# Alternating Current Chapter 7 Class 12 Physics NCERT Textbook PDF

NCERT Solutions for Class 12 Physics Chapter 7′ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 12 Physics Chapter 7 Exercise Solution’ using the download button.

### Chapter 7: 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 the 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 that are exploited in many devices of daily use.

For example, whenever we tune our radio to a favorite station, we are taking advantage of a special property of ac circuits – one of many that you will study in this chapter.

#### Representation Of Ac Current And Voltage By Rotating Vectors— Phasors

In the previous section, we learned 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 the 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
that 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 voltage and current phasors and their relationship at time t for the case of an ac source connected to a resistor i.e., corresponding to the circuit shown in Fig. 7.1.

The projection of voltage and current phasors on the 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.

### NCERT Solutions 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?

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 }I=RV

Substituting values, we getI = \frac{ 220 }{ 100 }I=100220​ = 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 }VRMS​ = V = \frac{ V _{0}}{ \sqrt{ 2 }}V=2​V0​​

Substituting the values, we getV = \frac{  300 }{ \sqrt{ 2 }}V=2​ 300​ = 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 equationI _{ 0 } = \sqrt{ 2 } II0​=2​I

Substituting the values, we get\sqrt{ 2 } \times 10 = 14.1 A2​×10=14.1A

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  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:I = \frac{ V }{ X _{ L }} = \frac{ 220 }{ 2\pi \times 50 \times 44 \times 10 ^{ – 3 }}I=XLV​=2π×50×44×10–3220​ = 15.92 A

Therefore, the RMS value of the current in the network is 15.92 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.