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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.

respectively, since they mean, literally, alternating current voltage and alternating

current current. Still, the abbreviation ac to designate an electrical quantity

displaying simple harmonic time dependance has become so universally accepted

that we follow others in its use. Further, voltage – another phrase commonly

used means potential difference between two points.

Physics

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

v = vm sin ω t (7.1)

where vm is the amplitude of the oscillating potential

difference and ω is its angular frequency.

1943) Yugoslov scientist,

inventor and genius. He

NICOLA TESLA (1836 – 1943)

rotating magnetic field,

which is the basis of

practically all alternating

current machinery, and

which helped usher in the

age of electric power. He FIGURE 7.1 AC voltage applied to a resistor.

also invented among other

things the induction motor, To find the value of current through the resistor, we

the polyphase system of ac

power, and the high apply Kirchhoff’s loop rule ∑ ε (t ) = 0 , to the circuit

frequency induction coil shown in Fig. 7.1 to get

(the Tesla coil) used in radio

vm sin ω t = i R

and television sets and

other electronic equipment. vm

The SI unit of magnetic field or i = sin ω t

R

is named in his honour.

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

i = im sin ω t (7.2)

where the current amplitude im is given by

vm

im = (7.3)

R

Equation (7.3) is just 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

FIGURE 7.2 In a pure values at the same time. Clearly, the voltage and current are in

resistor, the voltage and phase with each other.

current are in phase. The We see that, like the applied voltage, the current varies

minima, zero and maxima sinusoidally and has corresponding positive and negative values

occur at the same

respective times. during each cycle. Thus, the sum of the instantaneous current

values over one complete cycle is zero, and the average current

234 is zero. The fact that the average current is zero, however, does

Alternating Current

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 i 2R and depends on i 2

(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.

The instantaneous power dissipated in the resistor is

p = i 2 R = im2 R sin2 ω t (7.4)

The average value of p over a cycle is*

p = < i 2 R > = < i m2 R sin2 ω t > [7.5(a)]

where the bar over a letter(here, p) denotes its average

George Westinghouse

value and <......> denotes taking average of the quantity

2 (1846 – 1914) A leading

inside the bracket. Since, i m and R are constants, proponent of the use of

p = i m2 R < sin2 ωt > [7.5(b)] alternating current over

direct current. Thus,

Using the trigonometric identity, sin ω t = 2

he came into conflict

1/2 (1– cos 2ωt ), we have < sin2 ωt > = (1/2) (1– < cos 2ωt >) with Thomas Alva Edison,

and since < cos2ωt > = 0**, we have, an advocate of direct

1 curr ent. Westinghouse

< sin 2 ω t > = was convinced that the

2

technology of alternating

Thus, current was the key to

1 2 the electrical future.

p= im R [7.5(c)] He founded the famous

2

Company named after him

To express ac power in the same form as dc power and enlisted the services

(P = I2R), a special value of current is defined and used. of Nicola Tesla and

It is called, root mean square (rms) or effective current other inventors in the

(Fig. 7.3) and is denoted by Irms or I. development of alternating

current motors and

apparatus for the

transmission of high

tension current, pioneering

in large scale lighting.

peak current im by I = i m / 2 = 0.707 im.

T

1

T ∫0

* The average value of a function F (t ) over a period T is given by F (t ) = F (t ) dt

1 T

1 ⎡ sin 2ω t ⎤ T 1

** < cos 2ω t > = T ∫ cos 2ω t dt = T ⎢⎣ =

2ω ⎥⎦ 0 2ω T

[ sin 2ω T − 0] = 0 235

0

Physics

It is defined by

1 2 i

I = i2 = im = m

2 2

= 0.707 im (7.6)

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

1 2

P = p= im R = I 2 R (7.7)

2

Similarly, we define the rms voltage or effective voltage by

vm

V= = 0.707 vm (7.8)

2

From Eq. (7.3), we have

v m = i mR

vm im

or, = R

2 2

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

vm = 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 )

Example 7.1 A light bulb is rated at 100W for a 220 V supply. Find

(a) the resistance of the bulb; (b) the peak voltage of the source; and

(c) the rms current through the bulb.

Solution

(a) We are given P = 100 W and V = 220 V. The resistance of the

bulb is

V 2 ( 220 V )

2

R= = = 484Ω

P 100 W

(b) The peak voltage of the source is

EXAMPLE 7.1

vm = 2V = 311 V

(c) Since, P = I V

P 100 W

I = = = 0.450A

V 220 V

236

Alternating Current

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 vm and im of these

oscillating quantities. Figure 7.4(a) shows the FIGURE 7.4 (a) A phasor diagram for the

voltage and current phasors and their circuit in Fig 7.1. (b) Graph of v and

relationship at time t1 for the case of an ac source i versus ωt.

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., vm sinωt and im sinωt,

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.

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.

Figure 7.5 shows an ac source connected 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 ac circuit. Let

the voltage across the source be v = vm sinωt. Using

the Kirchhoff’s loop rule, ∑ ε (t ) = 0 , and since there

is no resistor in the circuit,

di

v −L =0 (7.10)

dt

where the second term is the self-induced Faraday FIGURE 7.5 An ac source

emf in the inductor; and L is the self-inductance of connected to an inductor.

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. 237

Physics

the inductor. The negative sign follows from Lenz’s law (Chapter 6).

Combining Eqs. (7.1) and (7.10), we have

d i v vm

= = sin ω t (7.11)

dt L L

Interactive animation on Phasor diagrams of ac circuits containing, R, L, C and RLC series circuits:

Equation (7.11) implies that the equation for i(t), the current as a

function of time, must be such that its slope di/dt is a sinusoidally varying

quantity, with the same phase as the source voltage and an amplitude

given by vm/L. To obtain the current, we integrate di/dt with respect to

time:

di vm

∫ dt dt = L ∫ sin(ωt )dt

and get,

vm

i =− cos( ωt ) + constant

ωL

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

⎛ π⎞

− cos(ω t ) = sin ⎜ ω t − ⎟ , we have

⎝ 2⎠

⎛ π⎞

http://www.phys.unsw.edu.au/~jw/AC.html

i = i m sin ⎜ ωt − ⎟ (7.12)

⎝ 2⎠

vm

where im = is the amplitude of the current. The quantity ω L is

ωL

analogous to the resistance and is called inductive reactance, denoted

by XL:

XL = ω L (7.13)

The amplitude of the current is, then

vm

im = (7.14)

XL

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

238 (7.12), respectively and as shown in Fig. 7.6(b).

Alternating Current

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

⎡T π /2 ⎤

voltage by one-fourth of a period ⎢ =

ω ⎥⎦

. You have seen that an

⎣4

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

⎛ π⎞

p L = i v = im sin ⎜ ω t − ⎟ ×vm sin ( ωt )

⎝ 2⎠

= −im vm cos ( ωt ) sin ( ωt )

i m vm

=− sin ( 2ωt )

2

So, the average power over a complete cycle is

i m vm

PL = − sin ( 2ω t )

2

i m vm

=− sin ( 2ω t ) = 0,

2

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.

220 V. Find the inductive reactance and rms current in the circuit if

the frequency of the source is 50 Hz.

