NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Physics
Chapter : 7
Chapter Name : Alternating Current

Q7.1 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?

Answer. Resistance of the resistor, R = 100 Ω
Supply voltage, V = 220 V
Frequency, v = 50 Hz

(a) The rms value of current in the circuit is given as:
V
I =
R
220
= = 2.20A
100

(b) The net power consumed over a full cycle is given as:
P = VI

= 220 × 2.2 = 484W

Page : 266 , Block Name : Exercise

Q7.2 (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?

Answer. (a) Peak voltage of the ac supply, V 0 = 300v

Rms voltage is given as:
V0
V =
√2

300
= = 212.1V
√2

(b) The rms value is current is given as:
I = 10A

Now, peak current is given as:
I0 = √2I

= 10√2 = 14.1A

Page : 266 , Block Name : Exercise

Q7.3 A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the
current in the circuit.

Answer. Inductance of indicator, L = 44 mH = 44 × 10 H
−3

Supply voltage, V = 220 V
Frequency, v = 50 Hz
Angular frequency, ω = 2πv
Inductive reactance, x = ωL = 2πvL = 2π × 50 × 44 × 10
L
−3
Ω

Rms value of current is given as:
v
I =
XL

220
= −3
= 15.92A
2π×50×44×10

Hence, the rms value of current in the circuit is 15.92 A.

Page : 266 , Block Name : Exercise

Q7.4 A 60 µF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the
current in the circuit.

Answer. Capacitance of capacitor, C = 60μF = 60 × 10 −6
F

Supply voltage, V = 110V
Frequence, v = 60Hz
Angular frequency, ω = 2πv
Capacitive reactance X =
1
c
ωC
1
=
2πvC
1 −1
= −6
Ω
2×3.14×60×60×10

Rms value of current is given as:
v
I =
Xc

−6
= 110 × 2 × 3.14 × 60 × 10 × 60 = 2.49A

Hence, the rms value of current s 2.49 A.

Page : 266 , Block Name : Exercise

Q7.5 In question 3 and 4, what is the net power absorbed by each circuit over a complete cycle.
Explain your answer.

Answer. In the inductive circuit,
Rms value of current, I = 15.92A
Rms value of voltage, V = 220V
Hence, the net power absorbed can be obtained by the relation,
P = V I cos Φ
where,

Φ = Phase difference between V and I
For a pure inductive circuit, the phase difference between alternating voltage and current is 90°
i.e., Φ= 90°.
Hence, P = 0 i.e., the net power is zero.
In the capacitive circuit,
Rms value of current, I = 2.49 A
Rms value of voltage, V = 110 V
Hence, the net power absorbed can be obtained as:
P = V I cos ϕ

For a pure capacitive circuit, the phase difference between alternating voltage and current is 90°
i.e., Φ= 90°.
Hence, P = 0 i.e., the net power is zero.

Page : 266 , Block Name : Exercise

Q7.6 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?

Answer. Inductance, L = 2.0 H
Capacitance, C = 32μF = 32 × 10 F −6

Resistance, R = 10Ω
Resonant frequency is given by the relation,
1
ωr =
√LC

1 1 −1
= = −3
= 125s
√2×32×10 −6 8×10

Now. Q-value of the circuit is given as:
1 L
Q = √
R C

1 2 1 1
= √ −6
= × −3
= 25
10 32×10 10 4×10

Hence, the Q-value of this circuit is 25.

Page : 266 , Block Name : Exercise

Q7.7 A charged 30 µF capacitor is connected to a 27 mH inductor. What is the angular frequency of
free oscillations of the circuit?

Answer. Capacitance, C = 30μF = 30 × 10 −6
F

Inductance, L = 27mH = 27 × 10 H
−3

Angular frequency is given as:
1
ωr =
√LC

1 1 3
= = −4
= 1.11 × 10 rad/s
√27×10 −3 −6 9×10
×30×10

Hence, the angular frequency of free oscillations of the circuit is 1.11 × 10 rad/s.
3

Page : 266 , Block Name : Exercise

Q7.8 Suppose the initial charge on the capacitor in question 7 is 6 mC. What is the total energy
stored in the circuit initially? What is the total energy at later time?

