Chapter Seven

Alternating Current

7.1 Introduction

 We have so far considered direct current (dc) sources and circuits with dc sources. These currents do not change direction with time. But voltages and currents that vary with time are very common. 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 (ac voltage) and the current driven by it in a circuit is called the alternating current (ac current)*. Today, most of the electrical devices we use require ac 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 ac voltage over dc voltage is that ac 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 circuits 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 ac circuits – one of many that you will study in this chapter.

7.2 AC Voltage Applied to a Resistor

 Figure 7.1 shows a resistor connected to a source ε of ac voltage. The symbol for an ac 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 ac voltage, be given by $\mathrm{V} = {\mathrm{V}}_{\mathrm{m}} \sin \mathrm{\omega t}$                                                    (7.1)

where v_m is the amplitude of the oscillating potential difference and ω is its angular frequency.

 

Figure 7.1 AC voltage applied to a resistor.

To find the value of current through the resistor, we apply Kirchhoff’s loop rule $\sum _{}\epsilon \left(t\right)=0$(refer to Section 3.13), to the circuit shown in Fig. 7.1 to get ${\mathrm{V}}_{\mathrm{m}} \sin \mathrm{\omega t} = \mathrm{iR}$

or $\mathrm{i} = \frac{{\mathrm{V}}_{\mathrm{m}}}{\mathrm{R}} \sin \mathrm{\omega t}$

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

$\mathrm{i} = {\mathrm{i}}_{\mathrm{m}} \sin \mathrm{\omega t}$                                                         (7.2)

where the current amplitude im is given by

${\mathrm{i}}_{\mathrm{m}} = \frac{{\mathrm{V}}_{\mathrm{m}}}{\mathrm{R}}$                                                         (7.3)

Figure 7.2 In a pure resistor, the voltage and current are in phase. The minima, zero and maxima occur at the same respective times.

 Equation (7.3) is Ohm’s law, which for resistors, works equally well for both ac and dc voltages. The voltage across a pure resistor and the current through it, given by Eqs. (7.1) and (7.2) are plotted as a function of time in Fig. 7.2. 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 ac current passes through a resistor.

George Westinghouse (1846 – 1914) A leading proponent of the use of alternating current over direct current. Thus, he came into conflict with Thomas Alva Edison, an advocate of direct current. Westinghouse was convinced that the technology of alternating current was the key to
the electrical future. He founded the famous Company named after him and enlisted the services of Nicola Tesla and other inventors in the development of alternating current motors and apparatus for the transmission of high tension current, pioneering in large scale lighting.

 The instantaneous power dissipated in the resistor is- $\mathrm{P} = {\mathrm{i}}^2\mathrm{R} = {\mathrm{i}}_{\mathrm{m}}^2 {\sin }^2\mathrm{\omega t} \mathrm{R}$                                                         (7.4)

The average value of p over a cycle is*

                                                         [7.5(a)]

where the bar over a letter (here, p) denotes its average value and <......> denotes taking average of the quantity inside the

bracket
$Since, i_m^2 and R are cons\tan ts,$

                                                       [7.5(b)]

 Using the trigonometric identity, sin² ωt = 1/2 (1– cos 2ωt), we have < sin² ωt > = (1/2) (1– < cos 2ωt >) and since < cos2ωt > = 0**, we have,

Thus,

$\overline{\mathrm{p}} = \frac{1}{2}{\mathrm{i}}_{\mathrm{m}}^2\mathrm{R}$                                                          [7.5(c)]

To express ac power in the same form as dc power (P = I₂R), a special value of current is defined and used. It is called, root mean square (rms) or effective current (Fig. 7.3) and is denoted by I_rms or I.

It is defined by

$\mathrm{I} = \sqrt{\overline{{\mathrm{i}}^2}} = \sqrt{\frac{1}{2}{\mathrm{i}}_{\mathrm{m}}^2} = \frac{{\mathrm{i}}_{\mathrm{m}}}{\sqrt{2}}$= 0.707 i_m                                          (7.6)

In terms of I, the average power, denoted by P is

$\mathrm{P} = \overline{\mathrm{P}} = \frac{1}{2}{\mathrm{i}}_{\mathrm{m}}^2\mathrm{R} = {\mathrm{I}}^2\mathrm{R}$                                                          (7.7)

Similarly, we define the rms voltage or effective voltage by

$\mathrm{V} =\frac{{\mathrm{v}}_{\mathrm{m}}}{\sqrt{2}}= 0.707 {\mathrm{v}}_{\mathrm{m}}$                                                           (7.8)

From Eq. (7.3), we have

v_m  =  i_mR

or, $\frac{{\mathrm{v}}_{\mathrm{m}}}{\sqrt{2}} = \frac{{\mathrm{i}}_{\mathrm{m}}}{\sqrt{2}}\mathrm{R}$

or, V = IR                                            (7.9)

Equation (7.9) gives the relation between ac current and ac voltage and is similar to that in the dc case. This shows the advantage of introducing the concept of rms values. In terms of rms values, the equation for power [Eq. (7.7)] and relation between current and voltage in ac circuits are essentially the same as those for the dc case.