Solution The inductive reactance,

X L = 2 π ν L = 2 × 3.14 × 50 × 25 × 10 –3 W

EXAMPLE 7.2

= 7.85Ω

The rms current in the circuit is

V 220 V

I = = = 28A

X L 7.85 Ω 239

Physics

0-1 Current i through the coil entering at A 1-2 Current in the coil is still positive but is

increase from zero to a maximum value. Flux decreasing. The core gets demagnetised and

lines are set up i.e., the core gets magnetised. the net flux becomes zero at the end of a half

With the polarity shown voltage and current cycle. The voltage v is negative (since di/dt is

are both positive. So their product p is positive. negative). The product of voltage and current

ENERGY IS ABSORBED FROM THE is negative, and ENERGY IS BEING

SOURCE. RETURNED TO SOURCE.

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

2-3 Current i becomes negative i.e., it enters 3-4 Current i decreases and reaches its zero

at B and comes out of A. Since the direction value at 4 when core is demagnetised and flux

of current has changed, the polarity of the is zero. The voltage is positive but the current

magnet changes. The current and voltage are is negative. The power is, therefore, negative.

both negative. So their product p is positive. ENERGY ABSORBED DURING THE 1/4

ENERGY IS ABSORBED. CYCLE 2-3 IS RETURNED TO THE SOURCE.

Alternating Current

Figure 7.8 shows an ac source ε generating ac voltage v = vm sin ωt

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 FIGURE 7.8 An ac source

capacitor is alternately charged and discharged as connected to a capacitor.

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

q

v= (7.15)

C

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

capacitor are equal,

q

vm sin ω t =

C

dq

To find the current, we use the relation i =

dt

d

i =

dt

(vm C sin ω t ) = ω C vm cos(ω t )

π

Using the relation, cos(ω t ) = sin ⎛⎜ ω t + ⎞⎟ , we have

⎝ 2⎠

⎛ π⎞

i = im sin ⎜ ω t + ⎟ (7.16)

⎝ 2⎠

where the amplitude of the oscillating current is im = ωCvm. We can rewrite

it as

vm

im =

(1/ ω C )

Comparing it to im= vm/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 Xc,

Xc= 1/ωC (7.17)

so that the amplitude of the current is

vm

im = (7.18) 241

XC

Physics

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.

FIGURE 7.9 (a) A Phasor diagram for the circuit

A comparison of Eq. (7.16) with the

in Fig. 7.8. (b) Graph of v and i versus ωt. equation of source voltage, Eq. (7.1) shows that

the current is π/2 ahead of voltage.

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

pc = i v = im cos(ωt)vm sin(ωt)

= imvm cos(ωt) sin(ωt)

i m vm

= sin(2ωt ) (7.19)

2

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

i m vm i v

PC = sin(2ωt ) = m m sin(2ωt ) = 0

2 2

since <sin (2ωt)> = 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.

your observations for dc and ac connections. What happens in each

case if the capacitance of the capacitor is reduced?

Solution When a dc source is connected to a capacitor, the capacitor

gets charged and after charging no current flows in the circuit and

EXAMPLE 7.3

the lamp will not glow. There will be no change even if C is reduced.

With ac source, the capacitor offers capacitative reactance (1/ω C )

and the current flows in the circuit. Consequently, the lamp will shine.

Reducing C will increase reactance and the lamp will shine less brightly

than before.

Find the capacitive reactance and the current (rms and peak) in the

circuit. If the frequency is doubled, what happens to the capacitive

reactance and the current?

EXAMPLE 7.4

1 1

XC = = = 212 Ω

2 π ν C 2π (50Hz)(15.0 × 10 −6 F)

The rms current is

242

Alternating Current

0-1 The current i flows as shown and from the 1-2 The current i reverses its direction. The

maximum at 0, reaches a zero value at 1. The plate accumulated charge is depleted i.e., the capacitor is

A is charged to positive polarity while negative charge discharged during this quarter cycle.The voltage gets

q builds up in B reaching a maximum at 1 until the reduced but is still positive. The current is negative.

current becomes zero. The voltage vc = q/C is in phase Their product, the power is negative.

with q and reaches maximum value at 1. Current THE ENERGY ABSORBED DURING THE 1/4

and voltage are both positive. So p = vci is positive. CYCLE 0-1 IS RETURNED DURING THIS QUARTER.

ENERGY IS ABSORBED FROM THE SOURCE

DURING THIS QUAR TER CYCLE AS THE

CAPACITOR IS CHARGED.

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

2-3 As i continues to flow from A to B, the capacitor 3-4 The current i reverses its direction at 3 and flows

is charged to reversed polarity i.e., the plate B from B to A. The accumulated charge is depleted

acquires positive and A acquires negative charge. and the magnitude of the voltage vc is reduced. vc

Both the current and the voltage are negative. Their becomes zero at 4 when the capacitor is fully

product p is positive. The capacitor ABSORBS discharged. The power is negative.ENERGY

ENERGY during this 1/4 cycle. ABSORBED DURING 2-3 IS RETURNED TO THE

SOURCE. NET ENERGY ABSORBED IS ZERO.

Physics

V 220 V

I = = = 1.04 A

X C 212 Ω

The peak current is

EXAMPLE 7.4 i m = 2I = (1.41)(1.04 A ) = 1.47 A

This current oscillates between +1.47A and –1.47 A, and is ahead of

the voltage by π/2.

If the frequency is doubled, the capacitive reactance is halved and

consequently, the current is doubled.

Example 7.5 A light bulb and an open coil inductor are connected to

an ac source through a key as shown in Fig. 7.11.

FIGURE 7.11

The switch is closed and after sometime, an iron rod is inserted into

the interior of the inductor. The glow of the light bulb (a) increases; (b)

decreases; (c) is unchanged, as the iron rod is inserted. Give your

answer with reasons.

Solution As the iron rod is inserted, the magnetic field inside the coil

EXAMPLE 7.5

magnetizes the iron increasing the magnetic field inside it. Hence,

the inductance of the coil increases. Consequently, the inductive

reactance of the coil increases. As a result, a larger fraction of the

applied ac voltage appears across the inductor, leaving less voltage

across the bulb. Therefore, the glow of the light bulb decreases.

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 = vm sin ωt.

If q is the charge on the capacitor and i the

current, at time t, we have, from Kirchhoff’s loop

rule:

di q

L +iR + =v (7.20)

dt C

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

FIGURE 7.12 A series LCR circuit phasors and in the second method, we solve

connected to an ac source. Eq. (7.20) analytically to obtain the time–

244 dependence of i .

Alternating Current

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 = im 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.

Let I be the phasor representing the current in the circuit as given by

Eq. (7.21). Further, let VL, VR, VC, and V represent the voltage across the

inductor, resistor, capacitor and the source, respectively. From previous

section, we know that VR is parallel to I, VC is π/2

behind I and VL is π/2 ahead of I. VL, VR, VC and I

are shown in Fig. 7.13(a) with apppropriate phase-

relations.

The length of these phasors or the amplitude

of VR, VC and VL are:

vRm = im R, vCm = im XC, vLm = im XL (7.22)

The voltage Equation (7.20) for the circuit can

be written as

vL + vR + vC = v (7.23)

The phasor relation whose vertical component

gives the above equation is FIGURE 7.13 (a) Relation between the

phasors VL, VR, VC, and I, (b) Relation

VL + VR + VC = V (7.24) between the phasors VL, VR, and (VL + VC)

This relation is represented in Fig. 7.13(b). Since for the circuit in Fig. 7.11.