Answer. Capacitance of the capacitor, C = 30μF = 30 × 10 F
−6

Inductance of the inductor, L = 27mH = 27 × 10 H −3

Charge on the capacitor, Q = 6mC = 6 × 10 C −3

Total energy stored in the capacitor can be calculated by the relation,
2
1 Q
E =
2 C
2
−3
(6×10 )
1
= × −6
2 30×10
6
= = 0.6J
10

Total energy at a later time will remain the same because energy is shared between the capacitor
and the inductor.

Page : 266 , Block Name : Exercise

Q7.9 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?

Answer. At resonance, the frequency of the supply power equals the natural frequency of the given
LCR circuit.
Resistance, R = 20Ω
Inductance, L = 1.5H
Capacitance, C = 35μF = 30 × 10 F −6

AC supply voltage to the LCR circuit, V = 200V
Impedance of the circuit is given by the relation,
2
1
Z = √R + (ωL −
2
)
ωC

1
ωL =
ωC

At resonance,
∴ Z = R = 20Ω

Current in the circuit can be calculated as:
V
I =
Z
200
= = 10A
20

Hence, the average power transferred to the circuit in one complete cycle = V I
= 200 × 10 = 2000W

Page : 266 , Block Name : Exercise

Q7.10 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.]

Answer. The range of frequency (v) os a radio is 800 kHz to 1200 kHz.
Lower tuning frequency, v = 800kHz = 800 × 10 Hz
3
1

Upper tuning frequency, v = 1200kHz = 1200 × 10 Hz
2
3

Effective inductance of circuit L = 200μH = 200 × 10 H
−6

Capacitance of variable capacitor for V is given as: 1

1
C1 =
2
ω L
1

Where,
ω1 = Angular frequency for capacitor C1
3 −1
= 2πv1 = 2π × 800 × 10 rads
1
∴ C1 =
2
3 −6
(2π × 800 × 10 ) × 200 × 10

−10
= 1.9809 × 10 F = 198.1pF

Capacitance of variable capacitor for v 2

1
c2 =
2
ω L
2

Where,
ω2 = Angular frequency for capacitor C2
3 −1
= 2πv2 = 2π × 1200 × 10 rads
1
∴ C2 =
2
3 −6
(2π × 1200 × 10 ) × 200 × 10

= 88.04pF

Hence, the range of the variable capacitor is from 88.04 pF to 198.1 pF.

Page : 266 , Block Name : Exercise

Q7.11 Figure shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H,
C = 80µF, R = 40 Ω.

(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 combination is zero at the resonating frequency.

Answer. Inductance of the inductor, L = 5.0H
Capacitance of the capacitor, C = 80μH = 80 × 10 −6
F

Resistance of the resistor, R = 40Ω
Potential of the variable voltage source, V = 230V
(a) Resonance angular frequency is given as:
1
ωR =
√LC
3
1 10
= = = 50rad/s
√5×80×10 −6 20

Hence, the circuit will come in resonance for a source frequency of 50 rad/s.

(b) Impedance of the circuit is given by the relation,
2
1
Z = √R
2
+ (ωL − )
ωC

At resonance,
1
ωL =
ωC

∴ Z = R = 40Ω

Amplitude of the current at the resonating frequency is given as: I
V0
0 =
Z

Where,
V = Peak voltage
0

= √2V
√2V
∴ I0 =
Z

√2×230
= = 8.13A
40

Hence, at resonance, the impedance of the circuit is 40 2 and the amplitude of the current is 8.13
A.

(c) Rms potential drop across the inductor,
(VL ) = I × ωR L
rms

Where,
I = rms current
I0 √2V 230
= = = A
√2 √2Z 40

230
∴ (VL ) = × 50 × 5 = 1437.5V
max 40

Potential drop across the capacitor,
1
(Vc ) = I ×
ms
ωR C

230 1
= × = 1437.5V
−6
40 50 × 80 × 10

Potential drop across the resistor,
(VR ) = IR
rms

230
× 40 = 230V
40

Potential drop across the LC combination,
1
VLC = I (ωR L − )
ωR C

1
ωR L =
ωR C

At resonance,
∴ VLC = 0

Hence, it is proved that the potential drop across the LC combination is zero at resonating
frequency.