It is customary to measure and specify rms values for ac quantities. For example, the household line voltage of 220 V is an rms value with a peak voltage of

$v_m=\sqrt{2}V = (1.414)(220 V) = 311 V$

In fact, the I or rms current is the equivalent dc current that would produce the same average power loss as the alternating current. Equation (7.7) can also be written as

P = V² / R = I V (since V = I R)

7.3 Representation of AC Current and Voltage by Rotating Vectors — Phasors

 In the previous section, we learnt that the current through a resistor is in phase with the ac 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 ac circuit, we use the notion of phasors. The analysis of an ac circuit is facilitated by the use of a phasor diagram. A phasor* is a vector which rotates about the origin with angular speed ω, as shown in Fig. 7.4. 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. Figure 7.4(a) shows the voltage and current phasors and their relationship at time t₁ for the case of an ac source connected to a resistor i.e., corresponding to the circuit shown in Fig. 7.1. The projection of voltage and current phasors on vertical axis, i.e., v_msinωt and i_msinωt, respectively represent the value of voltage and current at that instant. As they rotate with frequency ω, curves in Fig. 7.4(b) are generated.

Figure 7.4 (a) A phasor diagram for the circuit in Fig 7.1. (b) Graph of v and i versus ωt.

From Fig. 7.4(a) 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.

 * Though voltage and current in ac circuit are represented by phasors – rotating vectors, they are not vectors themselves. They are scalar quantities. It so happens that the amplitudes and phases of harmonically varying scalars combine mathematically in the same way as do the projections of rotating vectors of corresponding magnitudes and directions. The rotating vectors that represent harmonically varying scalar quantities are introduced only to provide us with a simple way of adding these quantities using a rule that we already know.

7.4 AC Voltage Applied to an Inductor

$\mathrm{i} = -\frac{{\mathrm{v}}_{\mathrm{m}}}{\mathrm{\omega L}}\cos \left(\mathrm{\omega t}\right) + \mathrm{constant}$

The integration constant has the dimension of current and is time-independent. Since the source has an emf 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.

Using

$- \cos \left(\mathrm{\omega t}\right) = \sin \left(\mathrm{\omega t} - \frac{\mathrm{\pi }}{2}\right)$, we have

$\mathrm{i} = {\mathrm{i}}_{\mathrm{m}} \sin \left(\mathrm{\omega t} - \frac{\mathrm{\pi }}{2}\right)$                                                                    (7.12)

where${\mathrm{i}}_{\mathrm{m}} = \frac{{\mathrm{v}}_{\mathrm{m}}}{\mathrm{\omega L}}$ is the amplitude of the current.  

The quantity ω L is analogous to the resistance and is called inductive reactance, denoted by X_L:

X_L = ω L                                                      (7.13)

The amplitude of the current is, then

${\mathrm{i}}_{\mathrm{m}} = \frac{{\mathrm{v}}_{\mathrm{m}}}{{\mathrm{X}}_{\mathrm{L}}}$                                                                    (7.14)

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 of Eqs. (7.1) and (7.12) 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. Figure 7.6 (a) shows the voltage and the current phasors in the present case at instant t₁. 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 given by Eqs. (7.1) and (7.12), respectively and as shown in Fig. 7.6(b).

Figure 7.6 (a) A Phasor diagram for the circuit in Fig. 7.5. (b) Graph of v and i versus ωt.

We see that the current reaches its maximum value later than the voltage by one-fourth of a period $\left[\frac{\mathrm{T}}{4}=\frac{{\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\mathrm{\omega }}\right]$. You have seen that an inductor has reactance that limits current similar to resistance in a dc circuit. Does it also consume power like a resistance? Let us try to find out.

 The instantaneous power supplied to the inductor is

$$\begin{gathered} {\mathrm{p}}_{\mathrm{L}} = \mathrm{iv} = {\mathrm{i}}_{\mathrm{m}} \sin \left(\mathrm{\omega t} - \frac{\mathrm{\pi }}{2}\right)\times {\mathrm{v}}_{\mathrm{m}} \sin \left(\mathrm{\omega t}\right) \\ = - {\mathrm{i}}_{\mathrm{m}}{\mathrm{v}}_{\mathrm{m}} \cos \left(\mathrm{\omega t}\right) \sin \left(\mathrm{\omega t}\right) \\ = -\frac{{\mathrm{i}}_{\mathrm{m}}{\mathrm{v}}_{\mathrm{m}}}{2} \sin \left(2\mathrm{\omega t}\right) \end{gathered}$$

So, the average power over a complete cycle is

since the average of sin (2ωt) over a complete cycle is zero.

