate the approximate plate area required for a capacitance of 0.0003 1.1F, the maximum working voltage being 10,000 V. '[147 X 10-3 m2] . 7. A capacitor consist of two metal plates, each 10cm square placed parallel and 3 mm apart. The space between the plates is occupied by a plate of insulating material 3 mm thick. The capacitor is charged to 300 V. (a) the metal plates are isolated from the 300 V supply and the insulating plate is removed. What is expected to happen to the voltage between the plates ? (b) if the metal plates are moved to a distance of 6 mm apart, what is the further effect on the voltage between them. Assume throughout that the insulation is perfect.. [300 £, ; 600 £, ; where £, is the relative permittivity of the insulating material] 8. A parallel-plate capacitor has an effecting plate area of 100 cm2(each plate) separated by a dielectric 0.5 mm thick. Its capacitance is 4421.11.1F and it is raised to a potential differences of 10kV. Calculate from first principles (a) potentialgradientin the dielectric (b) electricfluxdensityin thedielectric . (c) the relative permittivity of the dielectric material. [(a) 20 kV/mm (b) 442 JiC/m2(c) 2.5] 9. A parallel-plate capacitor with fixed dimensions has air as dielectric. It is connected to supply of p.d. V volts and then isolated. The air is then replaced by a dielectric medium of relative permittivity 6. Calculate the change in magnitude of each of the following quantities. (a) the capacitance (b) the charge (c) the p.d. between the plates (d) the displacement in the dielectric (e) the potential gradient in ~e dielectric. [(a) 6 : 1 increase (b) no change (c) 6 : 1 decrease (d) no change (e) 6 : 1 decrease] 5.8. Cylindrical Capacitor A single-core cable or cylindrical capacitor consisting two co-axial cylinders of radii a and b metres, is shown in Fig. 5.14. Let the charge per metre length of the cable on the outer surface of the inner cylinder be + Q coulomb and on the inner surface of the outer cylinder be - Q coulomb. For all practical purposes, the charge + Q coulomb/metre on the surface of the inner cylinder can be supposed to be located along its axis. Let E, be the relative permittivity of the medium between the two cylinders. The outer cylinder is earthed. Now, let us find the value of electric intensity at any point distant x metres from the axis of the inner cylinder. As shown in Fig. 5.15, consider and imaginary co-axial cylinder of radius x metres and length one metre between the two given cylinders. The electric field between the two cylinders is redial as shown. T Total flux coining out radially from the curved surface of this 1m imaginary cylinder is Q coulomb. Area of the curved surface L =21t X x I =2 1t X m2. Hence, the value of electric flux density on the surface of the imaginary cylinder is D = flux in coulon;b ='If=Q C/m2 :. D = JL C/m2 area in metre A A 21tX The value of electric intensity is D 208 E = LV/m 21tEoE, x or E = Now, EOE, = -Edx V = ra_E dx= ra_ Qdx Jb . Jb 21tEoE,x dV or Electrical Technology - - Fig. 5.14 Capacitance 209 = - Q ra dx = - Q lIog x la 2nEOE,Jb x 2nEOE, b = -Q (logea -loge b)=2 - Q loge( a b )= Q loge( E. nEoE, nEoE, 2nEoE, b ) Q 2nEoE, 2nEoE, ( ( b ) (b )) V = ( b ) :' C= ( b\ F/m .: loge ~ =2.3IoglO ~ loge - 2.3logio- I a a The capacitance of I metre length of this cable is C = 2nEo E, I . F 2.3 log10~~ In case the capacitor has compound dielectric, the relation becomes C = 2nEoi F L loge(~)/E, . The capacitance of I kIn length of the cable in f..LF can be found by putting I =I kIn in the above expression. C = 2n x 8.854 x 10-12x E, X1000 F/km = 0.024 E, f..LF/km 2.3 log 10(~) loglO(~) 5.9. Potential Gradient in a Cylindrical Capacitor It is seen from Art. 5.8 that in a cable capacitor Q 21tEoE, x where x is the distance from cylinder axis to thepoint under consideration. Now E = g :. g =... Q Vim ...(i) nEoE, x From Art. 5.8, we find that V = 2 Q loge( !2.) or . nEoE, a E = Vim Substituting this value of Q in (i) above, we get 2nE 10V V V g = 0 , Vim or g = Vim or g = b volt/metre IOge(~)X2nEoE, x xloge(~) 2.3x~oglO(~) . Obviously,potentialgradientvariesinverselyas x. Minimum value of x =a, hence maximum value of potential gradient is gmax = V .(b\ Vim ...(ii) 2.3a loglO-a Similarly, g_ = 2.3b loglo- V (b) Vim a Note. The above relation may be used to obtain most economical dimension while designing a cable. As seen, greater the value of permissible maximum stress Emax'smaller the cable may be for given value of V. However, Emaxis dependent on the dielectric strength of the insulating material used. If V and Emaxare fixed, then Eq. (ii) above may be written as E V ( b ) V . b kIa b kIa max = ( b ) or alogh ~ = y-" -;;=e or =ae a logh - max e a 210 Electrical Technology For most economical cable db/da =0 .. ~~ = 0 =ekJa + a (- kIa2)ekJa or a =k = VlEmaxand b = ae = 2.718a Example 5.8. A cable is 300 km long and has a conductor of 0.5 em in diameter with an insulation covering of 0.4 em thickness. Calculate the capacitance of the cable if relativepermittivity of Insulation is 4.5. (Elect. Engg. A.M.Ae. S.I. June 1987) Solution. Capacitance of a cahle ;s C = O'02( :) ~ FIkm loglo -a Here, a =0.5/2 =0.25 cm ; b =0.25 + 0.4 =0.65 cm ; bla =0.6510.25 =2.6 ; log~o6 =0.415 C = 0.024 X 4.5 =0.26 0.415 Total capacitance for 300 kIn is =300 x 0.26 =78 Ii F. .. Example 5.9. /n a concentric cable capacitor. the diameters of the inner and outer cylinders are 3 and /0 mm re.lpectivelv. /fErfor insulation is 3.find its capacitance per metre. Ap.d. (~l600 volts is applied hetween the two conductors. Calculate the values (~lthe electric force and electricflux density: (a) at the surface (~linnerconductor (b) at the inner surface (~t'outer conductor. Solution. a =1.5 mm; b =5 mm; :. bla =5/1.5 =10/3 ; 10glO(1~) =0.523 C = 21tEo Er I _ 21tx 8.854 X 10-12 X 3 X 1 =138.8 X 10-12F = 138.8 pF 2.3 log10(~) 2.3X 0.523 (a) D = QI21ta Now Q = CV =138.8 X 10-12 X 600 =8.33 X 10-9 C D = 8.33 X 10-8/21t x 1.5 X 10-3 =8.835 Ii C/m2 E = DI£oEr= 332.6 Vim 8.33 x 10-8 2 2 (b) D = ~ Clm =2.65 Ii C/m ; E =DI£oEr =99.82 Vim. 21tx 5x 10- Example 5.10. The radius of the copper core of a .I'ingle-corerubber-insulated cable is 2.25 mm. Calculate the radius (if the lead sheath which covers the rubber insulation and the cahle capacitance per metre. A voltage of 10kV may be applied between the core and the lead sheath with a safety factor of 3. The rubber insulation has a relative pennittivity of 4' and hreakdown field strength of /8 x / Of>Vim. Solution. As shown in Art. 5.9, gmax= V ( b ) 2.3 a loglO - . a 6 Now, gmax= Emax= 18 x 10 Vim; V = breakdown voltage x 4 Safety factor = 10 x 3 =30,000 V 6 30,000 b 2 b 2 :. 8x 10 = -3 ( b ) :. ~= .lor =2.1x2.5=4.72mm 2.3 x 2.25 x 10 x loglO -a C = 21tEoErI _ 21tx 8.854 X10-12x 4 x 1_ 3 X 10-9 F 2.31og10(~) 2.3 log 10(2.1) Capacitance 5.10. Capacitance Between Two Parallel Wires 211 This case is of practical importance in overhead transmission lines. The simplest system is 2-wire system (either d.c. or a.c.). In the case of a.c. system, if the transmission line is long and voltage high, the charging current drawn by the line due to the capacitance between conductors is appreciable and affects its performance considerably. With reference to Fig. 5.16, let d = distance between centres of the wires A and B r = radius of each wire ($; d) Q = charge in coulomblmetre of each wire* Now, let us consider electric intensity at any point P between conductors A and B. Electric intensity at p* due to charge + Q coulomblmetre on A is = Q Vim 2 1tEoE, x Electric intesity at P due to charge - Q coulomblmetre on B is Q - - - .--V/m - 21tEo E, (d - x) Total electric intensity at PE = __J? ( L+ - L2 1tEo E, x d - x ) Hence, potential difference between the two wires is f d-' Q f d-' ( I 1 ) V = E.dx =-- - .