Page : 266 , Block Name : Exercise

Q7.12 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?

Answer. Inductance of the inductor , L = 20mH = 20 × 10 −3
H

Capacitance of the capacitor, C = 50μF = 50 × 10 F −6

Initial charge on the capacitor, Q = 10mC = 10 × 10 C −3

(a) Total energy stored initially in the circuit is given as:

2
1 Q
E =
2 C
2
−3
(10 × 10 )
= = 1J
−6
2 × 50 × 10

Hence, the total energy stored in the LC circuit will be conserved because there is no resistor
connected in the circuit.

(b) Natural frequency of the circuit is given by the relation,
1
v =
2π√LC
1
=
√ −3 −6
2π 20×10 ×50×10
3
10
= = 159.24Hz
2π

Natural angular frequency,
1
ωr =
√LC

1 1 3
= = = 10 rad/s
√ −3 −6 √ −6
20×10 ×50×10 10

Hence, the natural frequency of the circuit is 10 rad/s
3

(c) (i) For time period ( T =
1

v
=
1

159.24
= 6.28ms ), total charge on the capacitor at time
′ 2π
Q = Q cos t
T

t,
For energy stored is electrical, we can write Q' = Q.
Hence, it can be inferred that the energy stored in the capacitor is completely electrical
at time, t = 0,
T 3T
, T, ,….
2 2

(ii) Magnetic energy is the maximum when electrical energy, Q' is equal to 0.
Hence, it can be inferred that the energy stored in the capacitor is completely magnetic at time,
T 3T 5T
t = , , ……
4 4 4

(d)
Q
1
Charge on the capacitor when total energy is equally shared between the capacitor and the
=

inductor at time t.
When total energy is equally shared between the inductor and capacitor, the energy
stored in the capacitor = ( (maximum energy).
1

2
′ 2
2 2
(Q ) Q Q
1 1 1 1
⇒ = ( ) =
2 C 2 2 C 4 C

′ Q
Q =
√2

But Q' = Q cos
2π
t
T
Q 2π
= Q cos t
√2 T

2π 1 π
cos t = = cos(2n + 1) ; where n = 0, 1, 2, … .
T √2 4

T
t = (2n + 1)
8

where n=0.1, 2. .....
Hence, total energy is equally shared between the inductor and the capacity at time,
T 3T 5T
t = , , ……
8 8 8

(e) If a resistor is inserted in the circuit, then total initial energy is dissipated as heat energy in the
circuit. The resistance damps out the LC oscillation.

Page : 267 , Block Name : Additional Exercise

Q7.13 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?

Answer. Inductance of the indicator, L = 0.50H
Resistance of the resistor , R = 100Ω
Potential of the supply voltage, V = 240V
Frequency of the supply, v = 50 Hz
(a) Peak voltage is given as:
V0 = √2V

= √2 × 240 = 339.41V

Angular frequency of the supply,
ω = 2πv

= 2π × 50 = 100nrad/s

Maximum current in the circuit is given as:
V0
I0 =
√ R 2 + ω 2 L2

339.41
= = 1.82A
√(100)2 + (100π)2 (0.50)2

(b) Equation for voltage is given as:
V = V0 cos ωt

Equation for current is given as:
I = I0 cos(ωt − Φ)

Where,
Φ = Phase difference between voltage and current

At time, t = 0.
V = V (voltage is maximum)
0

ϕ
For ωt − Φ = 0 i.e., at time t = ω

I = I (current is maximum)
0
ϕ
Hence, the time lag between maximum voltage and maximum current is ω
.
Now, phase angle ϕ is given by the relation,

ωL
tan ϕ =
R
2π×50×0.5
= = 1.57
100

∘ 57.5π
ϕ = 57.5 = rad
180
57.5π
ωt =
180
57.5
t =
180×2π×50
−3
= 3.19 × 10 s

= 3.2ms

Hence, the time lag between maximum voltage and maximum current is 3.2 ms.