Thus, the average power supplied to an inductor over one complete cycle is zero.

Figure 7.7 explains it in detail.

One complete cycle of voltage/current. Note that the current lags the voltage.

0-1 Current i through the coil entering at A increase from zero to a maximum value. Flux lines are set up i.e., the core gets magnetised. With the polarity shown voltage and current are both positive. So their product p is positive. ENERGY IS ABSORBED FROM THE SOURCE.

1-2 Current in the coil is still positive but is decreasing. The core gets demagnetised and the net flux becomes zero at the end of a half cycle. The voltage v is negative (since di/dt is negative). The product of voltage and current is negative, and ENERGY IS BEING RETURNED TO SOURCE.

2-3 Current i becomes negative i.e., it enters at B and comes out of A. Since the direction of current has changed, the polarity of the magnet changes. The current and voltage are both negative. So their product p is positive. ENERGY IS ABSORBED.

7.5 AC Voltage Applied to a Capacitor

 Figure 7.8 shows an ac source ε generating ac voltage v = v_msinωt connected to a capacitor only, a purely capacitive ac circuit.

When a capacitor is connected to a voltage source in a dc 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 dc 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 ac source, as in Fig. 7.8, 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

(7.15)

From the Kirchhoff’s loop rule, the voltage across the source and the capacitor are equal,

To find the current, we use the relation

Using the relation, $\cos \omega t=\sin \left(\omega t+\frac{\pi }{2}\right)$, we have

$i=i_m\sin \left(\omega t+\frac{\pi }{2}\right)$(7.16)

where the amplitude of the oscillating current is i_m = (ωC)v_m. We can rewrite it as

Comparing it to i_m= v_m/R for a purely resistive circuit,  we find that (1/ωC) plays the role of resistance. It is called capacitive reactance and is denoted by X_c,

X_c= 1/ωC (7.17)

so that the amplitude of the current is

(7.18)

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 of Eq. (7.16) with the equation of source voltage, Eq. (7.1) shows that the current is π/2 ahead of voltage.

Figure 7.9 (a) A Phasor diagram for the circuit in Fig. 7.8. (b) Graph of v and i versus ωt.

Figure 7.9(a) shows the phasor diagram at an instant t₁. Here the current phasor I is π/2 ahead of the voltage phasor V as they rotate counterclockwise. Figure 7.9(b) shows 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.

The instantaneous power supplied to the capacitor is

p_c = i v = i_m cos(ωt)v_m sin(ωt)

            = i_m v_m cos(ωt) sin(ωt)

(7.19)

So, as in the case of an inductor, the average power

since ωt)> = 0 over a complete cycle. Figure 7.10 explains it in detail.

Thus, we see that in the case of an inductor, the current lags the voltage by π/2 and in the case of a capacitor, the current leads the voltage by π/2.

                           One complete cycle of voltage/current. Note that the current leads the voltage.

 

 

0-1 The current i flows as shown and from the maximum at 0, reaches a zero value at 1. The plate A is charged to positive polarity while negative charge q builds up in B reaching a maximum at 1 until the current becomes zero. The voltage vc = q/C is in phase with q and reaches maximum value at 1. Current and voltage are both positive. So p = vci is positive.

Energy is absorbed from THE SOURCE during this quarter cycle AS THE CAPACITOR IS CHARGED.

1-2 The current i reverses its direction. The accumulated charge is depleted i.e., the capacitor is discharged during this quarter cycle.The voltage gets reduced but is still positive. The current is negative. Their product, the power is negative.

the Energy absorbed during THE 1/4 CYCLE 0-1 IS RETURNED DURING this QUARTER.

 

2-3 As i continues to flow from A to B, the capacitor is charged to reversed polarity i.e., the plate B acquires positive and A acquires negative charge. Both the current and the voltage are negative. Their product p is positive. The capacitor ABSORBS ENERGY during this 1/4 cycle.

 

3-4 The current i reverses its direction at 3 and flows from B to A. The accumulated charge is depleted and the magnitude of the voltage vc is reduced. vc becomes zero at 4 when the capacitor is fully discharged. The power is negative.ENERGY ABSORBED DURING 2-3 IS RETURNED TO THE SOURCE. NET ENERGY ABSORBED IS ZERO.

Figure 7.10 Charging and discharging of a capacitor.

 

7.6 AC Voltage Applied to a Series LCR Circuit

 Figure 7.12 shows a series LCR circuit connected to an ac source ε. As usual, we take the voltage of the source to be v = v_m sin ωt.

