NCERT Solutions for Class 12 Physics Chapter 7 AC: Download PDF for FREE

EXERCISES

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

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

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

Sol. Given,

R = 100 ohms

V = 220V

f = 50 Hz

(A) Irms = vrms / R

Substitute,

Irms = 220 / 100 = 2.2A

(B) Performance = VI

Or power = 220 x 2.2

Or power = 484 W

7.2. (a) The peak voltage of an AC power supply is 300 V. What is the effective voltage?

(b) The effective value of the current in an AC circuit is 10 A. What is the peak current?

Sol. (A) vrms = Vpeak / 1.414

vrms = 300 / 1.414

Or vrms = 212.13V

(B) The above identity is used for the current one

Ipeak = 1.414 x Irms

Or Ipeak = 1.414 x 10 = 14.14 A

7.3. A 44mH inductor is connected to a 220V 50Hz AC power supply. Determine the effective value of the current in the circuit.

Sol. Given,

L = 44 mH

V = 220V

f = 50 Hz

Irms is given by= V / XL

Determination of inductive reactance

XL= 2 x 3.14 x 50 x 44 x 10-³

XL= 13.82 ohms

Therefore,

Irms=220 / 13.82

Or Irms=15.92A

Also Read: NCERT Books for CBSE Class 12 – Latest Edition

7.4. A 60 μF capacitor is connected to a 110 V, 60 Hz AC power supply. Determine the effective value of the current in the circuit.

Sol. Given,

C = 60 microfarads

V = 110 volts

f = 60 hertz

Irms = V/Xc

Now

Xc = 1 / (2 x 3.14 x 60 x 60 x 10-⁶)

Xc = 44.248 ohms

Thus

Irms = 110 / 44.248 = 2.488 A

7.5. In Exercises 7.3 and 7.4, what is the net power consumed by each circuit over a complete cycle? Explain your answer.

Sol. (A) In the case of the inductive network:

The effective current value is I = 15.92 A

The effective value of the voltage is V = 220 V

Therefore, the total power consumed can be derived by the following equation: P = VI cos Φ

In a purely inductive circuit, the phase difference between an alternating voltage and an alternating current is 90°, i.e. Φ = 90°.

Therefore P = 0

Therefore, the total power consumed by the circuit is zero.

(B) In the case of the capacitive network, we know that the value of the effective current is given by I = 2.49 A

The value of the effective voltage results from V = 110 V

Thus, the total power consumed can be derived from the following equation: P = VI Cos Φ

The phase difference between alternating voltage and alternating current is 90°, i.e. Φ = 90°. Therefore P = 0

Therefore, the net power consumed by the circuit is zero.

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

Sol. Given,

C = 30 microF

L = 27 mH

Angular frequency of free oscillations = 1/√LC

Results of substitution

Angular frequency = 1111.11 /s

7.7. A serial LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a 200 V AC variable frequency power supply. If the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in a complete cycle?

Sol. Given,

R = 20 ohms

L = 1.5 Henry

C = 35 microfarads

V = 200 volts

natural frequency

=1/√LC

= 138 /s

At natural frequency

Z = R

So I = V / R = 200 / 20 = 10 A

Therefore

P = I²R

Or P = 10 x 10 x 20 = 2000 W

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

(a) Determine the source frequency that causes the circuit to resonate.

(b) Find the impedance of the circuit and the current amplitude at the resonant frequency.

(c) Determine the effective potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonance frequency.

Sol. Given,

V = 230V

L = 5H

C = 80 μF

R = 40 ohms

(A) Source frequency at resonance=1/√LC

Solve by entering the appropriate values

= 50 rad/s

(B) If there is resonance,

Impedance, Z = Resistance, R

So Z = R = 40 ohms

Now effective value of the current,

I = V / R

Or I = 230 / 40

Therefore I = 5.75 A

Amplitude of this current value = 1.414 x I

= 1.414 x 5.75

= 8.13 A

(C) Now the RMS potential drops are taken into account

Over the resistance

vR= IR

= 5.75 x 40

= 230V

About capacity

vC= IXC

= 1437.5V

About the inductance

vL= IXL

=5.75 x 50 x 5

=1437.5V

About LC combination

vLC= I(XL-XC)

= 0 (at resonance frequency)

To download these solutions in PDF format, click on the following link: