NCERT Solutions for Class 12 Physics Chapter 7 – Alternating Current

Question 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?

Answer:

Resistance of the resistor, = 100 Ω

Supply voltage, V = 220 V

Frequency, ν = 50 Hz

(a) The rms value of current in the circuit is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 6654e5d8

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

P = VI

= 220 × 2.2 = 484 W

Question 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?

Answer:

(a) Peak voltage of the ac supply, V0 = 300 V

Rms voltage is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m1ae8ba23

(b) Therms value of current is given as:

I = 10 A

Now, peak current is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m70957934

Question 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.

Answer:

Inductance of inductor, = 44 mH = 44 × 10−3 H

Supply voltage, V = 220 V

Frequency, ν = 50 Hz

Angular frequency, ωNS 12 11 08 Sravana 12 Physics 7 26 NRJ html m1493eef7

Inductive reactance, XL = ω LNS 12 11 08 Sravana 12 Physics 7 26 NRJ html 62dccbc7

Rms value of current is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 4edeb52a

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

Question 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.

Answer:

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

Supply voltage, = 110 V

Frequency, ν = 60 Hz

Angular frequency, ωNS 12 11 08 Sravana 12 Physics 7 26 NRJ html m1493eef7

Capacitive reactance NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 77cb6f18

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 5af1565c

Rms value of current is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 94d31c9

Hence, the rms value of current is 2.49 A.

Question 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.

Answer:

In the inductive circuit,

Rms value of current, I = 15.92 A

Rms value of voltage, V = 220 V

Hence, the net power absorbed can be obtained by the relation,

P = VI 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 ve obtained as:

P = VI Cos Φ

For a purely 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.

Question 7.6:

Obtain the resonant frequency ωof a series LCR circuit with = 2.0 H, = 32 μF and = 10 Ω. What is the Q-value of this circuit?

Answer:

Inductance, = 2.0 H

Capacitance, C = 32 μF = 32 × 10−6 F

Resistance, R = 10 Ω

Resonant frequency is given by the relation,

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 3f42c08f

Now, Q-value of the circuit is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m3baa8350

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

Question 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?

Answer:

Capacitance, C = 30μF = 30×10−6F

Inductance, L = 27 mH = 27 × 10−3 H

Angular frequency is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 4929193e

Hence, the angular frequency of free oscillations of the circuit is 1.11 × 103 rad/s.

Question 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?

Answer:

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

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

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

The total energy stored in the capacitor can be calculated by the relation,

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m37539a8

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

Question 7.9:

A series LCR circuit with = 20 Ω, = 1.5 H and = 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 = 20 Ω

Inductance, L = 1.5 H

Capacitance, C = 35 μF = 30 × 10−6 F

AC supply voltage to the LCR circuit, V = 200 V

The impedance of the circuit is given by the relation,

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 28bb43fa

At resonance,NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 1ee720b1

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 3bb42138

Current in the circuit can be calculated as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 15510633

Hence, the average power transferred to the circuit in one complete cycle= VI

= 200 × 10 = 2000 W.

Question 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 radio wave.]

Answer:

The range of frequency (ν) of radio is 800 kHz to 1200 kHz.

Lower tuning frequency, ν1 = 800 kHz = 800 × 103 Hz

Upper tuning frequency, ν2 = 1200 kHz = 1200× 103 Hz

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

The capacitance of variable capacitor for νis given as:

C1NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 57ef940a

Where,

ω1 = Angular frequency for capacitor C1

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 796b12d3

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 44ae19ab

The capacitance of variable capacitor for ν2,

C2 NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 3c1a1915

Where,

ω2 = Angular frequency for capacitor C2

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 76fff642

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 599d10be

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

Question 7.11:

Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. = 5.0 H, = 80μF, = 40 Ω

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 64fd4bc1

(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:

The inductance of the inductor, = 5.0 H

Capacitance of the capacitor, C = 80 μF = 80 × 10−6 F

Resistance of the resistor, R = 40 Ω

Potential of the variable voltage source, V = 230 V

(a) Resonance angular frequency is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m1eca9398

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,

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 28bb43fa

At resonance,

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 1329ac37

The amplitude of the current at the resonating frequency is given as: NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 27f27743

Where,

V0 = Peak voltage

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 5dd185a5

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

(c) Rms potential drop across the inductor,

(VL)rms = I × ωRL

Where,

I = rms current

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html me32d57a

The potential drop across the capacitor,

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m4fca7fbe

The potential drop across the resistor,

(VR)rms = IR

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 28fee272× 40 = 230 V

The potential drop across the LC combination,

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 27053e34

At resonance,NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m208894a0

VLC= 0

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

Page No 267:

Question 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 = 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 = 20 mH = 20 × 10−3 H

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

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

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

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m1e69108e

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,

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m33157fb7

Natural angular frequency,

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 496ccca4

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 73a82d5d

Hence, the natural frequency of the circuit is 103 rad/s.