--+ dx , 2 1tEoE,' x d - x Q d-, Q d - r V = -- Ilog x-log (d -x)1 =--Iog - - 2 1tEoE, e e '1t EoE, e r p ~ ~ ~ I;-x -18d (d-X) :I Fig. 5.16 . .. towards B. . .. towards B.. Now C =Q/V :. C = 1tEo E, _ 1tEoE, = 1tEo E, F/m (approx.) loge (d - r r) 2.3 10glO(d ~ r) 2.3 log 10(~) 1t EoE, The capacitance for a length of I metres C = ( d ) F 2.3 loglo -r The capacitance per kilometre is 1t x 8.854 X 10-12 X E, X 100 X 106 C = =11 m F/k 2 31 ( 4-) =0.0121 E, . oglO r ( d ) 10glO - r Example 5.11. The conductors of a two-wire transmission line (4 km long) are spaced 45 em between centre. IF each conductor has a diameter of 1.5 em, calculate the capacitance of the line. Solution. Formula used C = 1tEoE, F ( d 2.3 10glO -;:) Here 1= 4000 metres; r = 1.5/2 cm; d = 45 cm ; E,= 1- for air .. d 45x 2 =60 -;:= 1.5 1tx 8.854 X 10-12 x 4000 = 0.0272 x 10~ F C = 2.3 10glO60 * If charge on A is + Q, then on B will be- Q. 212 Electrical Technology [or C = 4 x 0.0121 =0.0272 ~ F] 10glO60 5.11. Capacitors in Series With reference of Fig. 5.17, let CI' C2, C3 = Capacitances of three capacitors V\, V2' V3 = p.ds.acrossthreecapacitors. V = applied voltage across combination C = combined or equivalent or join~ capacitance. In series combination, charge on all +Q. capacitors is the same but p.d. across each is different. .. Q+ Q+ . 'Q+ f+llc, I Jle, I +Ij VI-I- v2-+- v3 v V = V) + V2 + V3 Q Q Q Q - = -+-+- C CI C2 C3 1 1 1 1 - = -+-+- C C) C2 C3 For a changing applied voltage, dV dV) dV2 dV3 - = -+-+- dt dt dt dt We can also find values of V)' V2 and V3 in terms of V. Now, Q = C) V) = C2 V2 = C3 V3 =cv C) C2 C3 C)C2C3 C)C2+ C2C3+ C3C) 1:C)C2 C C2 C3 C) V) = CVor V)=V -=V.- C) 1:C)C2 C) C3 C) C2 V2 = V. ~ r ) 2 C and V3=V .tee) 2 or or Fig. 5.17 Fig. 5.18 where C = .. Similarly, 5.12. Capacitors in Parallel In this case, p.d. across each is the same but charge on each is different (Fig. 5.18). :. Q =QI + Q2 + Q3 or CV =C) V + C2V + C3V or C =C) + C2 + C3 For such a combination, dV/dt is the same for all capacitors. Example 5.12. Find the Ceqof the circuit shown !n Fig. 5./9. All capacitances are in fl F. (Basic Circuit Analysis Osmania Univ. JanJFeb. 1992) Solution. Capacitance between C and D =4 + 1112=14/3~ F. ~ m e ~ Capacitance between A and B i.e. Ceq =3 + 211 14/3 =4.4 fl F e 3 4 2 Example5.13. Twocapacitorsofacapacitance eq- - - 4 flF and 2 flF respectively, are Joined in series with a batten' of e.m.f /00 V. The connections are broken and the like terminals of the capacitors are then Joined. Find the final charge on each capacitor. . Solution. When joined in series, let VI and V2be the voltages across the capacitors. Then as charge across each is the same. :. 4 x VI = 2V2 o B D Fig. 5.19 Capacitance 213 .. VI + 2V) =100 :. V) =100/3 V and V2=200/3 V .. Q) =Q2 =(200/3) x 2 =(400/3) ~ C .. Total charge on both capacitors =800/3 ~ C Whenjoined in parallel, a redistribution of charge takes place because both capacitors arereduced to a common potential V. Total charge =800/3 ~ C; total capacitances =4 + 2 =6 ~ F 800 400 V = -=-volts 3x6 9 Q) = (400/9) x 4 = 1600/9 = 178 ~ C Q2 = (400/9) x 2 =800/9 = 89 ~ C (approx.) Three capacitorrs A, B, C have capacitances /0,50 and 25 Il F respectively. Calculate (i) charge on each when connected in parallel to a 250 V supply (ii) total capacitance and (iii) p.d. across each when connected in series. (Elect. Technology, Gwalior Univ. 1989) Solution. (i) Parallel connection is shown in Fig. 5.20 (a). Each capacitor has a p.d. of 250 V across it. QI =C)V =10x 250=2500~ C; Q2=50 x 250= 12,500~ C C3=25 x 250 =6,750 ~ C. (ii) C = C) + C2+ C3= 10 + 50 + 25 = 85 ~ F (iii) Series connection is shown in Fig. 5.20 (b). Here charge on F'i 0 ~ 0' e:oct'capa<;i'?ris the same and is equal to that on the equivalent Q Q Q smg1e capacitor. no. V -+- V_ V lIC = lIC) + lIC2 + lIC3 ; C =25/4 ~ F (V,I ) 2 3 Q = CV =25 x 250/4 =1562.5~ F 250 V Q = C) VI ; V) =1562.5/10=156.25 V Fig. 5.20 V2 = 1562.5/25 =62.5 V; V3= 1562.5/50 =31.25 V. Example 5.15. Find the charges on capacitors in Fig. 5.2/ and the p.d. across them. Solution. Equivalent capacitance between points A and B is .. Hence (a) C2 + C3 = 5 + 3 =8 ~ F Capacitanceof the wholecombination(Fig.5.21) 8x2 C = 8+ 2 1.6~ F rVI I ~ V~c~ ~ ~ Q. ~~F IB 2pF .~ C3 ~ Fig. 5.21 Charge on the combination is Q) = CV =100 x 1.6=160 ~ C V) =~= 160=80V; V2=100-80= 20 V C) 2 QI = C2V2=3x 1O-6x20=60~C Q3 = C3V2=5x 1O-6x20=100~C Example 5.16. Two capacitors A and B are connected in series across a /00 V suppl\ and It IS observed that the p.d.s. across them qre 60 V and 40 V respectiveh. A capacitor of 2 IlF capacitanct is nOH'connected in parallel with A and the p.d. across B rises to 90 volts. Calcualte the capacitanCt of A and B in microfarads. Solution. Let C. and C2 fl F be the capacitances of the two capacitors. SiI.1cethey are connected in series [Fig. 5.22 (a)], the charge acorss each in the same. :. 60 C. =40 C2 or ClC2 =2/3 ...(i) In Fig.5.22 (b) is showna capacitorof 2 fl F connectedacrosscapacitorA. Theircombined capacitance=(C. + 2) fl F :. (C. + 2) 10 = 90 C2 or C./C2 =2/3 Putting the value of C2=3Cp from (i) in (ii) we get C, +2 3Cl2 = C, = C2 = 214 or Electrical Technology ...(ii) 9 :. C. + 2 =13.5 C1 2/12.4 =0.16 fl F and (3/2) x 0.16 =0.24 fl F ~ IC' ~ BC; 60V ~I'" 40V IOOV ~ 2PF C1 ~ A B JOV ~I'" 90V WOV Fig. 5.22 Example 5.17. Three capacitors of2 fl F, 5 fl F and 10 fl F have breakdown voltage of200 V, 500 Vand 100 V respectively. The capacitors are connected in series and the applied direct voltage to the circuit is gradually increased. Which capacitor will breakdownfirst? Determine the total applied voltage and total energy stored at the point of breakdown. [Bombay Univeristy 2001] Solution. C1of2 flF, C2of5 flF, and C3of 10flF are connected in series. If the equivalent single capacitor is C, lIC = lIC, + lIC2 + l/C3, which gives C =1.25 fl F If V is the applied voltage, V. = Vx ClC. = Vx (1.25/2) = 62.5 % of V V2 = V x (C/C2) =C x (1.25/5) =25 % of V . V3 = V x (CIC3) = V x (1.25/10) = 12.5 % of V Fig. 5.23 . If VI =200 volts, V =320 volts and V2 =80 volts, V3 =40 volts. It means that, first capacitor C1 will breakdown first. Energy stored= 112CV2= il2 x 1.25 x 10- 6 x 320 x 320 =0.064 Joule Example 5.18. Amultipleplate capacitor has 10plates, each of area 10square cm and separation between 2 plates is J mm with air as dielectric. Detennine the energy stored when voltage of 100 volts is applied across the capacitor. [Bombay University 2001] (a) + v (b) Solution. Number of plates, n = 10 Energy stored (n -I) EO_ 9 x 8.854 x 10-12x lOx 10-4 =79.7 pF d lxlO-3 .2 . = 112x 79.7 x 10- x 100 x 100=0.3985 f.1J C = Capacitance 215 Example 5.19. Determine the capacitance between the points A and B in figure 5.24 (a). All capacitor values are in ~F. A B ~ o '/ "-< 10 10 D Fig. 5.24 (a) Solution. Capacitances are being dealt with in this case. For simplifying this, Delta to star transformation is necessary. Formulae for this transformation are known if we are dealing with resistors or impedances. Same formulae are applicable to capacitors provided we are aware that capacitive reactance is dependent on reciprocal of capacitance. Further steps are given below: C 25 ~ 20 \ >', T~~ IO~IOD A B Fig. 5.24 (b) Fig. 5.24 (c) Reciprocals of capacitances taken first : Between B-C- 0.05, Between B-D - 0.10 BetweenC-D- 0.05,Sumof thesethree=0.20 For this delta, star-transformation is done: Between N-C : 0.05 x 0.0510.20 =0.0125, its reciprocal = 80 ~ F Between N-B : 0.05 x 0.1010.20 =0.025, its reciprocal = 40 ~ F Between N-D : 0.05 x 0.10/0.20 =0.025, its reciprocal= 40 ~ F This is markedon Fig.5.24 (c). With series-parallelcombinationof capacitances,furthersimplificationgivesthe finalresult. CAB = 16.13 ~ F Note: Alternatively, with ADB as the vertices and C treated as the star point, star to delta transformation can be done. The results so obtained agree with previous effective capacitance of 16.14 ~ F. Example 5.20. (a) A capacitor of 10 pF is connected to a voltage source (~f 100 V. If the distance between the capacitor plates is reduced to 50 % while it remains, connected to the 100 V supply. Find the new values of charge. energy stored and potential as well as potential gradient. Which of these quantities increased hy reducing the distance and why? [Bombay University 2000] Solution. (i) C =10 pF (ii) C =20 pF, distance halved Charge = 1000 p Coul Charge =2000 p-coul Energy =1/2 cv2 =0.05 ~ J Energy =0.10 ~ J Potentialgradientin the secondcase willbe twiceof earliervalue. 