 # NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current

NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current is provided here to assist the students to understand the concepts thoroughly. Chapter 7 Physics Class 12 NCERT solutions comprehensively give answers to the questions asked in the textbook. Physics Class 12 Chapter 7 Alternating Current notes can be prepared to save the time of revision by studying Alternating Current Class 12 NCERT PDF.

For students of Class 12 who are preparing for their Class 12 CBSE board exams as well as competitive exams, it is important to work on NCERT Class 12 solutions for Physics Chapter 7 Alternating Current. We have provided NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current for students to sharpen their skills in the topic of Alternating Current. With the help of NCERT Class 12 Physics Solutions for Chapter 7 Alternating Current PDF, a student can learn the topic with the help of questions and solved answers.

## Class 12 Physics NCERT Solutions for Chapter 7 Alternating Current

Alternating Current is a crucial chapter in CBSE Class 12. Students must be well versed in this chapter to score well in their examinations. Also, NCERT solutions are one of the best tools to prepare for Class 12 Physics.

#### Concepts involved in NCERT Class 12 Physics Chapter 7 Alternating Current

1. Introduction
2. Ac Voltage Applied To A Resistor
3. Representation Of Ac Current And Voltage By Rotating Vectors — Phasors
4. Ac Voltage Applied To An Inductor
5. Ac Voltage Applied To A Capacitor
6. Ac Voltage Applied To A Series Lcr Circuit
1. Phasor-diagram solution
2. Analytical solution
3. Resonance
7. Lc Oscillations
8. Transformers

### Some key points of Class 12 Physics Alternating Current are given below:

• When a value is given for AC voltage or current, it is ordinarily the rms value.
• The power rating of an element used in ac circuits refers to its average power rating.
• There are no power losses associated with pure capacitances and pure inductances in an AC circuit.
• The power consumed in an AC circuit is never negative.
• A transformer (step-up) changes a low-voltage into a high-voltage.
• The power factor in an RLC circuit is a measure of how close the circuit is to expending the maximum power.
• In an AC circuit while adding voltages across different elements one should take care of their phases properly.

### Class 12 Physics NCERT Solutions Alternating Current Important Questions

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?

Given :

The resistance R of the resistor is 100 Ω

The source voltage  V is 220 V

The frequency of the supply is  50 Hz.

a) To determine the RMS value of the current in the connection, we use the following relation :

$I = \frac{ V }{ R }$

Substituting values, we get

$I = \frac{ 220 }{ 100 }$ = 2.20 A

Therefore, the RMS value of the current in the connection is 2.20 A.

b) The total power consumed over an entire cycle can be calculated using the following formula :

P = V x I

Substituting values in the above equation, we get

= 220 x 2.2 = 484 W

Therefore, the total power consumed is 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?

a) The peak voltage of the AC supply is V 0 = 300 V.

We know that,

$V _{RMS }$ = $V = \frac{ V _{0}}{ \sqrt{ 2 }}$

Substituting the values, we get

$V = \frac{ 300 }{ \sqrt{ 2 }}$ = 212.2 V

The RMS voltage of the AC supply is 212.2 V.

b) The RMS value of the current in the circuit is I = 10 A

We can calculate the peak current from the following equation

$I _{ 0 } = \sqrt{ 2 } I$

Substituting the values, we get

$\sqrt{ 2 } \times 10 = 14.1 A$

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.

As given :

The inductor connected to the AC supply has an inductance of  L = 44 m H = 44 × 10 – 3 H

The magnitude of the source voltage V is 220 V

The frequency of the source is ν = 50 Hz

The angular frequency of the source  is given by ω = 2 π ν

The Inductive reactance X L  can be calculated as follows:

ω L = 2 π ν L = 2π × 50 × 44 × 10 – 3 Ω

To find the RMS value of current we use the following relation:

à  $I = \frac{ V }{ X _{ L }} = \frac{ 220 }{ 2\pi \times 50 \times 44 \times 10 ^{ – 3 }}$ = 15.92 A

Therefore, the RMS value of the current in the network 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.