Page : 267 , Block Name : Additional Exercise

Q7.14 Obtain the answers (a) to (b) in question 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?

Answer. Inductance of the inductor, L = 0.5Hz
Resistance of the resistor, R = 100Ω
Potential of the supply voltages, V = 240V
Frequency of the supply, V = 10kHz = 10 Hz
4

Angular frequency ω = 2πv = 2π × 10 rad/s 4

(a) Peak voltage, V 0 = √2 × V = 240√2V

Maximum current, I
V0

0 =
√R +ω2 L2
2

240√2 −2
= = 1.1 × 10 A
2
√(100)2 +(2π×104 ) ×(0.50)2

(b) For phase difference Φ, we have the relation:
ωL
tan ϕ =
R
4
2π×10 ×0.5
= = 100π
100

∘ 89.82π
ϕ = 89.82 = rad
180
89.82π
ωt =
180
89.82π
t = = 25μs
4
180×2π×10

It can be observed that I is very small in this case. Hence, at high frequencies, the inductor
0

amounts to an open circuit.
In a dc circuit, after a steady state is achieved, ω = 0. Hence, inductor L behaves like a pure
conducting object.

Page : 267 , Block Name : Additional Exercise

Q7.15 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?

Answer. Capacitance of the capacitor, C = 100μF = 100 × 10
−6
F

Resistance of the resistor, R = 40Ω
Supply voltage, V = 110V
(a) Frequency of oscillations, V = 60Hz
Angular frequency, ω = 2πv = 2π × 60rad/s
For a RC circuit, we have the relation for impedance as:
2 1
Z = R +
2 2
ω C

Peak voltage, V = V √2 = 110√2V 0

Maximum current is given as:
V0
I0 =
Z
V0
=
2 1
√R +
2 2
ω C

110√2
=
1
(40)2 +
√ 2
2 −1
(120π) ×(10 )

110√2
= = 3.24A
8
10
√1600+
2
(120π)

(b) In a capacitor circuit, the voltage lags behind the current by a phase angle of ϕ. This angle is
given by the relation:
1

ωC 1
∴ tan ϕ = =
R ωCR
1
= = 0.6635
−4
120π×10 ×40
−1 ∘
ϕ = tan (0.6635) = 33.56
33.56π
= rad
180
ϕ
∴ Time lag =
ω
33.56π −3
= = 1.55 × 10 s = 1.55ms
180×120π

Hence, the time lag between maximum current and maximum voltage is 1.55 ms.

Page : 267 , Block Name : Additional Exercise

Q7.16 Obtain the answers to (a) and (b) in question 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.

Answer. Capacitance of the capacitor, C = 100μF = 100 × 10 −6
F

Resistance of the resistor, R = 40Ω
Supply voltage, V = 110V
Frequency of the supply, v = 12kHz = 12 × 10 Hz 3

Angular Frequency, ω = 2πv = 2 × n × 12 × 10 03 3

3
= 24π × 10 rad/s

Peak voltage, V 0 = V √2 = 110√2V
V0
I0 =
2 1
√R +
2 2
ω C

Maximum current,
110√2
=
1
(40)2 +
√ 2
(24π×103 ×100×10−6 )

110√2
= = 3.9A
2
10
√1600+( )
24π

For an RC circuit, the voltage lags behind the current by a phase angle of ϕ given as:
1

ωC 1
tan ϕ = =
R ωCR
1
= 3 −6
24π×10 ×100×10 ×40
1
tan ϕ =
96π
∘
∴ ϕ = 0.2
0.2π
= rad
180
ϕ
∴ Time lag =
ω
0.2π −3
= 3
= 1.55 × 10 s = 0.04μs
180×24π×10

Hence, ϕ tends to become zero at a high frequencies. At a high frequency, capacitor C acts as a
conductor.
In a dc circuit, after the steady state is achieved, ω = 0. Hence, capacitor C amounts to an open
circuit.