216 Electrical Technology Example 5.20 (b). A capacitor 5 f..lF charged to 10 V is connected with another capacitor of 10f..lF charged to 50 V, so that the capacitors have one and the same voltage after connection. What. are the possible values of this common vaoltage ? [Bombay University 2000] B A B + SOY Fig. 5.25 (a) x Fig. 5.25 (c) Simplification A A A Fig. 5.25 (b). Initial charge represented by equiv-source Fig. 5.25 (d). Final condition Solution. The clearer proce?ure is discussed here. Initial charges held by the capacitors are represented by equivalent voltage sources in Fig. 5.25 (b). The circuit is simplified to that in Fig. 5.25 (c). This is the case of CI and C2connected in series and excited by a 40-V source. If C is the equivalent capacitance of this series-combination, lIC = lICI + C2 This gives C = 3.33 f..lF In Fig. (c), VCl = 40 X C/CI =40 x 3.33/5 =26.67 volts VSIand VS2are integral parts of CI and C2 in Fig. 5.25 (c), Voltage across CI =10+ 26.67=36.67(A w.r.to 0) Voltage acorss C2=50 - 13.33= 36.67,(B w.r.to 0) Thus, the final voltage across the capacitor is 36.67 volts. Note: If one of the initialvoltageson thecapacitorshappensto be the opposIteto the singleequivalent sourcevoltagein Fig. 5.25(c) willbe 60 volts. Proceedingsimilarly,with properCareaboutsigns,the final situationwillbe the commonvoltagewillbe 30 volts. 5.13. Cylindrical Capacitor with Compound Dielectric Such a capacitor is shown in Fig. 5.26 Let rI = radius of the core r2 = radius of inner dielectric £,1 r3 = radius of outer dielectric £r2 Obviously, there are two capacitors joined in series. Now CI = 0.024 £,1 f..lF/kmand C2= 0.024 £,2 f..lF1M . loglO(ri'i) loglO(rir2) I .-...! X 36.67 _c, + 1 - cl c, Capacitance Total capacitance of the cable is C = ,..CICC 2 1+ 2 Now for capacitors joined in series, charge is the same. .. Q = CIVI = C2V2 or V2 =..s. = Erl loglo (Tjlr2) V; C2 Erlloglo (r21'i) From this relation, V2and VIcan be found, . . . VI gmax10 lOner capacItor 2,3'i loglO(ri'i) (Art. 5.9) V2 Similarly, gmax for outer capacitor = 2,3 r2 loglO(rir2) .. 217 ~ ~:7~ I 1 1 I I 1 I I I 1 I 1 1 I I 1 1 I I 1 1 I I 1 1 I I 1 1 1 I 1 I 1-+ 1 1 ,+- --r.. ::t L 'y1-:j ) ~ I I - Fig. 5.26 Putting the values of CI and C2,we get grrutXI _ 0.024Er2 loglo (rir2) _ r2 loglO(r2/'i) . gmax I _ Er2 .r2 -- x --x .. --- grrutX2 loglO(rir2) 0.024 E~I 'i loglO (r2/'i) grrutX2 Erl .'i Hence, voltage gradient is inversely proportional to the permittivity and the inner radius of the insulating material. Example 5.21. A single-core lead-sheathed cable, with a conductor diameter of2 cm is designed to withstand 66 kV. The dielectric consists of two layers A and B having relativepermittivities of 3.5 and 3 respectively. The corresponding maximum permissible electrostatic stresses are 72 and 60 kV/cm. Find the thicknesses of the two layers. (Power Systems-I, M.S. Univ. Baroda, 1989) Solution. As seen from Art. 5.13. gmax I Er2 .r2 72 3x r2 1 4 - = - or -=- or r2= . cm gmax2 Erl .'i 60 3.5xl vlxJ2 Now, gmax = 23 I I . 'i oglor2'i whereVIis the r.m.s.valuesof the voltageacrossthe firstdielectric. V;xJ2 .. 72 = -3 I I I" or VI=17.1kV . x x oglO . V2 = 60 - 17.1=48.9kV . V2 X J2 2.3 r2 loglO (r3Ir2) .. 60 = 48.9 2.3 x 1.4loglo (r3Ir2) r; = 0.2531 =loglO(1.79) :. -1. =1.79 or r3=2.5 cm r2 Obviously, Now, grrutX2 = .. ...Art. 5.9 218 Electrical Technology Thickness of first dielectric layer = 1.4 - 1.0 =0.4em. Thickness of second layer =2.5 - 1.4=1.1 em. 5.14. Insulation Resistance of a Cable Capacitor In a cable capacitor, useful current flows along the axis of the core but there is always present some leakage of current. This leakage is radial i.e. at right angles to the flow of useful corrent. The resitance offered to this radial leakage of current is called insulation resistance of the cable. If cable dr length is greater, than leakage is also greater. It means that more current will leak. In other words, insulation resistance is decreased. Hence, we findthat insulation resistance is inversely 1 proportional to the cable leng~h. This insulation resistance is not to be confused with conductor resistance which is directly proportional to the cable length. Consider i metre of a single-core cable of inner-radius r, and 1 outer radius r2 (Fig. 5.27). Imagine an annular ring of radius 'r' and radial thickness 'dr'. JIf resistivity of insulating material is p, then resistance of the h . .. dR P dr ndr I 1 . . t ISnarrow nng IS =_2 =.r:..:::.: :. nsu atlOnresIstance . 1trx I 21trl of l metre length of cable is J f r2 Pdr P r2 dR = -"- - or R =-I log (r) I rl 21t rl 21t rl e rl p _2.3 P 2 rclloge (r2hj)- 2lclloglO (r2hj) Q . I I -{ I --J I I 41 I Fig. 5.27 R It should be noted (i) that R is inversely proportional to the cable length (ii) that R depends upon the ratio r!r/ and NOT on the thickness of insulator itself. Example 5.22. A liquid resistor consists of two concentric metal cylinders of diameters D =35 cm and d = 20 cm respectively with water of specific resistance p =8000 Q cm between them. The length of both cylinders is 60 cm. Calculate the resistance of the liquid resistor. (Elect. Engg. Aligarh Univ., 1989) Solution. rl =10em; r2 = 17.5em;loglo(1.75)=0.243 3 P = 8 x 10 Q - em; l =60 em. R . f h 1 . .d ' 2.3 x 8x 103 eSlstance0 t e IqUi resIstor R = x 0.243=11.85Q. 2rcx 60 Example 5.23. Two underground cables having conductor resistances of O.7 Q and 0.5 and insulation resistance of 300 M Q respectively arejoind (i) in series (ii) in parallel. Find the resultant conductor and insulation resistance. (Elect. Engineering, Calcutta Univ. 1987) Solution. (i) The conductor resistance will add like resistances in series. However, the leakage resistances will decrease and would be given by the reciprocal relation. Total conductor resistance =0.7 + 0.5 = 1.2 Q If R is the combined leakage resistance, then 1 1 1 R =300 + 600 :. R =200 M Q (ii) In this case, conductor resistance is =0.7 x 0.5/(0.7 + 0.5) = 0.3.Q (approx) Insulation resistance = 300+ 600 = 900M Q Capacitance 219 Example 5.24. The insulation resistanceof a kilometre of the cable having a conductor diameter of 1.5 cm and an insulation thickness of 1.5 cm is 500 M Q. What would be the insulation resistance if the thickness of the insulation were increased to 2.5 cm ? (Communication Systems, Hyderadad Univ. 1992) Solution. The insulation resistance of a cable is . 2.3 P 1 ( / ) First Case R = 21t 1 oglO r2 'i r( =1.5/2=0.75 cm ; r2=0.75 + 1.5=2.25 cm .. rirl =2.25/0.75 =3; loglo (3) =0.4771 :. 500=2.3 P x 0.4771 ...(i) 2 1t1 Second Case r( =0.75 cm - as before r2=0.75 + 2.5 =3.25 cm rirl =3.25/0.75 =4.333 ; 10gIO(4.333) =0.6368 Dividing Eq. (ii) by Eq. (i), we get .Jl.. =0.6368. R =500 x 0 6368/0 4771 =667 4 M Q 500 0.4771' .. . .. R = 2.4 P x 0.6368 21t 1 ...(ii) 5.15. Energy Stored in a Capacitor Charging of a capacitor always involves some expenditure of energy by the charging agency. This energy is stored up in the electrostatic field set up in the dielectric medium. On discharging the capacitor, the field collapses and the stored energy is relesed. To begin with, when the capacitor is uncharged, little work is done in transferring charge from one plate to another. But further instalments of charge have to be carried against the repulsive force due to the charge already collected on the capacitor plates. Let us find the energy spent in charging a capacitor of capacitance C to a voltage V. Suppose at any stage of charging, the p.d. across the plates is v. By definition, it is equal to the work done in shifting one coulomb from one plate to another. If 'dq' is charge next transferred, the work done is dW = v.dq Now q =Cv :. dq = C.dv :. dW =Cv.dv Total work done in giving V units of potential is r I 2 1 v I 2 W = Jo Cv.dv=C v2 0 :. W ="2CV 2 If C is in farads and V is in volts, then W = tcv2 joules =-4-QV joules = iC joules If Q is in coulombsand C is in farads,the energystoredis giveninjoules. Note: As seen from above, energy stored in a capacitor is E =:4 cv2 Now, for a capacitor of plate area.4. m2 and dielectric of thickness d metre, energy per unit volume of dielectric medium. =:.! CV2 ==.!EA ~ =:.! £ ( ~ ) 2 =:.! £ E2 =:.! DE =:D212£ joules/m3* 2 Ad 2 d . Ad 2 d 2 2 It will be noted that the formula 1-DE is similar to the expression t stress x strain which is used for calculating the mechanical energy stored per unit volume of a body subjected to elastic stress. * It is similar to the expression for the energy stored per unit volume of a magenetic field. 220 Electrical Technology Example 5.25. Since a capacitor can store charge just like a lead-acid battery, it can be used at least theoretically as an electrostatic battery. Calculate the capacitance of 12-V electrostatic battery which the same capacity as a 40 Ah, 12 V lead-acid battery. Solution. Capacity of the lead-acid battery =40 Ah =40 x 36 As =144000 Coulomb Energy stored in the battery = QV =144000 x 12 = 1.728 x 106 J Energy stored in an electrostatic battery = !