Given:

The capacitance of the capacitor in the circuit is C = 60 μ F or 60 × 10 – 6 F

The source voltage is V = 110 V

The frequency of the source is ν = 60 Hz

The angular frequency can be calculated using the following relation,

ω = 2πν

The capacitive reactance in the circuit is calculated as follows:

$X _{ C } = \frac{ 1 }{ \omega C } = \frac{ 1 }{ 2\pi \nu C} = \frac{ 1 }{ 2 \pi \times 60 \times 60 \times 10 ^{ – 6 }}$ Ω

Now, the RMS value of the current is determined as follows:

à  $I = \frac{ V }{ X_{ C }} = \frac{ 220 }{ 2\pi \times 60 \times 60 \times 10^{ – 6 }} = 2.49 A$

Therefore, the RMS 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

a) In the case of the inductive network, we know that

the RMS current value is I = 15.92 A

the RMS voltage value is V = 220 V

Therefore, the total power taken in can be derived by the following equation :

P = VI cos Φ

Here,

Φ is the phase difference between V and I

In case of a purely inductive circuit, the difference in the phase of an alternating voltage and an alternating current is 90°,

i.e., Φ = 90°.

Therefore, P = 0

i.e., the total power absorbed by the circuit is zero.

b) In the case of the capacitive network, we know that

The value of RMS current is given by, I = 2.49 A

The value of RMS voltage is given by, V = 110 V

Thus, the total power absorbed is derived from the following equation :

P = VI Cos Φ

For a purely capacitive circuit, the phase difference between alternating Voltage and alternating current is 90°

i.e., Φ = 90°.

Thus , P = 0

i.e., the net power absorbed by the circuit is zero.

Question 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?

Given:

The inductance of the inductor is L = 2.0 H

The capacitance of the capacitor, C = 32 μF = 32 × 10 – 6 F

The resistance of the resistor is  R = 10 Ω.

We know that , resonant frequency can be calculated by the following relation,

à  $\omega _{ r } = \frac{ 1 }{ \sqrt{ LC }} = \frac{ 1 }{ \sqrt{ 2 \times 32 \times 10 ^{ – 6 }}} = \frac{ 1 }{ 8 \times 10 ^{ – 3 }} = 125 \frac{ rad }{ sec }$

Now , Q – value of the circuit  can be calculated as follows  :

à  $Q = \frac{ 1 }{ R } \sqrt{\frac{ L }{ C }} = \frac{ 1 }{ 10 }\sqrt{\frac{ 2 }{ 32 \times 10 ^{ – 6 }}} = \frac{ 1 }{ 10 \times 4 \times 10 ^{ – 3 }} = 25$

Thus , the Q – Value of the above question 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?

Given Capacitance value of the capacitor , C = 30 μF = 30 × 10 – 6 F

Given Inductance value of the charged inductor, L = 27 mH = 27 × 10 – 3 H

Angular frequency is given as :

à  $\omega _{ r } = \frac{ 1 }{ \sqrt{ LC }}$

à  $\omega _{ r } = \frac{ 1 }{ \sqrt{ 27 \times 10 ^{ – 3 } \times 30 \times 10 ^{ – 6 }}} = \frac{ 1 }{ 9 \times 10 ^{ – 4 }} = 1.11 \times 10 ^{ 3 } \frac{ rad }{ sec }$

Therefore , the calculated angular frequency of free oscillation of the connection is 1.11 x 10 3 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 a later time?

Given Capacitance value 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

Total energy stored in the capacitor can be calculated as :

à  $E = \frac{ 1 }{ 2 } \times \frac{ Q ^{ 2 }}{ C } = \frac{ 1 }{ 2 }\frac{\left ( 6 \times 10^{ -3 } \right )^{ 2 }}{ 30 \times 10 ^{ – 6}} = \frac{ 6 }{ 10 } = 0.6 J$

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 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?

The supply frequency and the natural frequency are equal at resonance condition in the circuit.

Given Resistance of the resistor, R = 20 Ω

Given Inductance of the inductor, L = 1.5 H

Given Capacitance of the capacitor , C = 35 μF = 30 × 10 – 6 F

An AC source with a voltage of V = 200 V is connected to the LCR circuit,

We know that the Impedance of the above combination can be calculated by the following relation,

à  $Z = \sqrt{ R ^{ 2 } + \left ( X _{ L } – X _{ C } \right ) ^{ 2 }}$

At resonant condition in the circuit , X L = X C

Therefore , Z = R = 20 Ω

We know that Current in the network is given by the relation :

à  $I = \frac{ V }{ Z } = \frac{ 200 }{ 20 } = 10 A$

Therefore , the average power that is being transferred to the circuit in one full cycle :

V I = 200 x 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: The condition for tuning is that the natural frequency which is the frequency of free oscillations of the LC network must be of the same value as the radio wave frequency ]

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

Given that the Lower tuning frequency of the circuit is , ν 1 = 800 kHz = 800 × 10 3 Hz

Given that the Upper tuning frequency of the circuit is, ν 2 = 1200 kHz = 1200 × 10 3 Hz