(c) (i) For time period (TNS 12 11 08 Sravana 12 Physics 7 26 NRJ html 2c340176), the total charge on the capacitor at time tNS 12 11 08 Sravana 12 Physics 7 26 NRJ html m3e4aa06f

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 the time, t =NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m68236e41

(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 the time, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 19c1522c

(d) Q1 = 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 = NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m5a4d85ce(maximum energy).

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m46c0a141

Hence, the total energy is equally shared between the inductor and the capacity at the time, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 1086510a

(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.

Question 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?

Answer:

The inductance of the inductor, L = 0.50 H

Resistance of the resistor, R = 100 Ω

Potential of the supply voltage, V = 240 V

Frequency of the supply, ν = 50 Hz

(a) Peak voltage is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 2195bce8

Angular frequency of the supply,

ω = 2 πν

= 2π × 50 = 100 π rad/s

The maximum current in the circuit is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m554dec0d

(b) Equation for voltage is given as:

V = V0 cos ωt

The equation for current is given as:

I = I0 cos (ωt − Φ)

Where,

Φ = Phase difference between voltage and current

At time, = 0.

V = V0(voltage is maximum)

Forωt − Φ = 0 i.e., at timeNS 12 11 08 Sravana 12 Physics 7 26 NRJ html m6f68eee9,

I = I0 (current is maximum)

Hence, the time lag between the maximum voltage and the maximum current isNS 12 11 08 Sravana 12 Physics 7 26 NRJ html b211ca0.

Now, phase angle Φ is given by the relation,

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m6e91911c

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

Question 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?

Answer:

Inductance of the inductor, L = 0.5 Hz

Resistance of the resistor, R = 100 Ω

Potential of the supply voltages, V = 240 V

Frequency of the supply,ν = 10 kHz = 104 Hz

Angular frequency, ω = 2πν= 2π × 104 rad/s

(a) Peak voltage, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html mb4bc4f0

Maximum current, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m323b3600

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m124de14c

(b) For phase differenceΦ, we have the relation:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m7457eae3

It can be observed that I0 is very small in this case. Hence, at high frequencies, the inductor 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.

Question 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?

Answer:

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

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

(a) Frequency of oscillations, ν= 60 Hz

Angular frequency, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m4fa674fe

For a RC circuit, we have the relation for impedance as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 2b9a67c6

Peak voltage, V0 = NS 12 11 08 Sravana 12 Physics 7 26 NRJ html mc2bb075

The maximum current is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m75c0f886

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html maadb682

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m282048c8

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

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m3b3e7e41

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

Question 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 behavior with that of a capacitor in a dc circuit after the steady-state.

Answer:

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

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

Frequency of the supply, ν = 12 kHz = 12 × 103 Hz

Angular Frequency, ω = 2 πν= 2 × π × 12 × 10303

= 24π × 103 rad/s

Peak voltage, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html me15b5aa

Maximum current, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 14667538

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 47fb6dc5

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

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 1dda124a

Hence, Φ tends to become zero at 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.

Question 7.17:

Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, LC, and 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.

Answer:

An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where,

L = 5.0 H

C = 80 μF = 80 × 10−6 F

R = 40 Ω

Potential of the voltage source, V = 230 V

Impedance (Z) of the given parallel LCR circuit is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m218227f8

Where,

ω = Angular frequency

At resonance,NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m4f84ede

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m2f284cf6

Hence, the magnitude of is the maximum at 50 rad/s. As a result, the total current is minimum.

Rms current flowing through inductor L is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html ffddc7f

Rms current flowing through capacitor C is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 42ffc9f1

Rms current flowing through resistor R is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 41000482

Question 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.]

Answer:

Inductance, L = 80 mH = 80 × 10−3 H

Capacitance, = 60 μF = 60 × 10−6 F

Supply voltage, V = 230 V

Frequency, ν = 50 Hz

Angular frequency, ω = 2πν= 100 π rad/s

Peak voltage, V0 NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 4a0c7f81

(a) Maximum current is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 42b95761

The negative sign appears because NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m556e9450

The amplitude of maximum current, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 64f39bba

Hence, rms value of current, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m35004f75

(b) Potential difference across the inductor,

VL= I × ωL

= 8.22 × 100 π × 80 × 10−3

= 206.61 V

Potential difference across the capacitor,

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m6d50eb75

(c) Average power consumed by the inductor is zero as actual voltage leads the current byNS 12 11 08 Sravana 12 Physics 7 26 NRJ html m4c4df2e.

(d) Average power consumed by the capacitor is zero as voltage lags current byNS 12 11 08 Sravana 12 Physics 7 26 NRJ html m4c4df2e.

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

Question 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.