- CV 2 .. lxCx122 =1.728x 106 :. C=2.4xl04 F=24kF 2 Example 5.26. A capacitor-typestored-energywelderis to deliverthe same heat to a single weld as a conventional welder that draws 20 kVA at 0.8 pffor 0.0625 second/weld. lfC =2000 j..lF, find the voltage to which it is charged. . (Power Electronics,A.M.I.E.Sec B, 1993) Solution. The energy supplied per weld in a conventional welder is W = VA x cos q>x time =20,000 x 0.8 x 0.0625 = 1000 J Now, energy stored in a capacitor is (1/2) CV .. W = 1 CV2 or V =~2 W =J 2x 1000 = 1000 V 2 C 2000 x 10-6 Example 5.27. A parallel-plate capacitor is charged to 50 j..lCat 150 V. It is then connected to another capacitor of capacita'!ce 4 times the capacitance of the first capacitor. Find the loss of energy. (Elect. Engg. Aligarh Univ. 1989) Solution. C, =50/150 =1/3 ~; C2=4 x 1/3=4/3 ~ Before Joining E, = !- C,,,,2=!-x(~) 10-6x 1502=37.5xlO-4 J; E2=0 Total energy = 37.5 x 10- 4J After Joining When the two capacitors are connected in parallel, the cahrge of 50 /.!C gets redistributed and the two capacitors come to a common potential V. V _ total charge = 50 j..lC =30 V - totalcapacitance [(113)+ (4/3)] /.!F E1 = !- x (1/3)X10-6 x 302=1.5X 10-4 J E2 = ~ x (4/3) X 10-6 x 302 =6.0 X 10-4 J Total energy = 7.5 x 10- 4 J; Loss of energy =(37.5 - 7.5) x 10- 4 = 3 x 10- 2 J The energy is wasted away as heat in the conductor connecting the two capacitors. Example 5.28. An air-capacitor of capacitance 0.005 j..lF is connected to a direct voltage of 500 V, is disconnected and then immersed in oil with a relative permittivity of 2.5. Find the energy stored in the capacitor before and after immersion. (Elect. Technology: London Univ.) Solution. Energy before immersion is E1 = !- CV2 =!- x 0.005 X10-6 x 5002 = 625 x 10~ J When immersed in oil, its capacitance is increased 2.5 times. Since charge is constant, voltage must become 2.5 times. Hence, new capacitances is 2.5 x 0.005 =0.0125 ~ and new voltage is 500/2.5 = 200 V. Capacitance 221 E2 = !x 0.0125 X10-6 x (200)2= 250 x 10..{iJ Example 5.29. A parallel-plate air capacitor is charged to 100 V. Its plate separation is 2 mm and the area of each of its plates is 120 cm2. Calaculate and account for the increase or decrase of stored energy when plate separation is reduced to I mm (a) at constant voltage (b) at constant charge. Solution. Capacitance is the first case A -12 -4 C = ~ =S.S54x 10 x 120x 10 _ 53.1X10-12F 1 d 2xlO-3 Capacitance in the second case i.e. with reduced spacing C2 = S.S54XIO-12~~20XIO-4_106.2XIO-12F lxlO (a) When Voltage is Constant . 1 2 I 2 Change In stored energy dE = "2 C2V -"2 C2V = tx 1002x (106.2-53.1)xl0-12 = 26.55X 10-.'!J This represents an increase in the energy of the capacitor. This extra work has been done by the external supply source because charge has to be given to the capacitor when its capacitance increases, voltage remaining constant. (b) When Charge Remains Constant 1 Q2 2 Energy in the first case E1 = 2"c ; Energy in the second case, E2 = 1. ~ 1 2 C2 .. change in energy is dE = !Q2 (5;.1 -1O~.2)x 1012J 1 2 ( 1 1 ) x 12 = "2(C1\-I) 53.1 - 106.2 10 J = ~(53.1x 10-12)2xl04 x 0.0094x 1012 = 13.3 x 10- 8joules Hence, there is a decrease in the stored energy. The reason is that charge remaining constant, when the capacitance is increased, the,nvoltage must fall with a consequent decrease in stored energy I (E =- QV) 2 Example 5.30. A point charge of 100 IlC is embedded in an extensive mass of bakelite which has a relative permittivity of 5. Calculate the total energy contained in the electricfield outside a radial distance of (i) 100 m (ii) 10 m (iii) I m and (iv) 1 em. Solution. As per the Coulomb's law, the electric field intensity at any distance x from the point charge is given by E =Q/41t € x2. Let us draw a spherical shell of radius x as shown in Fig. Another spherical shell of radius (x + dx) has also been drawn. A differential volume of the space enclosed between the two shells is dv =4 1t i dx. As per Art. 5.15, the energy stored per unit volume of the electric field is (II2) DE. Hence, differential energy contained in the small volume is dW = _ 2 1 DE dv = _2 1 € E2 dv =_2 1 € ( Q 2 ] 2 4 1t x2 dx = _ SQ2 .~ 41t€x 1tE x 222 Electrical Technology Total energy of the electric field extending from x =R to x = 00is Q2 f = - 2 Q2 Q2 W = 81t£ RX dx=81t£R= (i) The energy contained in the electic field lying outside a radius of R =100m is (l00xlO-6)2 W = 12 =0.90J 8 1tx 8.854 x 10- x 5 x 100 (ii) For R =10 m, W =10 x 0.09 =0.09J (iii) For R =1 m, W = 100 x 0.09 = 9 J (iv) For R = 1 em, W = 10,000 x 0.09 = 900 J Example 5.31. Calculate the change in the stored energy of a parallel-plate capacitor if a dielectric slab of relative permittivity 5 is introduced between its two plates. Solution. Let A be the plate area, d the plate separation, E the electric field intensity and D the electric flux density of the capacitor. As per Art. 5.15, energy stored per unit volume of the field is =(112) DE. Since the space volume is d x A, hence, ( ) 2 1 1 2 1 VI WI = "2D}E,XdA="2£oEIxdA="2£odA d When the dielectric slab is introduced, ( ) 2 1 1 2 1 V2 W2 = "2 D2E2 XdA ="2 £ E2 X dA ="2 £0 £r dA d 1£ £ dA ( ~L ) 2 =!£ dA ( VI ) 2 _L:. w = WL 2 0 r £r d 2 0 d £r 2 Er It is seen that the stored energy is reduced by a factor of £r' Hence, change in energy is dW = WI-W2=\V;( I-1- ) =Wl( I-~ ) =W,x~:. dW = 0.8 Er 5 5 WI Example 5.32. When a capacitor C charges through a resistor Rfrom a d.c. source voltage E, determine the energy appearing as heat. [Bombay University, 2000] Solution. R-C series ciroit switched on to a d.c. Source of voltage E, at t =0, results into a current i (t), given by where i (t) t ~ WR = (EIR) e- 11't =RC Energy apperaring as heat in time ~t = i2R~t = Energy appearing as heat in time ~t =;2R~t 1 = .2 WR = R 0 I dt = R (E/R)2 1= (£-11't)2=.1 CE2 o 2 Note: Energy stored by the capacitor at the end of charging process = 1/2 CE Hence, energy received from the source = CF. If capacitance becomes (C - dC) due to the movement dx, then 2 2 2 Final stored energy =J.2 (C -.QdC) =!2 C 1- . 9_( IdC I=!2 C C SL(I + dC )if dC « C C \. / 2 ( ) 2 2 .. Change in stored eneroy =! SL I + dC _! SL=! SL dC "" 2 C C 2 C 2 C2 . 2 Equating Eq. (i) and (ii), we have F.dx = 4 ~2 . dC 2 F = lQ_ ~~=~V2 dC 2 C2 . dx 2 . dx 10A . dC _ 10A C = ~ .. dx --7 1 2 10A I ( V ) 2 I 2 F = - - V - - =- - 10A - newtons = - - 10A E newtons . . 2 . x2 2 x 2 This represents the force between the plates of a parallel-plate capacitor charged to a p.d. of V volts. The negaitve sign showns that it is a force of attraction. Example 5.33. A parallel-plate capacitor is made of plates 1 m square and has a separation of 1 mm. The space between the plates is filled with dielectric of lOr= 25.0. If 1 k V potential difference is applied to the plates. find theforce squeezing the plates together. (Electromagnetic Theory, A.M.LE. See B, 1993) Solution. As seen from Art. 5.16, F =- (1/2)EolOrAff newton · Now E = Vld = 1000/1 x 10-3= IO~ VIm .. F=_! Eo10 AE2 = - !x8.854x 10-12x 25x I x (l06)2=-1.I X 10-4 N 2 r 2 Capacitance 5.16. Force of Attraction Between Oppositely-charged Plates In Fig. 5.28 are shown two parallel conducting plates A and B carrying constant charges of + Q and - Q coulombs respectively. Let the force of attraction between the two be F newtons. If one of the plates is pulled apart by distance dx, then work done is =F x dx joules ...(i) Since the plate charges remain constant, no electrical energy comes into the arrangement during the movement dx. .. Work done =change in stored energy Initial stored energy =~ ~ joules Now Tutorial Porblems No. 5.2 223 B Q -+jdx.... Fig. 5.28 ...(ii) (': V =QIC) 1. Find the capacitance per unit length of a cylindrical capcitor of which the two conductors have radii 2.5 and 4.5 cm and dielectric consists of two layers whose cylinder of contanct is 3.5 cm in radius, the inner layer having a dielectric constant of 4 and the outer one of 6. [440 pF/m] 2. A parallel-plate capacitor, having plates 100 cm2 area, has three dielectrics I mm each and of permittivities 3. 