# NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current

** NCERT solutions for class 12 physics chapter 7 Alternating current- ** The supply that is received in our home is alternating in nature. Do you know what is the value of normal supply voltage that is coming in our home and what is the supply frequency? Solutions of NCERT class 12 physics chapter 7 alternating current will help you to clear such doubts. The alternating current in our daily application can be of a single phase that is utilized in our home or can be of three phases which are usually used in industries. Questions explained in the CBSE NCERT solutions for class 12 physics chapter 7 alternating current is based on single-phase alternating current. NCERT solutions are the basics for the CBSE board exam. In this chapter, you will study the concept of phasor diagrams. The knowledge of the mathematics chapter complex numbers will help you to understand the phasors and the problems in NCERT solutions for class 12 physics chapter 7 alternating current.

To solve the circuited related problem of NCERT class 12 chapter 7 alternating current you must refresh the Kirchhoff's voltage law and Kirchoff's current law that you have studied in current electricity chapter. The last topic of class 12 physics chapter 7 alternating current gives an introduction to single-phase transformers.

## **NCERT solutions for class 12 physics chapter 7 alternating current exercises: **

** Q7.1 (a) ** A resistor is connected to a , ac supply.

a)what is the RMS value of current?

** Answer: **

Given,

RMS voltage in the circuit

Resistance in the circuit

Now,

RMS current in the circuit:

Hence, the RMS value of current is 2.2A.

** Q7.1 (b) ** A resistor is connected to a , ac supply.

What is the rms value of current in the circuit?

** Answer: **

Given,

RMS Voltage in the circuit

Resistance in the circuit

Now,

RMS Current in the circuit:

Hence, the RMS value of current is 2.2A.

** Q7.1 (c) ** A resistor is connected to a , ac supply.

What is the net power consumed over a full cycle?

** Answer: **

Given,

Supplied RMS Voltage

Supplied RMS Current

The net power consumed over a full cycle:

Hence net power consumed is 484W.

** Q7.2 (a) ** The peak voltage of an ac supply is . What is the RMS voltage?

** Answer: **

Given

Peak Value of ac supply:

Now as we know in any sinusoidal function

Since our ac voltage supply is also sinusoidal

Hence RMS value of voltage os 212.13V.

** Q7.2 (b) ** The RMS value of current in an ac circuit is . What is the peak current?

** Answer: **

Given,

RMS value of current

Since Current is also sinusoidal (because only resistance is present in the circuit, not the capacitor and inductor)

Hence the peak value of current is 14.1A.

** Q7.3 ** A inductor is connected to , ac supply. Determine the RMS value of the current in the circuit.

** Answer: **

Given

Supply Voltage

Supply Frequency

The inductance of the inductor connected

Now

Inductive Reactance

RMS Value of the current :

Hence the RMS Value of current is 15.92A.

** Q7.4 ** A capacitor is connected to a , ac supply. Determine the rms value of the current in the circuit.

** Answer: **

Given,

Supply Voltage

Supply Frequency

The capacitance of the connected capacitor

Now,

Capacitive Reactance

RMS Value of current

Hence the RMS Value of current is 2.49A.

** Answer: **

As we know,

Power absorbed

Where is the phase difference between voltage and current.

for the inductive circuit is -90 degree and for the capacitive circuit is +90 degree.

In both cases (inductive and capacitive), the power absorbed by the circuit is zero because in both cases the phase difference between current and voltage is 90 degree.

This can be seen as The elements(Inductor and Capacitor) are not absorbing the power, rather storing it. The capacitor is storing energy in electrostatic form and Inductor is storing the energy in magnetic form.

** Q7.6 ** Obtain the resonant frequency of a series circuit with , and . What is the Q -value of this circuit?

** Answer: **

Given, in a circuit,

Inductance,

Capacitance,

Resistance,

Now,

Resonance frequency (frequency of maximum current OR minimum impedance OR frequency at which inductive reactance cancels out capacitive reactance )

Hence Resonance frequency is 125 per second.

Q-Value:

Hence Q - value of the circuit is 25.

** Answer: **

Given

Capacitance

Inductance

Now,

Angular Frequency

Hence Angular Frequency is

** Answer: **

Given

Capacitance

Inductance

Charge on the capacitor

Now,

The total energy stored in Capacitor :

Total energy later will be same because energy is being shared with capacitor and inductor and none of them loses the energy, they just store it and transfer it.

** Answer: **

Given,

Resistance

Inductance

Capacitance

Voltage supply

At resonance, supply frequency is equal to the natural frequency, and at the natural frequency, the total impedance of the circuit is equal to the resistance of the circuit

as inductive and capacitive reactance cancels each other. in other words,

As

Now,

Current in the circuit

Average Power transferred in the circuit :

Hence average power transferred is 2000W.