Given that the Effective inductance of the inductor in the circuit is  L = 200 μH = 200 × 10 – 6 H

We know that, Capacitance of variable capacitor for ν 1 can be calculated as follows :

à  $C _{ 1 } = \frac{ 1 }{ \omega _{ 1 }^{ 2 } L }$

Here the variables are ,

ω 1 = Angular frequency for capacitor C  1

à  ω 1 = 2 π ν 1

à  ω 1 = 2π × 800 × 10 3 rad/s

therefore,

$C _{ 1 } = \frac{ 1 }{ \left ( 2\pi \times 800 \times 10 ^{ 3 } \right )^{2} \times 200 \times 10 ^{ – 6 }} = 1.9809 \times 10 ^{ – 10 } F = 198 pF$

Variable capacitor for v 2 has capacitance of :

$C _{ 2 } = \frac{ 1 }{ \omega _{ 2 }^{ 2 } L }$

Here the variables are ,

ω 2 = Angular frequency for capacitor C  2

à  ω = 2 π ν 2

à  ω 2 = 2π × 1200 × 10 3 rad/s

Therefore,

$C _{ 2 } = \frac{ 1 }{ \left ( 2\pi \times 1200 \times 10 ^{ 3 } \right )^{2} \times 200 \times 10 ^{ – 6 }} = 0.8804 \times 10 ^{ – 10 } F = 88 pF$

Thus , the variable capacitor has a range from 88.04 pF to 198.1 pF.

Question 7.11:

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

L = 5.0 H,

C = 80 μF,

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.

Given that the Inductance of the inductor in the circuit is , L = 5.0 H

Given that the Capacitance of the capacitor in the circuit is , C = 80 μH = 80 × 10 – 6 F

Given that Resistance of the resistor in the circuit , R = 40 Ω

Value of Potential of the variable voltage supply, V = 230 V

(a) We know that the Resonance angular frequency can be obtained by the following relation :

à  $\omega _{ r } = \frac{ 1 }{ \sqrt{ LC }}$ $\omega _{ r } = \frac{ 1 }{ \sqrt{ 5 x 80 x 10 – 6 }}$ $\omega _{ r } = \frac{ 10 ^{ 3 }}{ 20 } = 50 rad/sec$

Thus, the circuit encounters resonance at a frequency of 50 rad/s.

(b) We know that the Impedance of the circuit can be calculated by the following relation :

à  $Z = \sqrt{ R ^{ 2 } + \left ( X _{ L } – X _{ C } \right ) ^{ 2 }}$

At resonant condition ,

X L = X C

Z = R = 40 Ω

At resonating frequency amplitude of the current can be given by the following relation :

à  $I _{ 0 } = \frac{ V _{ 0 }}{ Z }$

Where,

à  V 0 = peak voltage = $\sqrt{ 2 } V$

Therefore,

à  $I _{ 0 } = \frac{ \sqrt{ 2V }}{ Z } = \frac{ \sqrt{ 2 } \times 230 }{ 40 } = 8.13 A$

Thus, at resonant condition , the impedance of the circuit is calculated to be 40 Ω and the amplitude of the current is found to be  8.13 A

c) rms potential drop across the inductor in the circuit ,

à  ( V L ) rms = I x ω r L

Where,

$I _{ rms } = \frac{ I _{ 0 }}{ \sqrt{ 2 }} = \frac{ \sqrt{ 2 } V}{ \sqrt{ 2 } Z } = \frac{ 230 }{ 40 } = \frac{ 23 }{ 4 } A$

Therefore, ( V L ) rms

$\frac{ 23 }{ 4 } \times 50 \times 5 = 1437.5 V$

We know that the Potential drop across the capacitor can be calculated with the following relation :

à  $\left ( V _{ c } \right )_{ rms } = I \times \frac{ 1 }{ \omega _{ r } C } = \frac{ 23 }{ 4 } \times \frac{ 1 }{ 50 \times 80 \times 10 ^{ – 6 }} = 1437.5 V$

We know that the Potential drop across the resistor can be calculated with the following relation :

à   $\left ( V _{ R } \right )_{ rms } = IR = \frac{ 23 }{ 4 } \times 40 = 230 V$

Now, Potential drop across the LC connection can be obtained by the following relation :

à  V L C  = I ( X L − X C )

At resonant condition ,

à X L = X C

à V L C = 0

Therefore, it has been proved from the above equation that the potential drop across the LC connection is equal to zero at a frequency at which resonance occurs.

 Also Access NCERT Exemplar for Class 12 Physics Chapter 7 CBSE Notes for Class 12 Physics Chapter 7

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