4 and 6. If a peak voltage of 2,000 V is applied to the plates, calculate: (a) potential gradient across each dielectric (b) energy stored in each dielectric. [8.89 kV/cm; 6.67 kV/cm; 4.44 kV/cm ; 1047,786,524 x 10-7 joule] A I + + QLF F + -+- 224 Electrical Technology 3. The core and lead-sheath of a sin~le-core cable are separated by a rubber covering. The cross- secitional area of the core is 16 mm. A voltage of 10 kV is applied to the cable. What must be the thickness of the rubber insulation if the electric field strength in it is not to exceed 6 x 106VIm ? [2.5 mm (approx)] 4. A circular -=onductorof I cm diameter is surrounded by a oncentric conducting cylinder having an inner diameter of 2.5 cm. If the maximum electric stress in the dielectric is 40 kV/cm, calculate the potential difference between the conductors and also the minimum value of the electric stress. [18.4 kV ; 16 kV/cm] 5. A multiple capacitor has parallel plates each of area 12 cm2 and each separated by a mica sheet 0.2 mm thick. If dielectric constant for mica is 5, calculate the capacitance. [265.6 J1J1F] 6. A p.d. of 10 kV is applied to the terminals of a capacitor of two circular plates each having an area of 100 sq. cm. separated by a dielectric 1 mm thick. If the capacitance is 3 x 10-4 microfarad, calculate the electric flux density and the relative permittivity of the dielectric. I [D =3 x 10-4 C/m2, £,= 3.39] (City & Guilds, London) 7. Each electrode of a capacitor of the electrolytic type has an area of 0.02 sq. metre. The relative permittivity of the dielectric film is 2.8. If the capacitor has a capacitance of 10 J1F.estimate the thickness of the dielectric film. [4.95 x 10-8 m] (I.E.E. London) 5.17. Current-Voltage Relationships in a Capacitor The charge on a capacitor is given by the expression Q =CV. By differentiating this relation, we i = dQ =iL(CV)=c dV dt dt dt Following important facts can be deduced from the above relations : (i) since Q = CV, it means that the voltage across a capacitor is proportional to charge. not the current. (ii) a capacitor has the ability to store charge and hence to provide a short of memory. (iii) a capacitor can have a voltage across it even when there is no currentflowing. (iv) from i =c dV/dt, it is clear that current in the capacitor is present only when voltage on it chages with time. If dV/dt =0 i.e. when its voltage is constant or for d.c. voltage, i =O. Hence,the capacitorbehaveslike an opencircuit. . (v) from i = C dV/dt, we have dV/q.t = i/c. It shows that for a given value of (chargeor discharge) current i, rate of change in voltage is inversely proportional to capacitance. Larger the value of C, slower the rate of change in capacitive voltage. Also, capcitor voltage cannot change instantaneously. (vi) the aboveequationcan be put as dv=i.. dt C Intergrating the above, we get f dv =~f i .dt or dv =~ f~ i dt Example 5.34. The voltage across a 5 ~F capacitor changes uniformlyfr~m 10 to 70 V in 5 ms. Calculate (i) change in capacitor charge (ii) charging current. Solution. Q = CV :. dQ =C . dVand i =C dV/dt (i) dV = 70 - 10 =60 V, :. dQ =5 x 60 =300 J1C. (ii) ; = C. dV/dt =5 x 60/5 =60 mA Example5.35. An unchargedcapacitorofO.OJF is chargedfirst by a currentof2 mAfor 30 secondsand theyby a currentof 4 mAfor 30 seconds. Findthefinal voltagein it. Solution. Sincethe capacitoris initiallyuncharged,we willusetheprincipleof Superposition. VI= 0.~1I:°2XIO-3 .dt=l00x2xlO-3x30=6V 1 r30 -3 -3 V2= O.OIJo 4xIO .dt=IOOx4xlO x30=12V;:. V=V1+V2=6+12=18V get Capacitance 225 Example 5.36. The voltage across two series-connected 10 11F capacitors changes unifonnly from 30 to /50 V in 1 ms. Calculate the rate of change of voltage for (i) each capacitor and (ii) combination. Solution. For series combination V - V Cz - V d V - V CI _ 2V I - --an z- . - CI+ Cz 3 CI+ Cz 3 When V =30 V VI = VI3 =30/3 = 10 V ; Vz=2V13=2 x 30/3 =20 V When V =150 V VI = 150/3=50 V and Vz=2 x 150/3=100 V ( ' ) . d~ _ (50-10)_ 40kV ' .dVz_(loo-20)V_ I .. d - 1 - ,s, d -. 80 kV/s t ms t dV (150- 30) (ii) - = - 120 kV/s dt I ms It is seenthatdV/dt=dV/dt + dV/dt. 5.18. Charging of a Capacitor In Fig. 5.29. (a) is shown an ammgement by which a capacitor C may be charged through a high resistanceR from a battery of V volts. The voltage across C can be measured by a suitable voltmeter. When switch S is connected to terminal (a), C is charged but when it is connected to b, C is short circuited through R and is thus discharged. As shown in Fig. 5.29. (b), switch S is shifted to a for charging the capacitor for the battery. The voltage across C does not rise to V instantaneously but builds up slowly i.e. exponentially and not linearly. Charging current ie is maximum at the start i.e. when C is uncharge, then it decreases exponentially and finally ceases when p.d. across capacitor plates becomes equal and opposite to the battery voltage V. At any instant during charging, let ve = p.d. across C; ie=charging current q = charge on capacitor paltes a S R C '~ b a R C b iCR-L-'C. Fig. 5.29 The applied voltage V is always equal to the sum of : (i) resistive drop (ieR) and (ii) voltage across capacitor (vC> .. V = ie R + ve ...(i) . dq d, dv dv .. Now Ie = dt =dt (Cve)= C d: :. V = ve + CR dte ...(11) dVe _ dt or ----- V - ve CR Integrating both sides, we get J - d Ve =_..l J dt; :. loge(V- ve)=--C t +K ...(iii) V - Ve CR 'R where K is the constant of integration whose value can be found from initial known conditions. We know that at the start of charging when t = 0, ve= O. I -v « A ::r: U V rn 0 226 Electrical Technology Substituting these values in (iii), we get loge V =K Hence, Eq. (iii) becomes loge (V - ve)= ~~ + loge V V-v -( I log ~ - - =- - where A. - CR - time constant e V - CR A. -- V - v(' -In.. V (I -10.. -'---"- = e or v = - e ) ... V e This gives variation with time of voltage across the capacitor plates and is shown in Fig. 5.27.(a) or ...(iv) Q~---------------- q t t 0 (a) (b) Fig. 5.30 Now ve = q/C and V =Q/C Equation (iv) becomes 9..=Q (1- e-I/A.):. q =Q (1- e-In..) ...(v) c c We find that increase of charge, like growth of potential, follows an exponential law in which the steady value is reached after infinite time (Fig. 5.30 b). Now, ie =dq/dt. Differentiating both sides of Eq. (v), we get dq = i =Q!L(I_e-In..)=Q ( +~e-In.. m ) e m A. Q -In.. CV -In.. = T e = CR e (": Q =CV and A. =CR) . V -In.. . I -In.. ( . ) .. 'e = R' e or 'e = 0 e ... VI where 10= maximum current = VIR Exponentially rising curves for ve and q are shown in Fig. 5.30 (a) and (b) respectively. Fig. 5.30 (c) shows the curve for exponentially decreasing charging current. It should be particularly Capacitance 227 noted that ic decreases in magnitude only but its direction of flow remains the same i.e. positive. As charging continues, charging current decreases according to equation (vI) as shown in Fig. 5.30 (c). It becomes zero when t = 00(though it is almost zero in about 5 time constants). Under steady-state conditions, the circuit appears only as a capacitor which means it acts as an open-circuit. Similarly, it can be proved that vRdecreases from its initial maximum value of Vto zero exponentially as given by the relation vR = Ve-tlA.. 5.19. Time Constant (a) Just at the start of charging, p.d. across capacitor is zero, hence from (il) putting Vc=0, we get V = CR dvc dt .. initial rate of rise of voltage across the capacitor is* = ( dvc ') =.Y... =~ volt/second dt )=0 CR II. If this rate of rise were maintained, then time taken to reach voltage V would have been V + VICR =CR. This time is known as time constant (A) of the circuit. Hence, time constant of an R-C circuit is defined as the time during which voltage across capacitor would have reached its maximum value V had it maintained its initial rate of rise. (b) In equation (iv) if t =A., then -If}.. -If}.. -I ) ( 1 ) V ( 1 ) v =V (1- e )=V (1- e )=V(1- e =V 1- - = I - - =0632 V c e 2.718' Hence, time constant may be defined as the time during which capacitor voltage acually rises to 0.632 of its final steady value. (c) From equaiton (vI), by putting t = A,we get ic = loe-)J')..=loe-I=Ir/l.718=.0.3710 Hence, the constant of a circuit is also the time during which the charging current falls to 0.37 of its initial maximum value (orfalls by 0.632 of its initial value). 