VC and VL are always along the same line and in

opposite directions, they can be combined into a single phasor (VC + VL)

which has a magnitude ⏐vCm – vLm⏐. Since V is represented as the

hypotenuse of a right-traingle whose sides are VR and (VC + VL), the

pythagorean theorem gives:

+ (vCm − v Lm )

2

vm2 = v Rm

2

Substituting the values of vRm, vCm, and vLm from Eq. (7.22) into the above

equation, we have

vm2 = (im R )2 + (i m X C − im X L )2

= im2 ⎡⎣R 2 + ( X C − X L )2 ⎤⎦

vm

or, i m = [7.25(a)]

R + ( X C − X L )2

2

in an ac circuit:

vm

im = [7.25(b)]

Z

Physics

Since phasor I is always parallel to phasor VR, the phase angle φ

is the angle between VR and V and can be determined from

Fig. 7.14:

vCm − v Lm

tan φ =

v Rm

Using Eq. (7.22), we have

XC − X L

tan φ = (7.27)

R

Equations (7.26) and (7.27) are graphically shown in Fig. (7.14).

FIGURE 7.14 Impedance This is called Impedance diagram which is a right-triangle with

diagram. 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 XC > XL, φ is positive and the circuit is predominantly capacitive.

Consequently, the current in the circuit leads the source voltage. If

XC < 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 ω t

for the case XC > XL.

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

FIGURE 7.15 (a) Phasor diagram of V and I. solution. Additionally, we do have a

(b) Graphs of v and i versus ω t for a series LCR transient solution which exists even for

circuit where XC > XL. 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.

The voltage equation for the circuit is

di q

L + Ri + = v

dt C

= vm sin ωt

246 the voltage equation becomes

Alternating Current

d2 q dq q

L +R + = vm sin ω t (7.28)

dt 2 dt C

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 = qm sin (ω t + θ ) [7.29(a)]

dq

so that = qm ω cos(ω t + θ ) [7.29(b)]

dt

d2 q

and = −qm ω 2 sin(ω t + θ ) [7.29(c)]

dt 2

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

qm ω [ R cos(ω t + θ ) + ( X C − X L )sin(ω t + θ )] = vm sin ωt (7.30)

where we have used the relation Xc= 1/ωC, XL = ω L. Multiplying and

2

⎡R (X − X L ) ⎤

qm ω Z ⎢ cos(ω t + θ ) + C sin(ω t + θ )⎥ = vm sin ω t (7.31)

⎣Z Z ⎦

R

Now, let = cos φ

Z

(XC − X L )

and = sin φ

Z

XC − X L

so that φ = tan −1 (7.32)

R

Substituting this in Eq. (7.31) and simplifying, we get:

qm ω Z cos(ω t + θ − φ ) = vm sin ω t (7.33)

Comparing the two sides of this equation, we see that

vm = qm ω Z = im Z

where

im = qm ω [7.33(a)]

π π

and θ − φ = − or θ = − + φ [7.33(b)]

2 2

Therefore, the current in the circuit is

dq

i = = qm ω cos(ω t + θ )

dt

= im cos(ωt + θ )

or i = imsin(ωt + φ ) (7.34)

vm vm

where im = = [7.34(a)]

Z R + ( X C − X L )2

2

−1 XC − X L

and φ = tan 247

R

Physics

Thus, the analytical solution for the amplitude and phase of the current

in the circuit agrees with that obtained by the technique of phasors.

7.6.3 Resonance

An interesting characteristic of the series RLC circuit is the phenomenon

of resonance. The phenomenon of resonance is common among systems

that have a tendency to oscillate at a particular frequency. This frequency

is called the system’s natural frequency. If such a system is driven by an

energy source at a frequency that is near the natural frequency, the

amplitude of oscillation is found to be large. A familiar example of this

phenomenon is a child on a swing. The the swing has a natural frequency

for swinging back and forth like a pendulum. If the child pulls on the

rope at regular intervals and the frequency of the pulls is almost the

same as the frequency of swinging, the amplitude of the swinging will be

large (Chapter 14, Class XI).

For an RLC circuit driven with voltage of amplitude vm and frequency

ω, we found that the current amplitude is given by

vm vm

im = =

Z R + ( X C − X L )2

2

(

ω0, Xc = XL, and the impedance is minimum Z = R + 0 = R . This

2 2

)

frequency is called the resonant frequency:

1

X c = X L or = ω0 L

ω0 C

1

or ω 0 = (7.35)

LC

At resonant frequency, the current amplitude is maximum; im = vm/R.

Figure 7.16 shows the variation of im with ω in

a RLC series circuit with L = 1.00 mH, C =

1.00 nF for two values of R: (i) R = 100 Ω

and (ii) R = 200 Ω. For the source applied vm =

⎛ 1 ⎞

100 V. ω0 for this case is ⎜⎝ ⎟ = 1.00×106

LC ⎠

rad/s.

We see that the current amplitude is maximum

at the resonant frequency. Since im = vm / R at

resonance, the current amplitude for case (i) is

twice to that for case (ii).

FIGURE 7.16 Variation of im with ω for two Resonant circuits have a variety of

cases: (i) R = 100 Ω, (ii) R = 200 Ω, applications, for example, in the tuning

L = 1.00 mH. mechanism of a radio or a TV set. The antenna of

a radio accepts signals from many broadcasting

stations. The signals picked up in the antenna acts as a source in the

248 tuning circuit of the radio, so the circuit can be driven at many frequencies.

Alternating Current

But to hear one particular radio station, we tune the radio. In tuning, we

vary the capacitance of a capacitor in the tuning circuit such that the

resonant frequency of the circuit becomes nearly equal to the frequency

of the radio signal received. When this happens, the amplitude of the

current with the frequency of the signal of the particular radio station in

the circuit is maximum.

It is important to note that resonance phenomenon is exhibited by a

circuit only if both L and C are present in the circuit. Only then do the

voltages across L and C cancel each other (both being out of phase)

and the current amplitude is vm/R, the total source voltage appearing

across R. This means that we cannot have resonance in a RL or RC

circuit.

Sharpness of resonance

The amplitude of the current in the series LCR circuit is given by

vm

im =

2

⎛ 1 ⎞

R + ⎜ω L −

2

⎝ ω C ⎟⎠

immax = vm / R .

For values of ω other than ω0, the amplitude of the current is less

than the maximum value. Suppose we choose a value of ω for which the

current amplitude is 1/ 2 times its maximum value. At this value, the

power dissipated by the circuit becomes half. From the curve in

Fig. (7.16), we see that there are two such values of ω, say, ω1 and ω2, one

greater and the other smaller than ω0 and symmetrical about ω0. We may

write

ω1 = ω0 + Δω

ω2 = ω0 – Δω

The quantity (ω0 / 2Δω) is regarded as a measure of the sharpness of

resonance. The smaller the Δω, the sharper or narrower is the resonance.