If q is the charge on the capacitor and i the current, at time t, we have, from Kirchhoff’s loop rule:

(7.20)

FIGURE 7.12 A series LCR circuit connected to an ac source.

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 Eq. (7.20) analytically to obtain the time–dependence of i.

7.6.1 Phasor-diagram solution

 From the circuit shown in Fig. 7.12, we see that the resistor, inductor and capacitor are in series. Therefore, the ac current in each element is the same at any time, having the same amplitude and phase. Let it be

i = i_m sin(ωt+φ) (7.21)

where φ is the phase difference between the voltage across the source and the current in the circuit. On the basis of what we have learnt in the previous sections, we shall construct a phasor diagram for the present case.

Figure 7.13 (a) Relation between the phasors V_L, V_R, V_C, and I, (b) Relation between the phasors V_L, V_R, and (V_L + V_C) for the circuit in Fig. 7.11.

Let I be the phasor representing the current in the circuit as given by Eq. (7.21). Further, let V_L, V_R, V_C, and V represent the voltage across the inductor, resistor, capacitor and the source, respectively. From previous section, we know that V_R is parallel to I, V_C is π/2 behind I and V_L is π/2 ahead of I. V_L, V_R, V_C and I are shown in Fig. 7.13(a) with apppropriate phase-relations.

The length of these phasors or the amplitude of V_L, V_R, V_C are:

v_Rm = i_m R, v_Cm = i_m X_C, v_Lm = i_m X_L                  (7.22)

The voltage Equation (7.20) for the circuit can be written as

v_L + v_R + v_C = v                       (7.23)

The phasor relation whose vertical component gives the above equation is

V_L + V_R + V_C = V                       (7.24)

This relation is represented in Fig. 7.13(b). Since V_C  and V_L are always along the same line and in opposite directions, they can be combined into a single phasor (V_C  + V_L) which has a magnitude |v_Cm – v_Lm.| Since V is represented as the hypotenuse of a right-triangle whose sides are V_R and (V_C  + V_L), the pythagorean theorem gives:

Substituting the values of v_Rm, v_Cm, and v_Lm from Eq. (7.22) into the above equation, we have

or, [7.25(a)]

By analogy to the resistance in a circuit, we introduce the impedance Z in an ac circuit:

[7.25(b)]

where (7.26)

 Since phasor I is always parallel to phasor V_R, the phase angle φ is the angle between V_R and V and can be determined from
Fig. 7.14:

Using Eq. (7.22), we have

(7.27)

Equations (7.26) and (7.27) are graphically shown in Fig. (7.14). This is called Impedance diagram which is a right-triangle with Z as its hypotenuse.

Equation 7.25(a) gives the amplitude of the current and Eq. (7.27) gives the phase angle. With these, Eq. (7.21) is completely specified.

If X_C > X_L, φ is positive and the circuit is predominantly capacitive. Consequently, the current in the circuit leads the source voltage. If X_C < X_L, φ is negative and the circuit is predominantly inductive. Consequently, the current in the circuit lags the source voltage.

Figure 7.15 shows the phasor diagram and variation of v and i with ωt for the case X_C > X_L.

Thus, we have obtained the amplitude and phase of current for an LCR series circuit using the technique of phasors. But this method of analysing ac circuits suffers from certain disadvantages. First, the phasor diagram say nothing about the initial condition. One can take any arbitrary value of t (say, t₁, as done throughout this chapter) and draw different phasors which show the relative angle between different phasors.

The solution so obtained is called the steady-state solution. This is not a general solution. Additionally, we do have a transient solution which exists even for v = 0. The general solution is the sum of the transient solution and the steady-state solution. After a sufficiently long time, the effects of the transient solution die out and the behaviour of the circuit is described by the steady-state solution.

7.6.2 Analytical solution

The voltage equation for the circuit is

= v_msinωt

We know that i = dq/dt. Therefore, di/dt = d²q/dt². Thus, in terms of q, the voltage equation becomes

(7.28)

This is like the equation for a forced, damped oscillator, [see Eq. {14.37(b)} in Class XI Physics Textbook]. Let us assume a solution

q = q_m sin (ω t + θ) [7.29(a)]

so that [7.29(b)]

and [7.29(c)]

Substituting these values in Eq. (7.28), we get

= (7.30)

where we have used the relation X_c= 1/ωC, X_L = ω L. Multiplying and dividing Eq. (7.30) by , we have

$$\begin{gathered} q_m\omega Z\left[\frac{R}{Z}\cos \left(\omega t+\theta \right) +\frac{X_C-X_L}{Z}\sin \left(\omega t+\theta \right)\right]=v_m\sin \omega t \left(7.31\right) \\ Now, let \frac{R}{Z}=\cos \varphi \end{gathered}$$

and

$so that \varphi ={\tan }^{-1}\frac{X_C-X_L}{R} \left(7.32\right)$