5.20. Discharging of a Capacitor As shown in Fig. 5.31 (a), when S is shifted to b, C is discharged through R. It will be seen that the discharging current flows in a direction opposite to that the charging current as shown in Fig. 5.31 (b). Hence, if the direction of the charging current is taken positive, then that of the discharging current will be taken as negative. To begin with, the discharge current is maximum but then decreases exponentially till it ceases when capacitor is fully discharged. Fig. 5.31 Since battery is cut of the circuit, therefore, by putting V =0 in equaiton (il) of Art. 5.18, we get dv dv ( . dv,. ) o =CR---L =V or v =- CR ---L .: I. =C - dt c c dt ( dt It can also be found by differentiating Eq. (iv) with respect to time and then putting t = O. (a) (b) * R C t + A 0 -=- V 228 Electrical Technology dvc dt J dV 1 J t - - - or ~=-- dt:. log v =--+k .. Vc - CR Vc CR c c CR At the start of discharge, when t =0, Vc = :. loge V =0 + K; or loge V =K R C Putting this value above, we get loge Vc = - * + loge V or loge v/V =- tlA. S. .1 1 Q -In.. d . I -In.. lOliary, q = e an 'c=- oe It can be proved that - In.. vR = - Ve The fall of cap1\citorpotential and its dischargining current are shown in Fig. 5.32 (b). One practical application of the above charging and discharging of a capacitor is found in digital control circuits where a square-wave input is applied across an R-C circuit as shown in Fig. 5.32 (a). The different waveforms of the current and voltages are shown in Fig. 5.32 (b), (c), (d), (e). The sharp voltage pulses of VRare used for control circuits. Example 5.37. Calculate the current in and voltage drop across each element of the circuit shown in Fig. 5.33 (a) after switch S has been closed long enough for steady-state Fig. 5.32 conditions to prevail. Also, calculte voltage drop across the capacitor and the discharge current at the instant when S is opened. (a) /\. INPUTI (b) ~o A 2A 3)" 4A ~ I I (c) Vc I I o I I I I I I I I I l I I I I' I I I I I I I I (e) I or Vc V = - In.. Tl - In.. e or Vc= ve Solution. Under steady-state conditions, the capacitor becomes fully charged and draws no current. In fact, it acts like an open circuit with the result that no current flows through the l-Q resistor. The steady state current lss flows through loop ABCD only. A S 6 B C lCE -- AUA +1b,E 4 \ IOOV -\00 V 4 G D C F 0 C F (a) Fig. 5.33 (b) Hence, lss = 100/(6 + 4) = 10 A Drop V6 = 100 x 61(6+ 4) = 60 V V4 = 100 x 4/10 =40 V VI = 0 x 2 =0 V Voltage across th.ecapacitor = drop across B - C =40 V Capacitance 229 Switch Open When S is opened, the charged capacitor discharges through the loop l1CFE as shown in Fig. 5.33 (b). The discharge current is given by ID = 40/(4 + 1)=8 A As seen, it flows in a direction opposite to that of Iss' Example 5.38. (a) A capacitor is charged through a large non-rective resistance by a battery of constant voltage V. Derive an expression for the instantaneous charge on the capacitor. (b) For the above arrangement, if the capacitor has a capacitance of 10 ~ F and the resistance is 1M Q, calculate the time takenfor the capacitor to receive 90% of itsfinal charge. Also. draw the charge/time curve. Solution. (a) For this part, please refer to Art. 5.18. (b) A. =CR = 10 x 10-6 x 1 x 106=10 s; q =0.9 Q Now. q =Q (I - e-III) :. 0.9 Q =Q (1 - e-tllO)or e'110=10 .. 0.1 t loge e =loge 10 or 0.1 t =2.3 loglo 10 =2.3 or t =23 s The charge/time curve is similar to that shown in Fig. 5.27 (b). Example 5.39. A resistance R and a 4 ~F capacitor are connected in series across a 200 V. d.c. supply. Across the capacitor is a neon lamp that strikes (glows) at 120 V. Calcualte the value of R to make the lamp strike (glow) 5 seconds after the switch has been closed. (Electrotechnics-I.M.S. Univ. Baroda 1988) Solution. Obviously, the capacitor voltage has to rise 120 V in 5 seconds. -Sf).. Sf).. .. 120=200 (I - e ) or e =2.5 or 1= 5.464second. Now, A.= CR :. R = 5.464/4x 10-6= 1.366MQ Example 5.40. A capacitor of 0.1 ~F is chargedfrom a 100-Vbattery through a series resistance of 1,000 ohms. Find (a) the timefor the capacitor to receive 63.2 % of itsfinal charge. (b) the charge received in this time (c) thefinal rate of charging. (d) the rate of charging when the charge is 63.2% of thefinal charge. (Elect. Engineering, Bombay Univ. 1985) Solution. (a) As seen from Art. 5.18 (b), 63.2% of charge is received in a time equal to the time constant of the circuit. Time required =A. =CR =0.1 x 10- 6 x I()()(:)=0.1 x 10-3=10- 4 second (b) Final charge, Q = CV = 0.1 x 100 = 10 ~C Charge received during this time is =0.632 x 10=6.32 II C (c) The rate of charging at any time is given by Eq. (ii) of Art. 5.18. dV V-v dt = CR d v V _ 100 _ 6 v=O, Hence - d = CR - 6 3 10 Vis t 0.1 x 10- x 10 v = 0.632 V =0.632 x 100=63.2 volts dv 100-63.2 - = 368kV/s dt 10-4 Initially (d) Here .. Example 5.41. A series combination having R =2 M Q and C =0.01 ~F is connected across a d.c. voltage source of 50 V. Determine (a) capacitor voltage after 0.02 s, 0.04 s, 0.06 sand 1 hour (b) cha.'-gingcurrent after 0.02 s, 0.04 s. 0.06 sand 0.1 s. 230 Electrical Technology Solution. A. = CR =2 x 106x 0.01 x 10- 6=0.02 second 6 1m = VIR= 50/2x 10 = 25 ~. While solving this question, it should be remembered that (i) in each time constant, Vcincreases further by 63.2% of its balance value and (ii) in each constant, ic decreases to 37% its previous value. (a) (i) t =0.02 s . Since, initially at t = 0, vc=0 V and Ve= 50 V, hence, in one time constant Vc = 0.632 (50 - 0) =31.6 V (ii) t =0.()4 s This time equals two time-constants. :. Vc = 31.6+ 0.632(50- 31.6)=43.2 V (iii) t =0.06 s This time equals three time-constants. :. Vc = 43.2+ 0.632(50- 43.2)=47. 5 V Since in one hour, steady-state conditions would be established, Vcwould have achieved its maximum possible value of 50 V. (b) (i) t =0.02 s, ic = 0.37 x 25 =9.25 J1A (ii) t =0.4s, ic = 0.37x 9.25= 3.4J1A (iii) t =0.06s, ic = 0.37x 3.4= 1.26~ (iv) t = 0.1 s, This time equals 5 time constants. In this time, current falls almost to zero value. Example 5.42. A voltage as shown in Fig. 5.43 (a) is applied to a series circuit consisting of a resistance of 2 Q in series with a pure capacitor of 100~. Determine the voltage across the capacitor at t =0.5 millisecond. [Bombay University, 2000] JOVI see) o Fig. 5.34 (a) Solution. \0 \0 r I 7.18 c 0~6 t (millisee) (msee) Fig. 5.34 (b) t =RC =0.2 milli-second Between 0 and 0.2 m sec; v (t) =10 [1 - exp (- tit)] At t =0.2, v (t) =6.32 volts Between 0.2 and 0.4 m See, counting time from A indicating it as tI -------- 0.2 0.4 0.6 t (milli Capacitance 231 v (t) =6.32 exp (tlt) At point B, t) =0.2, V =2.325 Between 0.4 and 0.6 m Sec, time is counted from ~with variable as t2, v (t2) = 2.325 + (10 - 2.325) [I - exp (- tit)] At C, t2 = 0.2, V =7.716 volts. 5.21. Transient Relations During Capacitor Charging Cycle Whenever a circuit goes from one steady-state condition to another steady-state condition, it passes through a transient state which is of short duration. The first steady-state condition is called the initial condition and the second steady-state condition is called the final condition. In fact, transient condition lies in between the initial and final conditions. For example, when switch S in Fig. 5.35 (a) is not connected either to a or b, the RC circuit is in its initial steady state with no current and hence no voltage drops. When S is shifted to point a, current starts flowing through R and hence, transient voltages are developed across R and C till they achieve their final steady values. The period during which current and voltage changes take place is called transient condition. The moment switch S is shifted to point 'a' as shown in Fig. 5.35 (b), a charging current ieis set up which starts charging C that is initially uncharged. At the beginning of the transient state, ie is maximum because there is no potential across C to oppose the applied voltage V. It has maximum value =VIR =10' It produces maximum voltage drop across R = ieR = loR. Also, initially, Vc= 0, but as time passes, ie decreases gradually so does vRbut ve increases exponentially ti1l it reaches the final steady value of V. Although V is constant, vRand veare variable. However, at auy time V =vR + ve =i~ + ve' At the beginning of the transient state, ie=10, ve =0 but vR=V. At the end of the transient state, ie =0 hence, vR =0 but ve= V. \{, /-- (dvc!dvh=0 / VL-__J____________ / / / / / / / /, o (a) Fig. 5.35 The initialrates of change of ve' vRand ieare given by ( dV I V d: ,to = 1:' volt/second, ( dv R 1 loR V - = -=--volt/second dt =0 A. A. ( die I I V dt }=o = ~ where 10='R These are the initial rates of change. However, their rate of change at any time during the charging transient are given as under: , (b) v (c) Fig. 5.35 232 Electrical Technology dVe V -IrA..die _ dVR _ V -IrA. - - -e e dt - A. ' dt dt A. It is shown in Fig. 5.35 (c). It should be clearly understood that a negative rate of change means a decreasing rate of change. It does not mean that the concerned quantity has reversed its direction. 