To get an expression for Δω, we note that the current amplitude im is

(1/ 2 ) i max

m for ω1 = ω0 + Δω. Therefore,

vm

at ω1 , im =

2

⎛ 1 ⎞

R + ⎜ ω 1L −

2

⎝ ω 1C ⎟⎠

i mmax vm

= = 249

2 R 2

Physics

2

⎛ 1 ⎞

R + ⎜ ω 1L −

2

=R 2

ω1 C ⎟⎠

or

⎝

2

⎛ 1 ⎞

R 2 + ⎜ ω1 L − = 2R 2

ω1 C ⎟⎠

or

⎝

1

ω 1L − =R

ω 1C

which may be written as,

1

(ω 0 + Δω ) L − =R

(ω 0 + Δω )C

⎛ Δω ⎞ 1

ω 0 L ⎜1 + − =R

⎝ ω 0 ⎟⎠ ⎛ Δω ⎞

ω 0C ⎜1 +

⎝ ω 0 ⎟⎠

1

Using ω 02 = in the second term on the left hand side, we get

LC

⎛ Δω ⎞ ω0 L

ω 0 L ⎜1 + ⎟ − =R

⎝ ω0 ⎠ ⎛ Δω ⎞

⎜⎝1 + ω ⎟⎠

0

−1

⎛ Δω ⎞ ⎛ Δω ⎞ Δω

We can approximate ⎜1 + as ⎜1 − ⎟

ω 0 ⎟⎠

since ω <<1. Therefore,

⎝ ⎝ ω0 ⎠ 0

⎛ Δω ⎞ ⎛ Δω ⎞

ω 0 L ⎜1 + ⎟ − ω 0 L ⎜1 − =R

⎝ ω0 ⎠ ⎝ ω 0 ⎟⎠

2 Δω

or ω0 L =R

ω0

R

Δω = [7.36(a)]

2L

The sharpness of resonance is given by,

ω0 ω L

= 0 [7.36(b)]

2 Δω R

ω0 L

The ratio is also called the quality factor, Q of the circuit.

R

ω0L

Q= [7.36(c)]

R

ω0

250 From Eqs. [7.36 (b)] and [7.36 (c)], we see that 2 Δω = . So, larger the

Q

Alternating Current

value of Q, the smaller is the value of 2Δω or the bandwidth and sharper

is the resonance. Using ω 02 = 1/ L C , Eq. [7.36(c)] can be equivalently

expressed as Q = 1/ω0CR.

We see from Fig. 7.15, that if the resonance is less sharp, not only is

the maximum current less, the circuit is close to resonance for a larger

range Δω of frequencies and the tuning of the circuit will not be good. So,

less sharp the resonance, less is the selectivity of the circuit or vice versa.

From Eq. (7.36), we see that if quality factor is large, i.e., R is low or L is

large, the circuit is more selective.

connected in series to a 220 V, 50 Hz ac source. (a) Calculate the

current in the circuit; (b) Calculate the voltage (rms) across the

resistor and the capacitor. Is the algebraic sum of these voltages

more than the source voltage? If yes, resolve the paradox.

Solution

Given

R = 200 Ω, C = 15.0 μF = 15.0 × 10−6 F

V = 220 V, ν = 50 Hz

(a) In order to calculate the current, we need the impedance of the

circuit. It is

Z = R 2 + X C2 = R 2 + (2π ν C )−2

= 291.5 Ω

Therefore, the current in the circuit is

V 220 V

I = = = 0.755 A

Z 291.5 Ω

(b) Since the current is the same throughout the circuit, we have

VR = I R = (0.755 A)(200 Ω ) = 151 V

VC = I X C = (0.755 A)(212.3 Ω) = 160.3 V

The algebraic sum of the two voltages, VR and VC is 311.3 V which is

more than the source voltage of 220 V. How to resolve this paradox?

As you have learnt in the text, the two voltages are not in the same

phase. Therefore, they cannot be added like ordinary numbers. The

two voltages are out of phase by ninety degrees. Therefore, the total

of these voltages must be obtained using the Pythagorean theorem:

EXAMPLE 7.6

VR +C = VR2 + VC2

= 220 V

Thus, if the phase difference between two voltages is properly taken

into account, the total voltage across the resistor and the capacitor is

equal to the voltage of the source. 251

Physics

7.7 POWER IN AC CIRCUIT: THE POWER FACTOR

We have seen that a voltage v = vm sinωt applied to a series RLC circuit

drives a current in the circuit given by i = im sin(ωt + φ) where

vm ⎛ X − XL ⎞

im = and φ = tan −1 ⎜ C ⎟⎠

Z ⎝ R

Therefore, the instantaneous power p supplied by the source is

p = v i = (vm sin ω t ) × [im sin(ω t + φ )]

vm i m

=

2

[ cos φ − cos(2ω t + φ )] (7.37)

The average power over a cycle is given by the average of the two terms in

R.H.S. of Eq. (7.37). It is only the second term which is time-dependent.

Its average is zero (the positive half of the cosine cancels the negative

half). Therefore,

vm i m v i

P = cos φ = m m cos φ

2 2 2

= V I cos φ [7.38(a)]

This can also be written as,

P = I 2 Z cos φ [7.38(b)]

So, the average power dissipated depends not only on the voltage and

current but also on the cosine of the phase angle φ between them. The

quantity cosφ is called the power factor. Let us discuss the following

cases:

Case (i) Resistive circuit: If the circuit contains only pure R, it is called

resistive. In that case φ = 0, cos φ = 1. There is maximum power dissipation.

Case (ii) Purely inductive or capacitive circuit: If the circuit contains

only an inductor or capacitor, we know that the phase difference between

voltage and current is π/2. Therefore, cos φ = 0, and no power is dissipated

even though a current is flowing in the circuit. This current is sometimes

referred to as wattless current.

Case (iii) LCR series circuit: In an LCR series circuit, power dissipated is

given by Eq. (7.38) where φ = tan–1 (Xc – XL )/ R. So, φ may be non-zero in

a RL or RC or RCL circuit. Even in such cases, power is dissipated only in

the resistor.

Case (iv) Power dissipated at resonance in LCR circuit: At resonance

Xc – XL= 0, and φ = 0. Therefore, cosφ = 1 and P = I 2Z = I 2 R. That is,

maximum power is dissipated in a circuit (through R) at resonance.

EXAMPLE 7.7

low power factor implies large power loss in transmission. Explain.

(b) Power factor can often be improved by the use of a capacitor of

appropriate capacitance in the circuit. Explain.

252

Alternating Current

Solution (a) We know that P = I V cosφ where cosφ is the power factor.

To supply a given power at a given voltage, if cosφ is small, we have to

increase current accordingly. But this will lead to large power loss

(I2R) in transmission.

(b)Suppose in a circuit, current I lags the voltage by an angle φ. Then

power factor cosφ =R/Z.

We can improve the power factor (tending to 1) by making Z tend to R.

Let us understand, with the help of a phasor diagram (Fig. 7.17) how

this can be achieved. Let us resolve I into two components. Ip along

FIGURE 7.17

the applied voltage V and Iq perpendicular to the applied voltage. Iq

as you have learnt in Section 7.7, is called the wattless component

since corresponding to this component of current, there is no power

loss. IP is known as the power component because it is in phase with

the voltage and corresponds to power loss in the circuit.

EXAMPLE 7.7

It’s clear from this analysis that if we want to improve power factor,

we must completely neutralize the lagging wattless current Iq by an

equal leading wattless current I′q. This can be done by connecting a

capacitor of appropriate value in parallel so that Iq and I′ q cancel

each other and P is effectively Ip V.

