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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.

* The phrases ac voltage and ac current are contradictory and redundant,


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.

Nicola Tesla (1836 –


1943) Yugoslov scientist,
inventor and genius. He
NICOLA TESLA (1836 – 1943)

conceived the idea of the


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

GEORGE WESTINGHOUSE (1846 – 1914)


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.

FIGURE 7.3 The rms current I is related to the


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

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 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.

7.4 AC VOLTAGE APPLIED TO AN INDUCTOR


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.

* 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. 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.

Example 7.2 A pure inductor of 25.0 mH is connected to a source of


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.

240 FIGURE 7.7 Magnetisation and demagnetisation of an inductor.


Alternating Current

7.5 AC VOLTAGE APPLIED TO A CAPACITOR


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.

Example 7.3 A lamp is connected in series with a capacitor. Predict


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.

Example 7.4 A 15.0 μF capacitor is connected to a 220 V, 50 Hz source.


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

Solution The capacitive reactance is


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.

FIGURE 7.10 Charging and discharging of a capacitor. 243


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.

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 = 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

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


in an ac circuit:
vm
im = [7.25(b)]
Z

where Z = R 2 + ( X C − X L )2 (7.26) 245


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.

7.6.2 Analytical solution


The voltage equation for the circuit is
di q
L + Ri + = v
dt C
= vm sin ωt

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


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

dividing Eq. (7.30) by Z = R 2 + ( X c − X L ) , we have


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

with Xc = 1/ωC and XL = ω L . So if ω is varied, then at a particular frequency

(
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 ⎟⎠

and is maximum when ω = ω 0 = 1/ L C . The maximum value is

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 difference ω1 – ω2 = 2Δω is often called the bandwidth of the circuit.


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.

Example 7.6 A resistor of 200 Ω and a capacitor of 15.0 μF are


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

= (200 Ω)2 + (2 × 3.14 × 50 × 10 −6 F)−2

= (200 Ω)2 + (212 Ω)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

Example7.7 (a) For circuits used for transporting electric power, a


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.

Example 7.8 A sinusoidal voltage of peak value 283 V and frequency


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 7.9 Suppose the frequency of the source in the previous


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

(b) The impedance Z at resonant condition is equal to the resistance:

Z = R = 3Ω

The rms current at resonance is

V V ⎛ 283 ⎞ 1
= = = = 66.7A
Z R ⎜⎝ 2 ⎟⎠ 3

The power dissipated at resonance is


EXAMPLE 7.9

P = I 2 × R = (66.7)2 × 3 = 13.35 kW

You can see that in the present case, power dissipated


at resonance is more than the power dissipated in Example 7.8.
254
Alternating Current

Example 7.10 At an airport, a person is made to walk through the


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

FIGURE 7.19 The oscillations in an LC circuit are analogous to the oscillation of a


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

At t = 0, the switch is closed and the capacitor starts to discharge


[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.

TABLE 7.1 ANALOGIES BETWEEN MECHANICAL AND


ELECTRICAL QUANTITIES

Mechanical system Slectrical system

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:

Forced oscillations Driven LCR circuit

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

Replace each parameter in the above equation by the corresponding electrical


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

Example 7.11 Show that in the free oscillations of an LC circuit, the


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.

When an alternating voltage is applied to the primary, the resulting


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

εs = − N s (7.45)
dt
The alternating flux φ also induces an emf, called back emf in the
primary. This is

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

vs = − N s [7.45(a)]
dt


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

1. An alternating voltage v = vm sin ω t applied to a resistor R drives a


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

The average power loss over a complete cycle is given by


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 ]

Impedance Z [M L2 T –3 A–2] Ω Depends on elements


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

1. When a value is given for ac voltage or current, it is ordinarily the rms


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

total voltage across RC combination is VRC = VR2 + VC2 and not


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

(a) Determine the source frequency which drives the circuit in


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?