5.22. Transient Relations During Capacitor Discharging Cycle As shown in Fig. 5.36 (b), switch S has been shifted to b. Hence, the capacitor undergoes the discharge cycle. Just before the transient state starts, ie= 0, vR= 0 and ve= V. The moment transient state begins, ie has maximum value and decreases exponentially to zero 'at the end of the transient state. So does ve. However, during discharge, all rates of change have polarity opposite to that during charge. For example, dvIdt has a positive rate of change during charging and negative rate of change during discharging. ic o -1m (b) o -v (c) Fig. 5.36 Also, it should be noted that during discharge, ve maintains its original polarity whereas ie reverses its direction of flow. Consequently, during capacitor discharge, vR also reverses its direction. The various rates of change at any time during the discharge transients are as given in Art. dVe V -IrA.. die _ 10 -IrA..dVR _ V -IrA. dt = - A. e 'di-Te 'dt- A. e These are represented by the curves of Fig. 5.32. 5.23. Charging and Discharging of a capacitor with Initial Charge In Art. 5.18, we considered the case when the capacitor was initially uncharged and hence, had no voltage across it. Let us now consider the case, when the capacitor has an initial potential of Vo fless than V) which opposes the applied battery voltage Vas shown in Fig. 5.37 (a). As seen from Fig. 5.37 (b), the initial rate of rise of ve is now somewhat less than when the Capacitance 233 capacitor is initially uncharged. Since the capacitor voltage rises from an initial value of Voto the final value of V in one time constant, its initial rate of rise is given by ( dVc 1 V - Vo _ V - Vo dt =0 = ~-RC s 'J( ..~~ } l!o~'-\I< -=-v An (0) v -t (b) Fig. 5.37 The value of the capacitor voltage at any time during the charging cycle is given by Vc = (V-V~(1-e-tn..)+Vo ;r~~ -=-v A(! (0) v r-~-j --T.--------- II /1 I t o (b) Fig. 5.38 However, as shown in Fig. 5.38 (a), if the inital capacitor votlage is negative with respect to the battery votlage i.e. the capacitor votlage is seriesaiding the battery voltage, rate of change of Vcis steeper than in the previous case. It is so because as shown in Fig. 5.38 (b), in one time period, the voltage chage =V - (- V~ =(V + Yo)' Hence, the initial rate of change of voltage is given by ( dVc I V + Vo_ V + Vo dt 1=0 = -X--Ji(; The value of capacitor voltage at any time durign the charging cycle is given by . Vc = (V + Yo)(l - e-tn..)- Vo The time required for the capacitor voltage to attain any value of Vcduring the charging cycle is given by = 'AIn ( V - Vo ) =RC In [ V - Vo ) ... when Vois positive V - Vc V - vc t = 'AIn ( V + Vo ) =RC In [ V + Vo ) ... when Vois negative V - Vc V - vc Example 5.43. In Fig. 5.39, the capacitor is initially uncharged and the switch S is then closed. Find the vlaues of I, II' 12and the voltage at the point A at the start and finish of the transient state. 234 . Electrical Technology Solution. At the moment of closing the switch i.e. at the start of the transient state, the capacitor acts as a short-circuit. Hence, there is only a resitance of 2 Q in the circuit because 1Q resistance is shorted out thereby grounding point A. Hence, II =0; I:: 12= 12/2 = 6A. Obviously, VA = 0 V. At the end of the transient state, the capacitor acts as an open-circuit. Hence, 12 = 0 and I = 11= 12/(2 + 1) = 4 A. VA = 6 V. Example 5.44. Calculate the values of i2. i3' v2, v3, Va' Vcand VLof the network shown in Fig. 5.40 at the following times: (i) At time, t =0 + immediately after the switch S is closed; (ii) At time. t ~ 00 i.e. in the steadystate.. (NetworkAnalysisAMIE Sec.B Winter 1990) i Solution. (i) In this case the coil acts as an open 3 circuit, hence i2 = 0 ; v2 = 0 and vL = 20 V. Since a capacitor acts as a short circuit i3=20/(5 + 4) =9 =20/9 A. Hence, v3 = (20/9) x 4 = 80/9 V and Vc= O. (ii) Under steady state conditions, capacitor acts as an open circuit and coil as a short circuit. Hence, i2=20/ (5 + 7) =20/12 =5/3 A; v2 =7 x 5/3 A; v2=7 x 5/3 =35/ 3 V; vL=O. Also i3=0, v3=0 but Vc=20 V. Example 5.45. If in the RC circuit of Fig. 5.36; R =2 M Q, C =5 m F and V = 100 V. calculate (a) initial rate of change of capacitor voltage (b) initial rate of change of capacitor current (c) initial rate of change of voltage across the 2 M Q resistor (d) all of the above at t =80 s. I, 6NF 2 Fig. 5.39 s 20V 1H 3F IVe Fig. 5.40 (d) All the above rates of change would be zero because the transient disappears after about 5 A. =5 x 10 =50 s. Example 5.46. In Fig. 5.41 (a). the capacitor C is fully discharged. since the switch is in positIOn 2. At time t = 0, the switch is shifted to position I for 2 seconds. It is then returned 10 position 2 where it remains indefinite". Calculate (a) the maximum voltage to which the capacitor is charged when in position I. (b) charging time constant 1./ in position I. (c) discharging time contant A.2in position 2. (d) Vcand icat the end of I second in position I. (dV ) V 100 _100 =lOV/s Solution. (a) d: 1=0 = I"= 2x106x5xlO-6 106 " (di 1 10-_ VIR =_IOOl2xlO --5JlNs (b) -£... dt =0 A. = -- T 10 (dVR 1 - V =_100 =-lOV/s (c) - = A. 10 dt =0 Capacitance 235 (e) ve and ie at the instant the s.vitch is shifted to positon 2 at t =] second. (fJ veand ieafter a lapse of] second when inposition2. (g) sketch the wavefonns for Veand iefor the first 2 seconds of the above switching sequence. Solution. (a) We will ftrst ftnd the voltage available at tenninal 1. As seen the net battery voltage around the circuit =40 - 10 == 30 V. Drop across 30 K resistor =30 x 30/(30 + 60) =10V. Hence, potential of tenninal I with respect to ground G =40 - 10=30 V. Hence, capacitor will charge to a maximum voltage of 30 V when in position I. (b) Total resistance, R = [(30 K II60 K) + 10 K] = 30 K .. AI =RC =30 K x 10 J.1F=0.3 s (c) ~ =10 K x 1O!l F =0.1 s _,r I - 1103 (d) Vc= V (l - e ') = 30 (l - e . ) = 28.9 V i = V e-,Jl...1=30V e-1I0.03=lx00361=0036mA c R 30K .. (e) Vc =28.9 V at t =l+S at position 2 but ic =-28.9 V/lO K =- 2/89 mA at t = l+s in position 2. (fJ Vc =28.9 e-,Jl...2=28.9 e -110.1=0.0013 V =0 V. ic =28.9 e -tJl...2=- 2.89 e -1/0.1 =0.00013 mA ==O. The waveform of the capacitor voltage and charging current are sketched in Fig. 5.41 (b). I (s) -2.89 mA (e) (b) Fig. 5.41 Example 5.47. In the RC circuit of Fig. 5.42, R ::: 2 M Q and C = 5 Il F, the capaciTOr I charged to an initial potential of 50 V. When the switch is closed at t =0+.calculate (a) initial rate of change of capacitor voltage and (b) capacitor voltage after a lapse of 5 times the time constant i.e. 5A. 30K 40V -=- I 10K fQ) "l-*' 30V -=- :?60K (a) ic 1 30L . 0.036 mA 28.9 V '0 t (s) If the polarity of capacitor voltage is reversed, calculate (c) the values of the above quantities and (d) time for Vcto reach - IO V, 0 V and 95 V. ( dVc I V - V. Solution. (a) -;:j( 1=0 = T = V - Vo=100 - 50 = 5 Vis RC 10 (b) vc=(V- Vo)(l-e-1t?.)+ Vo = (l00 - 50) (l- e-S'AJ")= 50 = 49.7 + 50 = 99.7 V (c) When Vo = - 50 V,( dVc ) = V - (- yo) = V + Vo= ISO= 15 Vis dt 1=0 A A 10 Vc = (V - V~ (l - e-1t?.)+ Vo=[100 - (- 50)](1 - e-s) + (- 50) = 150(I-e-s)-50=99V. = Aln ( V - Vo ) = 10In [ 100 - (- 50) ] = 10In ( 150 ) = 3.1s V-vc 100-(-10) 110 = 10 In [ 100 - (- 50) ] =10In ( 150 ) = 4.055 s 100- (0) 100, t = 10 In [ 100- (- 50) J =10In ( ISO ) = 34 s 100 - 95 5 Example 5.48. The uncharged capacitor, if it is initially switched to position I of the switch for 2 sec and then switched to position 2 for the next two seconds. What will be the voltage on the capacitor at the end of this period ? Sketch the variation of voltage across the capacitor.[Bombay University 2001] Solution. Uncharged capacitor is switched to position I for 2 seconds. It will be charged to 100 volts instantaneously since resistance is not present in the charging circuit. After 2 seconds, the capacitor charged to 100 volts will get discharged through R-C circuit with a time constant of 't = RC=1500x 10-3=1.5sec. Counting time from instant of switching over to positon 2, the expression for voltage across the capacitor is V (t) =100 cxp (- t/'t) After 2 seconds in this position, v (t) = 100 exp (- 2/1.5) =26/36 Volts. Example 5.49. There are three passive elements in the circuit below and a voltage and a current are defined for each. Find the values of these six qualities at both t = 0- and t = 0+. [Bombay University, 2001] Solution. Current source 4 u (1)means a step function of 4 amp applied at t =O. Other current source of 5 amp is operative throughout. At t =0-, 5 amp source is operati'/e. This unidirectional constant current establishes a steady ('urrentof 5 amp through 30-ohm resistor and 3-H inductor. Note that positive VRmeans a rise from !ht to left. 236 (d) Electrical Technology s ~ ')(... "~A ;;v 1- ,:. ~m--1- -=-JOOV Fig. 5.42 2 + 100I'F 100V Fig. 5.43. . 