50 Hz is applied to a series LCR circuit in which

R = 3 Ω, L = 25.48 mH, and C = 796 μF. Find (a) the impedance of the

circuit; (b) the phase difference between the voltage across the source

and the current; (c) the power dissipated in the circuit; and (d) the

power factor.

Solution

EXAMPLE 7.8

(a) To find the impedance of the circuit, we first calculate XL and XC.

XL = 2 πνL

= 2 × 3.14 × 50 × 25.48 × 10–3 Ω = 8 Ω

1

XC =

2 πν C 253

Physics

1

= = 4Ω

2 × 3.14 × 50 × 796 × 10 −6

Therefore,

Z = R 2 + ( X L − X C )2 = 32 + (8 − 4)2

=5Ω

XC − X L

(b) Phase difference, φ = tan–1

R

⎛ 4 − 8⎞

= tan −1 ⎜ = −53.1°

⎝ 3 ⎟⎠

Since φ is negative, the current in the circuit lags the voltage

across the source.

(c) The power dissipated in the circuit is

P = I 2R

EXAMPLE 7.8

im 1 ⎛ 283 ⎞

Now, I = = ⎜ ⎟ = 40A

2 2⎝ 5 ⎠

Therefore, P = (40A )2 × 3 Ω = 4800 W

(d) Power factor = cos φ = cos 53.1° = 0.6

example can be varied. (a) What is the frequency of the source at

which resonance occurs? (b) Calculate the impedance, the current,

and the power dissipated at the resonant condition.

Solution

(a) The frequency at which the resonance occurs is

1 1

ω0 = =

−3

LC 25.48 × 10 × 796 × 10 −6

= 222.1rad/s

ω0 221.1

νr = = Hz = 35.4Hz

2π 2 × 3.14

Z = R = 3Ω

V V ⎛ 283 ⎞ 1

= = = = 66.7A

Z R ⎜⎝ 2 ⎟⎠ 3

EXAMPLE 7.9

P = I 2 × R = (66.7)2 × 3 = 13.35 kW

at resonance is more than the power dissipated in Example 7.8.

254

Alternating Current

doorway of a metal detector, for security reasons. If she/he is carrying

anything made of metal, the metal detector emits a sound. On what

principle does this detector work?

Solution The metal detector works on the principle of resonance in

ac circuits. When you walk through a metal detector, you are,

in fact, walking through a coil of many turns. The coil is connected to

EXAMPLE 7.10

a capacitor tuned so that the circuit is in resonance. When

you walk through with metal in your pocket, the impedance of the

circuit changes – resulting in significant change in current in the

circuit. This change in current is detected and the electronic circuitry

causes a sound to be emitted as an alarm.

7.8 LC OSCILLATIONS

We know that a capacitor and an inductor can store electrical and

magnetic energy, respectively. When a capacitor (initially charged) is

connected to an inductor, the charge on the capacitor and

the current in the circuit exhibit the phenomenon of

electrical oscillations similar to oscillations in mechanical

systems (Chapter 14, Class XI).

Let a capacitor be charged qm (at t = 0) and connected

to an inductor as shown in Fig. 7.18.

The moment the circuit is completed, the charge on

the capacitor starts decreasing, giving rise to current in

the circuit. Let q and i be the charge and current in the

circuit at time t. Since di/dt is positive, the induced emf

in L will have polarity as shown, i.e., vb < va. According to

Kirchhoff’s loop rule,

q di FIGURE 7.18 At the

−L =0 (7.39) instant shown, the current

C dt

is increasing so the

i = – (dq/dt ) in the present case (as q decreases, i increases). polarity of induced emf in

Therefore, Eq. (7.39) becomes: the inductor is as shown.

d2 q 1

+ q=0 (7.40)

dt 2 LC

d2 x

This equation has the form 2

+ ω 02 x = 0 for a simple harmonic

dt

oscillator. The charge, therefore, oscillates with a natural frequency

1

ω0 = (7.41)

LC

and varies sinusoidally with time as

q = qm cos ( ω 0 t + φ ) (7.42)

where qm is the maximum value of q and φ is a phase constant. Since

q = qm at t = 0, we have cos φ =1 or φ = 0. Therefore, in the present case, 255

Physics

q = qm cos(ω 0t ) (7.43)

The current i ⎛⎜ = −

dq ⎞

⎝ ⎟ is given by

dt ⎠

i = im sin(ω 0t ) (7.44)

where im = ω 0qm

Let us now try to visualise how this oscillation takes place in the

circuit.

Figure 7.19(a) shows a capacitor with initial charge qm connected to

an ideal inductor. The electrical energy stored in the charged capacitor is

1 qm2

UE = . Since, there is no current in the circuit, energy in the inductor

2 C

is zero. Thus, the total energy of LC circuit is,

1 qm2

U = UE =

2 C

block at the end of a spring. The figure depicts one-half of a cycle.

[Fig. 7.19(b)]. As the current increases, it sets up a magnetic field in the

inductor and thereby, some energy gets stored in the inductor in the

form of magnetic energy: UB = (1/2) Li2. As the current reaches its

maximum value im, (at t = T/4) as in Fig. 7.19(c), all the energy is stored

in the magnetic field: UB = (1/2) Li2m. You can easily check that the

maximum electrical energy equals the maximum magnetic energy. The

capacitor now has no charge and hence no energy. The current now

starts charging the capacitor, as in Fig. 7.19(d). This process continues

till the capacitor is fully charged (at t = T/2) [Fig. 7.19(e)]. But it is charged

with a polarity opposite to its initial state in Fig. 7.19(a). The whole process

just described will now repeat itself till the system reverts to its original

state. Thus, the energy in the system oscillates between the capacitor

256 and the inductor.

Alternating Current

The LC oscillation is similar to the mechanical oscillation of a block

attached to a spring. The lower part of each figure in Fig. 7.19 depicts

the corresponding stage of a mechanical system (a block attached to a

spring). As noted earlier, for a block of a mass m oscillating with frequency

ω0, the equation is

d2 x

+ ω 20 x = 0

dt 2

Here, ω 0 = k / m , and k is the spring constant. So, x corresponds to q.

In case of a mechanical system F = ma = m (dv/dt) = m (d2x/dt2 ). For an

electrical system, ε = –L (di/dt ) = –L (d2q/dt 2 ). Comparing these two

equations, we see that L is analogous to mass m: L is a measure of

resistance to change in current. In case of LC circuit, ω 0 = 1/ LC and

for mass on a spring, ω 0 = k / m . So, 1/C is analogous to k. The constant

k (=F/x) tells us the (external) force required to produce a unit

displacement whereas 1/C (=V/q ) tells us the potential difference required

to store a unit charge. Table 7.1 gives the analogy between mechanical

and electrical quantities.

ELECTRICAL QUANTITIES

Mass m Inductance L

Force constant k Reciprocal capacitance

1/C

Displacement x Charge q

Velocity v = dx/dt Current i = dq/dt

Mechanical energy Electromagnetic energy

1 1 1 q2 1 2

E = k x 2 + m v2 U = + Li

2 2 2 C 2

Note that the above discussion of LC oscillations is not realistic for two

reasons:

(i) Every inductor has some resistance. The effect of this resistance is to

introduce a damping effect on the charge and current in the circuit

and the oscillations finally die away.