Answer:

Average power transferred to the resistor = 788.44 W

Average power transferred to the capacitor = 0 W

Total power absorbed by the circuit = 788.44 W

Inductance of inductor, L = 80 mH = 80 × 10−3 H

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

Resistance of resistor, R = 15 Ω

Potential of voltage supply, V = 230 V

Frequency of signal, ν = 50 Hz

Angular frequency of signal, ω = 2πν= 2π × (50) = 100π rad/s

The elements are connected in series to each other. Hence, the impedance of the circuit is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 39d0a8cf

Current flowing in the circuit, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m3b1a3d50

Average power transferred to resistance is given as:

PRI2R

= (7.25)2 × 15 = 788.44 W

Average power transferred to capacitor, PC = Average power transferred to inductor, PL = 0

Total power absorbed by the circuit:

P+ PC + PL

= 788.44 + 0 + 0 = 788.44 W

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

Page No 268:

Question 7.20:

A series LCR circuit with = 0.12 H, = 480 nF, = 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 = 0.12 H

Capacitance, C = 480 nF = 480 × 10−9 F

Resistance, R = 23 Ω

Supply voltage, V = 230 V

Peak voltage is given as:

V0 = NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 7b6da420= 325.22 V

(a) Current flowing in the circuit is given by the relation, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 3ff8537d

Where,

I0 = maximum at resonance

At resonance, we have

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 2389a322

Where,

ωResonance angular frequency

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 3335dda1

∴Resonant frequency, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m1e382a90

And, maximum current NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m47041c44

(b) Maximum average power absorbed by the circuit is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m318ab1ab

Hence, resonant frequency (NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 756f72b1) is NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 28bb5301

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m53d4ecadNS 12 11 08 Sravana 12 Physics 7 26 NRJ html m53d4ecad

(c) The power transferred to the circuit is half the power at the resonant frequency.

Frequencies at which power transferred is half, = NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 1bf49612

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m6fb046c8

Where,

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html be47ee0

Hence, change in frequency, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 5b51d36c

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m668c3410

And, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 501f30bb

Hence, at 648.22 Hz and 678.74 Hz frequencies, the power transferred is half.

At these frequencies, current amplitude can be given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 1eb81134

(d) Q-factor of the given circuit can be obtained using the relation, NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 597f3c1c

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 688b97b7

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

Question 7.21:

Obtain the resonant frequency and Q-factor of a series LCR circuit with = 3.0 H, = 27 μF, and = 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 = 3.0 H

Capacitance, C = 27 μF = 27 × 10−6 F

Resistance, R = 7.4 Ω

At resonance, the angular frequency of the source for the given LCR series circuit is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m753b479a

Q-factor of the series:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 2c5014f6

To improve the sharpness of the resonance by reducing its ‘full width at half maximum’ by a factor of 2 without changingNS 12 11 08 Sravana 12 Physics 7 26 NRJ html 590db411, we need to reduce R to half i.e.,

Resistance =NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 19887939

Question 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 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?

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 dc signal will appear across capacitor C because for dc signals, the impedance of an inductor (L) is negligible while the impedance of a capacitor (C) is very high (almost infinite). 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 fluorescent 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.

Question 7.23:

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, V1 = 2300

Number of turns in primary coil, n1 = 4000

Output voltage, V2 = 230 V

Number of turns in secondary coil = n2

Voltage is related to the number of turns as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m5cc6fd50

Hence, there are 400 turns in the second winding.

Question 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 ms−1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g= 9.8 m s−2).

Answer:

Height of water pressure head, h = 300 m

Volume of water flow per second, V = 100 m3/s

Efficiency of turbine generator, n = 60% = 0.6

Acceleration due to gravity, g = 9.8 m/s2

Density of water, ρ = 103 kg/m3

Electric power available from the plant = η × gV

= 0.6 × 300 × 103 × 9.8 × 100

= 176.4 × 106 W

= 176.4 MW

Question 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.

Answer:

Total electric power required, P = 800 kW = 800 × 103 W

Supply voltage, V = 220 V

Voltage at which electric plant is generating power, V‘ = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wire lines carrying power = 0.5 Ω/km

Total resistance of the wires, R = (15 + 15)0.5 = 15 Ω

A step-down transformer of rating 4000 − 220 V is used in the sub-station.

Input voltage, V1 = 4000 V

Output voltage, V2 = 220 V

Rms current in the wire lines is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m7d9d38e9

(a) Line power loss = I2R

= (200)2 × 15

= 600 × 103 W

= 600 kW

(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 = IR = 200 × 15 = 3000 V

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 − 7000 V.

Question 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). Hence, explain why high voltage transmission is preferred?

Answer:

The rating of a step-down transformer is 40000 V−220 V.

Input voltage, V1 = 40000 V

Output voltage, V2 = 220 V

Total electric power required, P = 800 kW = 800 × 103 W

Source potential, V = 220 V

Voltage at which the electric plant generates power, V‘ = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wire lines carrying power = 0.5 Ω/km

Total resistance of the wire lines, R = (15 + 15)0.5 = 15 Ω

P = V1I

Rms current in the wire line is given as:

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 160c5af5

NS 12 11 08 Sravana 12 Physics 7 26 NRJ html m14440a2d

(a) Line power loss = I2R

= (20)2 × 15

= 6 kW

(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 × 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 = NS 12 11 08 Sravana 12 Physics 7 26 NRJ html 31024cc5

In the previous exercise, the power loss due to the same reason isNS 12 11 08 Sravana 12 Physics 7 26 NRJ html 74b2823a. Since the power loss is less for a high voltage transmission, high voltage transmissions are preferred for this purpose.

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