237 o - 150 Volts (Since right-terminal of Resistor is + ve) 5 amp 0, it represents the voltage between B and O. o 150 volts = VBO + (Voltage between A and B with due regards to sign). 0-(- 150) =+ 150 volts 30n A + 3H + SA 1/27F 4 ~ (t)A 1 Fig. 5.44 (a) At t = 0+, 4 amp step function becomes operative. Capacitive-voltage and Inductance-current cannot change abruptly. Hence iL (0+) = 5 amp VdO+) = 150 amp . VdO) =150 volts, with node A positive with respect to O. With these two values known, the waveforms for current sources are drawn in Fig. 5.44 (b). 5 amp 4 amp i 4 u (t)amp o 14u(t)amp time, t · Fig. 5.44 (b) Remaining four parameters are evaluated from Fig. 5.44 (c). VL = VB = VA - (30 x I) = 120 Volts iR =1 amp, VR =- 30 Volts ic =4 amp in downward direction. Additional Observation. After 4 amp SOUIceis operative, final conditions (at t tending to infinity) are as follows. Inductance carnes a total direct current of 9 amp, with vi. = o. Hence, VB = 0 iR = 5 amp.VR= - 150volts Vc = 150volts,i( = 0 Example5.50. Thevoltageasshownin Fig.5.45(a) isappliedacross- (i)A resistmof 2 ohms (ii) A capacitor of2 f. Find and sketch the current in each case up to 6 seconds. Capacitance At t = VR = iL = VL = ic = Vc = = B A lamp L -L 5=P 1127F C 5amp T 4amp r Fig. 5.44 (c) Fig. 5.45 (b) Current in a Resistor of 2 ohms iR =V (t)/2 amp 20 (Volt) i. <,,"p)1I f-- 2 3 4 -20 (Volt) I 5 (see) 6 Fig. 5.45 (c) Current thro 2-F capacitor, ic = C (dv/dt) Examplf' 5.51. Three capacitors 2 /-if. 3/-iF.and J1Fare connected in series and chargedJrom I(Ji' 'UP/II> Find the voltage across condensers. They are then disconnected from the ,/IJ'11. fI,J n i/lllected with all tlu + ve plates connected together and all the -ve plates connected t>" ,tli, J'TI,) [/ll t 'Jltages aa{lss the combinations and the charge on each capacitor after 1/1IIcti,J' .\\"J/nn "e'f,'ct in 'dation. [Bombay University, 1998] , ,Iution. The capacitors are connected in series. If C is the resultant capacitance. llC = IIC, + llC2=llC3, which gives C =(30/31) ~ V, = 900 x (30/31)12=435.5 volts V2 = 900 x (30/31)/3 =290.3 volts 238 ElectricalTechnology . 10 (Volt) lor I I \ 4 5 6 I 1 3 \ I I I I 2 time,t (see) 10 (Volt)r---:---------- Fig. 5.45(a) [BombayUniversity 1998] Solution. 5 L I I 5 6 I 2 3 time,t (see) - 5 ------------- Capacftance 239 V3 = 900 x (30/31)/5 =174.2 volts 8 A C1 2 ~f 1 C2 3 ~f C3 5 ~f II l' n-- +-V3 V + -~OO V Fig. 5.46 In series connection, charge held by each capacitor is same. If it is denoted by Q. -{j Q = 435x 2 x 10 =871 J.1coulombs Threecapacitorsholda totalchargeof (3 x 871)=2613 J.1coulombs Withparallelconnectionof thesethreecapacitors,equivalentcapacitance,C = C1+ C2+ C3= 1OJ.1F Since, Q' = C, 2613 x 1O-{j= 10 x 1O-{jx V' or V' = 261 volts. Charge on each capacitor after reconnection is as follows : QI' = C1 VI= 2 x lO~x 261 = 522.J.1-coulombs Q2' = C2 VI = 3 x 10 x 261 = 783 J.1-coulombs Q3' = C3V2= 5 x lO-{jx261 = 1305 J.1-coulombs Tutorial Problems No. 5.3 1. Foir the circuit shown in Fig. 5.47 calculate (;) equivalent capacitance and (ii) voltage drop across eachcapacitor. All capacitancevaluesarein IJF. [(i)6 JlF(ii) VAB = 50V, VBC=40V] 2. In the circuit of Fig. 5.48 find (i) equivalent capacitance (ii) drop across each capacitor and (iii) charge on each capacitor. All capacitance values are in IJF. [(i) 1.82JIF(ii) VI = 50V; V2= V3= 20V; V4= 40V (iii) QI = 200JlC;Q2= 160JlC;Q3= 40 JlC;Q4= 200JlC] 5 8 A~E~ .=..12V 30pF 6K S 2K lOOV Fig. 5.47 JlOV Fig.5.48 Fig. 5.49 Fig. 5.50 3. With switch in Fig. 5.49 closed and steady-state conditions established, calculate (i) steady-state current (iI) voltage and charge across capacitor (iii) what would be the discharge current at the instant of openingtheswitch? [(i) 1.5mA (ii) 9V; 270JlC (iii) 1.5mAl 4. Whenthe circuit of Fig. 5.50 is in steady state, what would be the p.d. across the capacitor? Also, find the discharge current at the instant S is opened. [8 V; 1.8A] 5. Find the time constant of the circuit shown in Fig. 5.51. [200 JlS] IK 6. A capacitor of capacitance 0.011JF is being charged by 1000 V d.c. supply through a resistor of 0.01 megaohm. Determine the voltage to whichthe capacitorhasbeenchargedwhenthechargingcurrenthas *0.1j1F 2K:E 2K decreased to 90 % of its initial value. Find also the time taken for the current to decrease to 90% of its initial value. [100 V, 0.1056 ms] 7. An 81JF capacitor is being charged by a 400 V supply through 0.1 Fig. 5.51 Electrical Technology mega-ohm resistor. How long will it take the capacitor to develop a p.d. of 300 V? Also what fraction of the final energy is stored in the capacitor? [1.11 Second, 56.3% of full energy] 8. An 10 f,IF'capacitor is charged from a 200 V battery 250 times/second and completely discharged through a 5 0 resistor during the interval between charges. Determine (a) the power taken from the battery. (b) the average value of the current in'5 0 resistor. [(a) 50 W (b) 0.5 A] 9. When a capacitor, charged to a p.d. of 400 V, is connected to a voltmeter having a resistance of 25 MO, the voltmeter reading is observed to have fallen to 50 V at the end of an interval of 2 minutes. Find the capacitance of the capacitor. [2.31 JlF] (App. Elect. London Univ.) 240 OBJECTIVE TESTS - 5 1. A capacitor consists of two (a) insulation separated by a dielectric (b) conductors separated by an insulator (c) ceramic plates and one mica disc (d) silver-coated insulators 2. The capacitance of a capacitor is NOT influenced by (a) plate thickness (b) plate area (c) plate separation (d) nature of the dielectric 3. A capacitor that stores a charge of 0.5 C at 10 volts has a capacitance of farad. (a) 5 (b) 20 (c) 10 (d) 0.05 4. If dielectric slab of thickness 5 mm and E,= 6 is inserted between the plates of an air capacitor with plate separation of 8 mm, its capacitance is (a) decreased (b) almostdoubled (c) almost halved (d) unaffected 5. In a cable capacitor, voltage gradient is maximum at the surface of the (a) sheath (b) conductor (c) insulator (d) earth 6. In Fig. 5.52 voltage across C1will be volt. (a) 100 (b) 200 (c) 150 (d) 300 W 300 V Fig. 5.52 7. The capacitanceof a cablecapacitordependson (a) core,diameter (b) insulationthickness (c) ratio of cylinder radii (d) potential difference 8. The insulation resistance of a cable capacitor depends on (a) applied voltage (b) insulationthickness (c) core diameter (d) ratio of inner and outer radii 9. The time constant of an R-C circuit is defined as the time during which capacitor charging current becomes percent of its...value. (a) 37, final (b) 63, final (c) 63, initial (d) 37, initial 10. The period during which current and voltage changes take place in a circuit is called ..... condition. (a) varying (b) permanent (c) transient (d) steady 11. In an R-C circuit connnected across a d.c. voltage source, which of the following is zero at the beginning of the transient state ? (a) drop across R (b) charging current (c) capacitor voltage (d) none of the above 12. When an R-C circuit is suddenly connected across a d.c. voltage source, the initial rate of charge of capacitor is (a) - [rIA (b) [riA (c) VIR (d) - VIA. 13. Which of the following quantity maintains the same polarity during charging and discharging of a capacitor? (a) capacitor voltage (b) capacitor current (c) resistive drop (d) none of the above 14. In a cable capacitorwith compound dielectric, voltage gradient is inverselyproportionalto (a) permittivity . (b) radius of insulating material (c) cable length (d) both (a) and (b) 15. While a capacitoris still connectedto a power source, the spacing'between its plates is halved. Which of its following quantity would remain constant ? (a) field strength (b) plate charge (c) potential difference (d) electric .flux density 16. After being disconnected from the power source, the spacing between the plates of a capacitor is halved. Which of the following quantity would be halved ? (a) plate charge (b) field strength (c) electric flux density (d) potential difference P'9( J'S( p.t( v'£( v'n J'n J'O( P'6 P'8 J'L Q'9 q'S q'" p'£ v"Z; q'( SH:iMSNV