(ii) Even if the resistance were zero, the total energy of the system would

not remain constant. It is radiated away from the system in the form

of electromagnetic waves (discussed in the next chapter). In fact, radio

and TV transmitters depend on this radiation. 257

Physics

TWO DIFFERENT PHENOMENA, SAME MATHEMATICAL TREATMENT

You may like to compare the treatment of a forced damped oscillator discussed in Section

14.10 of Class XI physics textbook, with that of an LCR circuit when an ac voltage is

applied in it. We have already remarked that Eq. [14.37(b)] of Class XI Textbook is exactly

similar to Eq. (7.28) here, although they use different symbols and parameters. Let us

therefore list the equivalence between different quantities in the two situations:

2 d2q dq q

m d x + b dx + kx = F cos ω d t L +R + = vm sin ω t

dt 2 dt 2

dt dt C

Displacement, x Charge on capacitor, q

Time, t Time, t

Mass, m Self inductance, L

Damping constant, b Resistance, R

Spring constant, k Inverse capacitance, 1/C

Driving frequency, ωd Driving frequency, ω

Natural frequency of oscillations, ω Natural frequency of LCR circuit, ω0

Amplitude of forced oscillations, A Maximum charge stored, qm

Amplitude of driving force, F0 Amplitude of applied voltage, vm

You must note that since x corresponds to q, the amplitude A (maximum displacement)

will correspond to the maximum charge stored, qm. Equation [14.39 (a)] of Class XI gives

the amplitude of oscillations in terms of other parameters, which we reproduce here for

convenience:

F0

A=

{m }

1/ 2

2

(ω − ω ) + ω d2b 2

2 2 2

d

quantity, and see what happens. Eliminate L, C, ω , and ω , using XL= ωL, XC = 1/ωC, and

0

ω02 = 1/LC. When you use Eqs. (7.33) and (7.34), you will see that there is a

perfect match.

You will come across numerous such situations in physics where diverse physical

phenomena are represented by the same mathematical equation. If you have dealt with

one of them, and you come across another situation, you may simply replace the

corresponding quantities and interpret the result in the new context. We suggest that

you may try to find more such parallel situations from different areas of physics. One

must, of course, be aware of the differences too.

258

Alternating Current

sum of energies stored in the capacitor and the inductor is constant

in time.

Solution Let q0 be the initial charge on a capacitor. Let the charged

capacitor be connected to an inductor of inductance L. As you have

studied in Section 7.8, this LC circuit will sustain an oscillation with

frquency

⎛ 1 ⎞

ω ⎜ = 2π ν = ⎟

⎝ LC ⎠

At an instant t, charge q on the capacitor and the current i are given

by:

q (t) = q0 cos ωt

i (t) = – q0 ω sin ωt

Energy stored in the capacitor at time t is

1 1 q 2 q02

UE = C V2 = = cos 2 (ωt )

2 2 C 2C

Energy stored in the inductor at time t is

1

UM = L i2

2

1

= L q02 ω 2 sin 2 (ωt )

2

=

q02

2C

sin 2 (ωt ) (∵ω 2

= 1/ LC )

Sum of energies

q02

U E +U M = ⎡cos2 ωt + sin 2 ωt ⎤⎦

2C ⎣

EXAMPLE 7.11

q02

=

2C

This sum is constant in time as qo and C, both are time-independent.

Note that it is equal to the initial energy of the capacitor. Why it is

so? Think!

7.9 TRANSFORMERS

For many purposes, it is necessary to change (or transform) an alternating

voltage from one to another of greater or smaller value. This is done with

a device called transformer using the principle of mutual induction.

A transformer consists of two sets of coils, insulated from each other.

They are wound on a soft-iron core, either one on top of the other as in

Fig. 7.20(a) or on separate limbs of the core as in Fig. 7.20(b). One of the

coils called the primary coil has Np turns. The other coil is called the

secondary coil; it has Ns turns. Often the primary coil is the input coil

and the secondary coil is the output coil of the transformer. 259

Physics

FIGURE 7.20 Two arrangements for winding of primary and secondary coil in a transformer:

(a) two coils on top of each other, (b) two coils on separate limbs of the core.

current produces an alternating magnetic flux which links the secondary

and induces an emf in it. The value of this emf depends on the number of

turns in the secondary. We consider an ideal transformer in which the

primary has negligible resistance and all the flux in the core links both

primary and secondary windings. Let φ be the flux in each turn in the

core at time t due to current in the primary when a voltage vp is applied

to it.

Then the induced emf or voltage εs, in the secondary with Ns turns is

dφ

εs = − N s (7.45)

dt

The alternating flux φ also induces an emf, called back emf in the

primary. This is

dφ

ε p = −N p (7.46)

dt

But εp = vp. If this were not so, the primary current would be infinite

since the primary has zero resistance(as assumed). If the secondary is

an open circuit or the current taken from it is small, then to a good

approximation

εs = vs

where vs is the voltage across the secondary. Therefore, Eqs. (7.45) and

(7.46) can be written as

dφ

vs = − N s [7.45(a)]

dt

dφ

v p = −N p [7.46(a)]

dt

From Eqs. [7.45 (a)] and [7.45 (a)], we have

vs N

= s (7.47)

260 vp N p

Alternating Current

Note that the above relation has been obtained using three

assumptions: (i) the primary resistance and current are small; (ii) the

same flux links both the primary and the secondary as very little flux

escapes from the core, and (iii) the secondary current is small.

If the transformer is assumed to be 100% efficient (no energy losses),

the power input is equal to the power output, and since p = i v,

ipvp = isvs (7.48)

Although some energy is always lost, this is a good approximation,

since a well designed transformer may have an efficiency of more than

95%. Combining Eqs. (7.47) and (7.48), we have

i p vs N

= = s (7.49)

is vp N p

Since i and v both oscillate with the same frequency as the ac source,

Eq. (7.49) also gives the ratio of the amplitudes or rms values of

corresponding quantities.

Now, we can see how a transformer affects the voltage and current.

We have:

⎛N ⎞ ⎛ Np ⎞

Vs = ⎜ s ⎟ V p and I s = ⎜ Ip

⎝ N s ⎟⎠

(7.50)

⎝ Np ⎠

That is, if the secondary coil has a greater number of turns than the

primary (Ns > Np), the voltage is stepped up(Vs > Vp). This type of

arrangement is called a step-up transformer. However, in this arrangement,

there is less current in the secondary than in the primary (Np/Ns < 1 and Is

< Ip). For example, if the primary coil of a transformer has 100 turns and

the secondary has 200 turns, Ns/Np = 2 and Np/Ns=1/2. Thus, a 220V

input at 10A will step-up to 440 V output at 5.0 A.

If the secondary coil has less turns than the primary(Ns < Np), we have

a step-down transformer. In this case, Vs < Vp and Is > Ip. That is, the

voltage is stepped down, or reduced, and the current is increased.

The equations obtained above apply to ideal transformers (without

any energy losses). But in actual transformers, small energy losses do

occur due to the following reasons:

(i) Flux Leakage: There is always some flux leakage; that is, not all of

the flux due to primary passes through the secondary due to poor

design of the core or the air gaps in the core. It can be reduced by

winding the primary and secondary coils one over the other.