Page : 267 , Block Name : Additional Exercise

Q7.17 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 speci ed in question 11 for this frequency.

Answer. An inductor ( L), a capacitor ( C), and a resistor ® is connected in parallel with each other
in a circuit where,
L = 5.0H
−6
C = 80μF = 80 × 10 F

R = 40Ω

Potential of the voltage source, V = 230V
Impedance (Z) of the given parallel LCR circuit is given as:
2
1 1 1
= √ + ( − ωC)
2
Z R ωL

Where,
ω =Angular Frequency

At resonance, − ωC = 0
1

ωL

1
∴ ω =
√LC

1
= = 50rad/s
√ −6
5×80×10

Hence, the magnitude of Z is the maximum at 50 rad/s. As a result, the total current is minimum.
Rms current owing through inductor L is given as:
V
IL =
ωL
230
= = 0.92A
50×5

Rms current owing through capacitor C is given as:
V
IC = = ωCV
1

ωC

−6
= 50 × 80 × 10 × 230 = 0.92A

Rms current owing through resistor R is given as:
V
IR =
R
230
= = 5.75A
40

Page : 267 , Block Name : Additional Exercise

Q7.18 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’.]

Answer. Inductance, L = 80mH = 80 × 10 −3
H

Capacitance, C = 60μF = 60 × 10 F −6

Supply voltage, V = 230V
Frequency, v = 50Hz
Angular frequency, ω = 2πv = 100nrad/s
Peak voltage, v = V √2 = 230√2V 0

(a) Maximum current is given as:
V0
I0 =
1
(ωL− )
ωC

230√3
=
−3 1
(100π×80×10 − )
100π×60×10−6

230√2
= = −11.63A
1000
(8π− )
6π

The negative sign appears because ωL <
1

ωC

Amplitude, rms of maximum current, |I | = 11.63A 0

Hence, rms value of current, I =
I0 −11.63
= = −8.22A
√2 √2

(b) Potential difference across the indicator,
VL = I × ωL
−3
= 8.22 × 100π × 80 × 10

= 206.61V

Potential difference across the capacitor,
1
Vc = I ×
ωC
1
= 8.22 × = 436.3V
−6
100π×60×10

(c) Average power consumed by the conductor is zero as actual voltage leads the current by π

2

(d) Average power consumed by the capacitor is zero as voltage lags current by π

2
.

(e) The total power absorbed (average over one cycle) is zero.

Page : 267 , Block Name : Additional Exercise

Q7.19 Suppose the circuit in question 18 has a resistance of 15 Ω. Obtain the average power
transferred to each element of the circuit, and the total power absorbed.

Answer. Average power transferred to the resistor = 788.44W
Average power transferred to the capacitor = 0W
Total power absorbed by the circuit = 788.44w
Inductance of inductor, L = 80mH = 80 × 10 H −3

Capacitance of capacitor, C = 60μF = 60 × 10 F −6

Resistance of resistor, R = 15Ω
Potential of voltage supply, V = 230V
Frequency of signal, v = 50Hz
Angular frequency of signal, ω = 2πv = 2π × (50) = 100πrad/s
The elements are connected in series to each other. Hence, impedance of the circuit is given as:
2
1
Z = √R
2
+ (ωL − )
ωC

2

−3 1
= √(15)
2
+ (100π (80 × 10 ) − )
−6
(100π×60×10 )

2 2
= √(15) + (25.12 − 53.08) = 31.728Ω

Current owing in the circuit, I =
V 230
= = 7.25A
Z 31.728

Average power transferred to resistance is given as:
2
PR = I R
2
= (7.25) × 15 = 788.44W

Average power transferred to capacitor, P C = Average power transferred yo inductor, P L

= 0

Total power absorbed by the circuit:
= PR + PC + PL

= 788.44 + 0 + 0 = 788.44W

Hence, the total power absorbed by the circuit is 788.44W.

