ALTERNATING CURRENT

               1. INTRODUCTION

The electricity which we are using at our home, offices, schools, to run many appliances such as fans, lights, projectors, computers, etc. are Alternating current (a c). The electric mains supply in our homes and offices is a voltage that varies like a sine function with time. Such a voltage is called alternating voltage (a c voltage) and the current driven by it in a circuit is called the alternating current (a c current). Today, most of the electrical devices we use require a c voltage. This is mainly because most of the electrical energy sold by power companies is transmitted and distributed as alternating current. The main reason for preferring use of a c voltage over d voltage is that a c voltages can be easily and efficiently converted from one voltage to the other by means of transformers. Further, electrical energy can also be transmitted economically over long distances. AC circuit exhibit characteristics which are exploited in many devices of daily use. For example, whenever we tune our radio to a favourite station, we are taking advantage of a special property of a c circuits. 

2.AC VOLTAGE APPLIED TO A RESISTOR

A resistor connected to a source ε of a c voltage. The symbol for an a c source in a circuit diagram is ~ . We consider a source which produces sinusoidally varying potential difference across its terminals. Let this potential difference, also called a c voltage, be given by 

              v =v m sin ω 

where v m is the amplitude of the oscillating potential difference and ω is its angular frequency. To find the value of current through the resistor, we apply Kirchhoff’s loop rule ∑ε( )t = 0, to the circuit to get 

                    v m sin ω t= i R

or 

                        i= v m sin ω t/ R

Since R is a constant, we can write this equation as

                        i= i m sin ω  

where the current amplitude i m is given by 

                       i m= v m / R

Ohm’s law, which for resistors, works equally well for both a c and d c voltages. The voltage across a pure resistor and the current through it, are plotted as a function of time. Note, in particular that both v and i reach zero, minimum and maximum values at the same time. Clearly, the voltage and current are in phase with each other. We see that, like the applied voltage, the current varies sinusoidally and has corresponding positive and negative values during each cycle. Thus, the sum of the instantaneous current values over one complete cycle is zero, and the average current is zero. The fact that the average current is zero, however, does not mean that the average power consumed is zero and that there is no dissipation of electrical energy. As you know, Joule heating is given by i²R and depends on i² (which is always positive whether i is positive or negative) and not on i. Thus, there is Joule heating and dissipation of electrical energy when an a c current passes through a resistor. 

3. REPRESENTATION OF AC CURRENT AND VOLTAGE BY ROTATING VECTORS — PHASORS 

We have learn above that the current through a resistor is in phase with the a c voltage. But this is not so in the case of an inductor, a capacitor or a combination of these circuit elements. In order to show phase relationship between voltage and current in an a c circuit, we use the notion of phasors. The analysis of an a c circuit is facilitated by the use of a phasor diagram. A phasor is a vector which rotates about the origin with angular speed, ω . The vertical components of phasors V and I represent the sinusoidally varying quantities v and i. The magnitudes of phasors V and I represent the amplitudes or the peak values v m and i m of these oscillating quantities. The  voltage and current phasors and their relationship at time t 1 for the case of an a c source connected to a resistor i.e., corresponding to the circuit. The projection of voltage and current phasors on vertical axis, i.e., v m sin ω t and i m  sin ω t, respectively represent the value of voltage and current at that instant. As they rotate with frequency ω, curves are generated.  We see that phasors V and I for the case of a resistor are  in the same direction. This is so for all times. This means that the phase angle between the voltage and the current is zero. 

4. AC VOLTAGE APPLIED TO AN INDUCTOR

Usually, inductors have appreciable resistance in their windings, but we shall assume that this inductor has negligible resistance. Thus, the circuit is a purely inductive a c circuit. Let the voltage across the source be v = v m sin ω t. Using the Kirchhoff’s loop rule, ∑ε ( ) t = 0 , and since there is no resistor in the circuit, 

                         v - L di/ d t = 9

where the second term is the self-induced Faraday e m f in the inductor; and L is the self-inductance of the inductor. The negative sign follows from Lenz’s law. The integration constant has the dimension of current and is time-independent. Since the source has an em f which oscillates symmetrically about zero, the current it sustains also oscillates symmetrically about  zero, so that no constant or time-independent component of the current exists. Therefore, the integration constant is zero. The dimension of inductive reactance is the same as that of resistance and its SI unit is ohm (Ω). The inductive reactance limits the current in a purely inductive circuit in the same way as the resistance limits the current in a purely resistive circuit. The inductive reactance is directly proportional to the inductance and to the frequency of the current. A comparison for the source voltage and the current in an inductor shows that the current lags the voltage by π/2 or  one-quarter (1/4) cycle. The voltage and the current phasors in the present case at instant t 1 . The current phasor I is π/2 behind the voltage phasor V. When rotated with frequency ω counter- clockwise, they generate the voltage and current. 

5. AC VOLTAGE APPLIED TO A CAPACITOR

An a c source ε generating a c voltage v = v m sin ω t  connected to a capacitor only, a purely capacitive a c circuit. When a capacitor is connected to a voltage source in a d c circuit, current will flow for the short time required to charge the capacitor. As charge accumulates on the capacitor plates, the voltage across them increases, opposing the current. That is, a capacitor in a d c circuit will limit or oppose the current as it charges. When the capacitor is fully charged, the current in the circuit falls to zero. When the capacitor is connected to an a c source. It limits or regulates the current, but does not completely prevent the flow of charge. The capacitor is alternately charged and discharged as the current reverses each half cycle. Let q be the charge on the capacitor at any time t. The instantaneous voltage v across the capacitor is 

                              v = q /C

From the Kirchhoff’s loop rule, the voltage across the source and the capacitor are equal. The dimension of capacitive reactance is the same as that of resistance and its SI unit is ohm (Ω). The capacitive reactance limits the amplitude of the current in a purely capacitive circuit in the same way as the resistance limits the current in a purely resistive circuit. But it is inversely proportional to the frequency and the capacitance. A comparison with the equation of source voltage, shows that the current is π/2 ahead of voltage. The phasor diagram at an instant t 1 . Here the current phasor I is π/2 ahead of the voltage phasor V as they rotate counterclockwise. The variation of voltage and current with time. We see that the current reaches its maximum value earlier than the voltage by one-fourth of a period. 

6. AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT

A series LCR circuit connected to an a c source ε. As usual, we take the voltage of the source to be v= v m sin ω t.  If q is the charge on the capacitor and i the current, at time t, we have, from Kirchhoff’s loop rule:

             L di/d t  +  i r + q/C = v

We want to determine the instantaneous current i and its phase relationship to the applied alternating voltage v. We shall solve this problem by two methods. First, we use the technique of phasors and in the second method, we solve analytically to obtain the time– dependence of i.