(ii) Resistance of the windings: The wire used for the windings has some

resistance and so, energy is lost due to heat produced in the wire

(I 2R). In high current, low voltage windings, these are minimised by

using thick wire.

(iii) Eddy currents: The alternating magnetic flux induces eddy currents

in the iron core and causes heating. The effect is reduced by having a

laminated core.

(iv) Hysteresis: The magnetisation of the core is repeatedly reversed by

the alternating magnetic field. The resulting expenditure of energy in

the core appears as heat and is kept to a minimum by using a magnetic

material which has a low hysteresis loss. 261

Physics

The large scale transmission and distribution of electrical energy over

long distances is done with the use of transformers. The voltage output

of the generator is stepped-up (so that current is reduced and

consequently, the I 2R loss is cut down). It is then transmitted over long

distances to an area sub-station near the consumers. There the voltage

is stepped down. It is further stepped down at distributing sub-stations

and utility poles before a power supply of 240 V reaches our homes.

SUMMARY

vm

current i = im sinωt in the resistor, i m = . The current is in phase with

R

the applied voltage.

2. For an alternating current i = im sinωt passing through a resistor R, the

average power loss P (averaged over a cycle) due to joule heating is

( 1/2 )i 2mR. To express it in the same form as the dc power (P = I 2R), a

special value of current is used. It is called root mean square (rms)

current and is donoted by I:

im

I = = 0.707 im

2

Similarly, the rms voltage is defined by

vm

V = = 0.707 vm

2

We have P = IV = I 2R

3. An ac voltage v = vm sinωt applied to a pure inductor L, drives a current

in the inductor i = im sin (ωt – π/2), where im = vm/XL. XL = ωL is called

inductive reactance. The current in the inductor lags the voltage by

π/2. The average power supplied to an inductor over one complete cycle

is zero.

4. An ac voltage v = vm sinωt applied to a capacitor drives a current in the

capacitor: i = im sin (ωt + π/2). Here,

vm 1

im = , XC =

XC ωC is called capacitive reactance.

The current through the capacitor is π/2 ahead of the applied voltage.

As in the case of inductor, the average power supplied to a capacitor

over one complete cycle is zero.

5. For a series RLC circuit driven by voltage v = vm sinωt, the current is

given by i = im sin (ωt + φ )

vm

where im =

R + ( XC − X L )

2 2

XC − X L

and φ = tan −1

R

R2 + ( X C − X L )

2

Z = is called the impedance of the circuit.

262

Alternating Current

P = V I cosφ

The term cosφ is called the power factor.

6. In a purely inductive or capacitive circuit, cosφ = 0 and no power is

dissipated even though a current is flowing in the circuit. In such cases,

current is referred to as a wattless current.

7. The phase relationship between current and voltage in an ac circuit

can be shown conveniently by representing voltage and current by

rotating vectors called phasors. A phasor is a vector which rotates

about the origin with angular speed ω. The magnitude of a phasor

represents the amplitude or peak value of the quantity (voltage or

current) represented by the phasor.

The analysis of an ac circuit is facilitated by the use of a phasor

diagram.

8. An interesting characteristic of a series RLC circuit is the

phenomenon of resonance. The circuit exhibits resonance, i.e.,

the amplitude of the current is maximum at the resonant

1

frequency, ω 0 = . The quality factor Q defined by

LC

ω0 L 1

Q= = is an indicator of the sharpness of the resonance,

R ω 0CR

the higher value of Q indicating sharper peak in the current.

9. A circuit containing an inductor L and a capacitor C (initially

charged) with no ac source and no resistors exhibits free

oscillations. The charge q of the capacitor satisfies the equation

of simple harmonic motion:

d2 q 1

+ q=0

dt 2 LC

1

and therefore, the frequency ω of free oscillation is ω 0 = . The

LC

energy in the system oscillates between the capacitor and the

inductor but their sum or the total energy is constant in time.

10. A transformer consists of an iron core on which are bound a

primary coil of N p turns and a secondary coil of N s turns. If the

primary coil is connected to an ac source, the primary and

secondary voltages are related by

⎛N ⎞

Vs = ⎜ s ⎟ V p

⎝ Np ⎠

and the currents are related by

⎛ Np ⎞

Is = ⎜ ⎟⎠ I p

⎝ Ns

If the secondary coil has a greater number of turns than the primary, the

voltage is stepped-up (Vs > Vp). This type of arrangement is called a step-

up transformer. If the secondary coil has turns less than the primary, we

have a step-down transformer.

263

Physics

Physical quantity Symbol Dimensions Unit Remarks

vm

rms voltage V [M L2 T –3 A–1] V V = , vm is the

2

amplitude of the ac voltage.

im

rms current I [ A] A I= , im is the amplitude of

2

the ac current.

Reactance:

Ω XL = ω L

2 –3 –2

Inductive XL [M L T A ]

Ω XC = 1/ ω C

2 –3 –2

Capacitive XC [M L T A ]

present in the circuit.

1

Resonant ωr or ω0 [T ]

–1

Hz ω0 = for a

frequency LC

series RLC circuit

ω0 L 1

Quality factor Q Dimensionless Q= = for a series

R ω0 C R

RLC circuit.

Power factor Dimensionless = cos φ , φ is the phase

difference between voltage

applied and current in

the circuit.

POINTS TO PONDER

value. The voltage across the terminals of an outlet in your room is

normally 240 V. This refers to the rms value of the voltage. The amplitude

of this voltage is

vm = 2V = 2(240) = 340 V

2. The power rating of an element used in ac circuits refers to its average

power rating.

3. The power consumed in an circuit is never negative.

4. Both alternating current and direct current are measured in amperes.

But how is the ampere defined for an alternating current? It cannot be

derived from the mutual attraction of two parallel wires carrying ac

264 currents, as the dc ampere is derived. An ac current changes direction

Alternating Current

with the source frequency and the attractive force would average to

zero. Thus, the ac ampere must be defined in terms of some property

that is independent of the direction of the current. Joule heating

is such a property, and there is one ampere of rms value of

alternating current in a circuit if the current produces the same

average heating effect as one ampere of dc current would produce

under the same conditions.

5. In an ac circuit, while adding voltages across different elements, one

should take care of their phases properly. For example, if VR and VC

are voltages across R and C, respectively in an RC circuit, then the

VR + VC since VC is π/2 out of phase of VR.

6. Though in a phasor diagram, voltage and current are represented by

vectors, these quantities are not really 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 as the law of vector addition.

7. There are no power losses associated with pure capacitances and pure

inductances in an ac circuit. The only element that dissipates energy

in an ac circuit is the resistive element.

8. In a RLC circuit, resonance phenomenon occur when X L = X C or

1

ω0 = . For resonance to occur, the presence of both L and C

LC

elements in the circuit is a must. With only one of these (L or C )

elements, there is no possibility of voltage cancellation and hence,

no resonance is possible.

9. The power factor in a RLC circuit is a measure of how close the

circuit is to expending the maximum power.

10. In generators and motors, the roles of input and output are

reversed. In a motor, electric energy is the input and mechanical

energy is the output. In a generator, mechanical energy is the

input and electric energy is the output. Both devices simply

transform energy from one form to another.