Page : 267 , Block Name : Additional Exercise

Q7.20 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?

Answer. Inductance , L = 0.12H
Capacitance, C = 480nF = 480 × 10 −9
F

Resistance, R = 23Ω
Supply voltage, V = 230V
Peak voltage is given as:
V0 = √2 × 230 = 325.22V

(a) Current owing in the circuit is given by the relation , I
V0

0 =
2
1
√R2 +(ωL− )
ωC

Where,
I = maximum at resonance
0

At resonance, we have
1
ωR L − = 0
ωR C

Where,
ωR= Resonance angular frequency
1
∴ ωR =
√LC

1
= = 4166.67rad/s
√0.12×480×10−9

Resonant frequency, v
ωR 4166.67
∴ R = = = 663.48Hz
2π 2×3.14

And, maximum current (I )
V0 325⋅22
0 = = = 14.14A
Max R 23

(b) Maximum average power absorbed by the circuit is given as:
1 2
(Pav ) = (I0 ) R
Max 2 Max

1 2
= × (14.14) × 23 = 2299.3W
2

Hence, resonant frequency ( vR
) is 663.48Hz

(c) The power transferred to the circuit is half the power at resonant frequency.
Frequencies at which power transferred is half, = ω ± Δω R

= 2π (vR ± Δv)

Where,
R
Δω =
2L
23
= = 95.83rad/s
2×0.12

Hence, change in frequency, Δv = 2π
1
Δω =
95.83

2π
= 15.26Hz

ν
∴ +Δv = 663.48 + 15.26 = 678.74Hz
R

And, v − Δv = 663.48 − 15.26 = 648.22Hz
R

Hence, at 648.22 Hz and 678.74 Hz frequencies, the power transferred is half.
At these frequencies, current amplitude can be given as:
′ 1
I = × (I0 )
Max
√2

14.14
= = 10A
√2

(d) Q-factor of the given circuit can be obtained using the relation, Q =
ωR L

R
4166.67×0.12
= = 21.74
23

Hence, the Q-factor of the given circuit is 21.74.

Page : 268 , Block Name : Additional Exercise

Q7.21 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.

Answer. Inductance, L = 3.0H
Capacitance, C = 27μF = 27 × 10 F −6

Resistance, R = 7.4Ω
At resonance, angular frequency of the source for the given LCR series circuit is given as:
1
ωr =
√LC
3
1 10 −1
= = = 111.11rads
√ −6 9
3×27×10

Q-factor of the series:
ωr L
Q =
R
111.11×3
= = 45.0446
7.4

To improve the sharpness of the resonance by reducing its ‘full width at half maximum’ by a factor
of 2 without changing ω , we need to reduce R to half i.e.,
r

Resistance =
R 7.4
= = 3.7Ω
2 2

Page : 268 , Block Name : Additional Exercise

Q7.22 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 uorescent tubes with ac mains? Why can we not use an
ordinary resistor instead of the choke coil?

Answer. (a) Yes; the statement is not true for rms voltage
It is true that in any ac circuit, the applied voltage is equal to the average sum of the
instantaneous voltages across the series elements of the circuit. However, this is not true for rms
voltage because voltages across different elements may not be in phase.

(b) High induced voltage is used to charge the capacitor.
A capacitor is used in the primary circuit of an induction coil. This is because when the circuit is
broken, a high induced voltage is used to charge the capacitor to avoid sparks.

(c) The de signal will appear across capacitor C because for de signals, the impedance of an
inductor (L) is negligible while the impedance of a capacitor (C) is very high (almost in nite).
Hence, a dc signal appears across C. For an ac signal of high frequency, the impedance of L is high
and that of C is very low. Hence, an ac signal of high frequency appears across L.

(d) If an iron core is inserted in the choke coil (which is in series with a lamp connected
to the ac line), then the lamp will glow dimly. This is because the choke coil and the iron core
increase the impedance of the circuit.

(e) A choke coil is needed in the use of uorescent tubes with ac mains because it reduces the
voltage across the tube without wasting much power. An ordinary resistor cannot be used instead
of a choke coil for this purpose because it wastes power in the form of heat.