11. A transformer (step-up) changes a low-voltage into a high-voltage.

This does not violate the law of conservation of energy. The

current is reduced by the same proportion.

12. The choice of whether the description of an oscillatory motion is

by means of sines or cosines or by their linear combinations is

unimportant, since changing the zero-time position transforms

the one to the other.

265

Physics

EXERCISES

7.1 A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.

(a) What is the rms value of current in the circuit?

(b) What is the net power consumed over a full cycle?

7.2 (a) The peak voltage of an ac supply is 300 V. What is the rms voltage?

(b) The rms value of current in an ac circuit is 10 A. What is the

peak current?

7.3 A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine

the rms value of the current in the circuit.

7.4 A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine

the rms value of the current in the circuit.

7.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each

circuit over a complete cycle. Explain your answer.

7.6 Obtain the resonant frequency ωr of a series LCR circuit with

L = 2.0H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?

7.7 A charged 30 μF capacitor is connected to a 27 mH inductor. What is

the angular frequency of free oscillations of the circuit?

7.8 Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC.

What is the total energy stored in the circuit initially ? What is the

total energy at later time?

7.9 A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected

to a variable-frequency 200 V ac supply. When the frequency of the

supply equals the natural frequency of the circuit, what is the average

power transferred to the circuit in one complete cycle?

7.10 A radio can tune over the frequency range of a portion of MW

broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective

inductance of 200 μH, what must be the range of its variable

capacitor ?

[Hint: For tuning, the natural frequency i.e., the frequency of free

oscillations of the LC circuit should be equal to the frequency of the

radiowave.]

7.11 Figure 7.21 shows a series LCR circuit connected to a variable

frequency 230 V source. L = 5.0 H, C = 80μF, R = 40 Ω.

FIGURE 7.21

resonance.

(b) Obtain the impedance of the circuit and the amplitude of current

at the resonating frequency.

(c) Determine the rms potential drops across the three elements of

the circuit. Show that the potential drop across the LC

266 combination is zero at the resonating frequency.

Alternating Current

ADDITIONAL EXERCISES

7.12 An LC circuit contains a 20 mH inductor and a 50 μF capacitor with

an initial charge of 10 mC. The resistance of the circuit is negligible.

Let the instant the circuit is closed be t = 0.

(a) What is the total energy stored initially? Is it conserved during

LC oscillations?

(b) What is the natural frequency of the circuit?

(c) At what time is the energy stored

(i) completely electrical (i.e., stored in the capacitor)? (ii) completely

magnetic (i.e., stored in the inductor)?

(d) At what times is the total energy shared equally between the

inductor and the capacitor?

(e) If a resistor is inserted in the circuit, how much energy is

eventually dissipated as heat?

7.13 A coil of inductance 0.50 H and resistance 100 Ω is connected to a

240 V, 50 Hz ac supply.

(a) What is the maximum current in the coil?

(b) What is the time lag between the voltage maximum and the

current maximum?

7.14 Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is

connected to a high frequency supply (240 V, 10 kHz). Hence, explain

the statement that at very high frequency, an inductor in a circuit

nearly amounts to an open circuit. How does an inductor behave in

a dc circuit after the steady state?

7.15 A 100 μF capacitor in series with a 40 Ω resistance is connected to a

110 V, 60 Hz supply.

(a) What is the maximum current in the circuit?

(b) What is the time lag between the current maximum and the

voltage maximum?

7.16 Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is

connected to a 110 V, 12 kHz supply? Hence, explain the statement

that a capacitor is a conductor at very high frequencies. Compare this

behaviour with that of a capacitor in a dc circuit after the steady state.

7.17 Keeping the source frequency equal to the resonating frequency of

the series LCR circuit, if the three elements, L, C and R are arranged

in parallel, show that the total current in the parallel LCR circuit is

minimum at this frequency. Obtain the current rms value in each

branch of the circuit for the elements and source specified in

Exercise 7.11 for this frequency.

7.18 A circuit containing a 80 mH inductor and a 60 μF capacitor in series

is connected to a 230 V, 50 Hz supply. The resistance of the circuit is

negligible.

(a) Obtain the current amplitude and rms values.

(b) Obtain the rms values of potential drops across each element.

(c) What is the average power transferred to the inductor?

(d) What is the average power transferred to the capacitor?

(e) What is the total average power absorbed by the circuit? [‘Average’

implies ‘averaged over one cycle’.]

7.19 Suppose the circuit in Exercise 7.18 has a resistance of 15 Ω. Obtain

the average power transferred to each element of the circuit, and

the total power absorbed. 267

Physics

7.20 A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected

to a 230 V variable frequency supply.

(a) What is the source frequency for which current amplitude is

maximum. Obtain this maximum value.

(b) What is the source frequency for which average power absorbed

by the circuit is maximum. Obtain the value of this maximum

power.

(c) For which frequencies of the source is the power transferred to

the circuit half the power at resonant frequency? What is the

current amplitude at these frequencies?

(d) What is the Q-factor of the given circuit?

7.21 Obtain the resonant frequency and Q-factor of a series LCR circuit

with L = 3.0 H, C = 27 μF, and R = 7.4 Ω. It is desired to improve the

sharpness of the resonance of the circuit by reducing its ‘full width

at half maximum’ by a factor of 2. Suggest a suitable way.

7.22 Answer the following questions:

(a) In any ac circuit, is the applied instantaneous voltage equal to

the algebraic sum of the instantaneous voltages across the series

elements of the circuit? Is the same true for rms voltage?

(b) A capacitor is used in the primary circuit of an induction coil.

(c) An applied voltage signal consists of a superposition of a dc voltage

and an ac voltage of high frequency. The circuit consists of an

inductor and a capacitor in series. Show that the dc signal will

appear across C and the ac signal across L.

(d) A choke coil in series with a lamp is connected to a dc line. The

lamp is seen to shine brightly. Insertion of an iron core in the

choke causes no change in the lamp’s brightness. Predict the

corresponding observations if the connection is to an ac line.

(e) Why is choke coil needed in the use of fluorescent tubes with ac

mains? Why can we not use an ordinary resistor instead of the

choke coil?

7.23 A power transmission line feeds input power at 2300 V to a step-

down transformer with its primary windings having 4000 turns. What

should be the number of turns in the secondary in order to get output

power at 230 V?

7.24 At a hydroelectric power plant, the water pressure head is at a height

of 300 m and the water flow available is 100 m3s–1. If the turbine

generator efficiency is 60%, estimate the electric power available

from the plant (g = 9.8 ms–2 ).

7.25 A small town with a demand of 800 kW of electric power at 220 V is

situated 15 km away from an electric plant generating power at 440 V.

The resistance of the two wire line carrying power is 0.5 Ω per km.

The town gets power from the line through a 4000-220 V step-down

transformer at a sub-station in the town.

(a) Estimate the line power loss in the form of heat.

(b) How much power must the plant supply, assuming there is

negligible power loss due to leakage?

(c) Characterise the step up transformer at the plant.

7.26 Do the same exercise as above with the replacement of the earlier

transformer by a 40,000-220 V step-down transformer (Neglect, as

before, leakage losses though this may not be a good assumption

any longer because of the very high voltage transmission involved).

268 Hence, explain why high voltage transmission is preferred?

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