Page : 268 , Block Name : Additional Exercise

Q7.23 A power transmission line feeds input power at 2300 V to a stepdown 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?

Answer. Input voltage, V = 2300
1

Number of turns in primary coil, n = 4000
1

Output voltage, V = 230V
2

Number of turns in secondary coil = n 2

Voltage is related to the number of turns as:

V1 n1
=
V2 n2

2300 4000
=
230 n2

4000×230
n2 = = 400
2300

Hence, there are 400 turns in the second winding.

Page : 268 , Block Name : Additional Exercise

Q7.24 At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water
ow available is 100m s . If the turbine generator ef ciency is 60%, estimate the electric power
3 −1

available from the plant (g = 9.8ms ). −2

Answer. Height of water pressure head, h = 300 m
Volume of water ow per second, V = 100m /s 3

Ef ciency of turbine generator, n = 60% = 0.6
Acceleration due to gravity, g = 9.8m/s 2

Density of water, ρ = 10 kg/m 3 3

Electric power available from the plant = η × hρgV
3
= 0.6 × 300 × 10 × 9.8 × 100
6
= 176.4 × 10 w

= 176.4MW

Page : 268 , Block Name : Additional Exercise

Q7.25 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 - 220V 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.

Answer. Total electric power required, P = 800kW = 800 × 10 W 3

Supply voltage, V = 220V
Voltage at which electric plant is generating power, V = 440V
Distance between the town and power generating station, d = 15km
Resistance of the two wire lines carrying power = 0.5Ω/km
Total resistance of the wires, R = (15 + 15)0.5 = 15Ω
A step-down transformer of rating 4000 – 220 V is used in the sub-station.
Input voltage, V = 4000V 1

Output voltage, V = 220V 2

Rms current in the wire lines is given as:
P
I =
V1

3
800×10
= = 200A
4000

(a) Line power loss = I R 2

2
= (200) × 15
3
= 600 × 10 W

= 600kW

(b) Assuming that the power loss is negligible due to the leakage of the current:
Total power supplied by the plant = 800 kW + 600 kW
= 1400 kW

(c) Voltage drop in the power line = I R = 200 × 15 = 3000V
Hence, total voltage transmitted from the plant = 3000 + 4000
= 7000 V
Also, the power generated is 440 V.
Hence, the rating of the step-up transformer situated at the power plant is 440 V - 7000V.

Page : 268 , Block Name : Additional Exercise

Q7.26 Do the same exercise as above with the replacement of the earlier transformer by a 40,000-
220V 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). Hence, explain
why high voltage transmission is preferred?

Answer. The rating of a step-down transformer is 40000V-220V.
Input voltage, V = 40000V 1

Output voltage, V = 220V 2

Total electric power required, P = 800kW = 800 × 10 W 3

Source potential, V = 220V
Voltage at which the electric plant generates power, V = 440V
Distance between the town and power generating station, d = 15 km
Resistance of the two wire lines carrying power = 0.5Ω/km
Total resistance of the wire lines, R = (15 + 15)0.5 = 15Ω
P = V1 I

Rms current in the wire line is given as:
P
I =
V1

3
800×10
= = 20A
40000

(a) Line power loss = I R 2

2
= (20) × 15

= 6kW

(b) Assuming that the power loss is negligible due to the leakage of current.

Hence, power supplied by the plant = 800 kW + 6kW = 806 kW
(c) Voltage drop in the power line = IR = 20 x 15 = 300 V
Hence, voltage that is transmitted by the power plant
= 300 + 40000 = 40300 V
The power is being generated in the plant at 440 V.
Hence, the rating of the step-up transformer needed at the plant is
440 V - 40300 V.
Hence, power loss during transmission = 600

1400
× 100 = 42.8%

In the previous exercise, the power loss due to the same reason is 6

806
× 100 = 0.744%

Since the power loss is less for a high voltage transmission, high voltage transmissions are
preferred for this purpose.

Page : 268 , Block Name : Additional Exercise