** Answer: **

Given,

Range of the frequency in which radio can be tune =

The effective inductance of the Circuit =

Now, As we know,

where is tuning frequency.

For getting the range of the value of a capacitor, let's calculate the two values of the capacitor, one maximum, and one minimum.

first, let's calculate the minimum value of capacitance which is the case when tuning frequency = 800KHz.

Hence the minimum value of capacitance is 198pF.

Now, Let's calculate the maximum value of the capacitor.

in this case, tuning frequency = 1200KHz

Hence the maximum value of the capacitor is 88.04pf

Hence the Range of the values of the capacitor is .

** Q7.11 (a) ** Figure shows a series LCR circuit connected to a variable frequency source. , , .

** (a) ** Determine the source frequency which drives the circuit in resonance.

** Answer: **

Given,

Variable frequency supply voltage = 230V

Inductance

Capacitance

Resistance

a) Resonance angular frequency in this circuit is given by :

Hence this circuit will be in resonance when supply frequency is 50 rad/sec.

** Q7.11 (b) ** Figure shows a series LCR circuit connected to a variable frequency source.

(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

** Answer: **

Given,

Variable frequency supply voltage = 230V

Inductance

Capacitance

Resistance

Now,

The impedance of the circuit is

at Resonance Condition

:

Hence, Impedance at resonance is 40 .

Now, at resonance condition, impedance is minimum which means current is maximum which will happen when we have a peak voltage, so

Current in the Resonance circuit is Given by

Hence amplitude of the current at resonance is 8.13A.

** Q7.11 ** Figure shows a series LCR circuit connected to a variable frequency source.

** Answer: **

Potential difference across any element =

Now

The potential difference across the capacitor:

The potential difference across the inductor

The potential difference across Resistor

Potential difference across LC combination

Hence at resonating, frequency potential difference across LC combination is zero.

**NCERT solutions for class 12 physics chapter 7 alternating current additional exercises: **

What is the total energy stored initially? Is it conserved during oscillations?

** Answer ** :

Given,

The inductance of the inductor:

The capacitance of the capacitor :

The initial charge on the capacitor:

Total energy present at the initial moment:

Hence initial energy in the circuit is 1J. Since we don't have any power-consuming element like resistance in the circuit, the energy will be conserved

What is the natural frequency of the circuit?

** Answer: **

Given,

The inductance of the inductor:

The capacitance of the capacitor :

The initial charge on the capacitor:

The natural angular frequency of the circuit:

Hence the natural angular frequency of the circuit is .

The natural frequency of the circuit:

Hence the natural frequency of the circuit is 159Hz.

** Q7.12 (c-i) ** An circuit contains a inductor and a capacitor with an initial charge of . The resistance of the circuit is negligible. Let the instant the circuit is closed be .

** ( c) ** At what time is the energy stored

** (i) ** completely electrical (i.e., stored in the capacitor)?

** Answer: **

at any instant, the charge on the capacitor is:

Where time period :

Now, when the total energy is purely electrical, we can say that

this is possible when

Hence Total energy will be purely electrical(stored in a capacitor) at

.

** Q7.12 (c-ii) ** An circuit contains a inductor and a capacitor with an initial charge of . The resistance of the circuit is negligible.

(C) Let the instant the circuit is closed be .

(ii) completely magnetic (i.e., stored in the inductor)?

** Answer: **

The stored energy will we purely magnetic when the pure electrical stored is zero. i.e. when the charge on the capacitor is zero, all energy will be stored in the inductor.

So, t for which charge on the capacitor is zero is

Hence at these times, the total energy will be purely magnetic.

** Q7.12 (d) ** An circuit contains a inductor and a capacitor with an initial charge of . The resistance of the circuit is negligible. Let the instant the circuit is closed be .

At what times is the total energy shared equally between the inductor and the capacitor?

** Answer: **

The energy will be shared equally when the energy in the capacitor is half of the maximum energy it can store.i.e.

From here, we got

So now, we know the charge on the capacitor, we can calculate the time for which

From here,

Hence for these times, the total energy will be shared equally between capacitor and inductor.

If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?

** Answer: **

If the resistance is added to the circuit, the whole energy will dissipate as heat eventually. energy will keep moving between the capacitor and inductor with reducing in magnitude in each cycle and eventually all energy will be dissipated.

** Answer: **

Given,

The inductance of the coil

the resistance of the coil

Supply voltage

Supply voltage frequency

Now, as we know peak voltage = (RMS Voltage)

Peak voltage

The impedance of the circuit :

Now peak current in the circuit :

Hence peak current is 1.82A in the circuit.

** Answer: **

Let the voltage in the circuit be

and

Current in the circuit be

Where is the phase difference between voltage and current.

V is maximum At

t = 0

is maximum At

Hence, the time lag between voltage maximum and the current maximum is

.

For phase difference we have

Hence time lag between the maximum voltage and the maximum current is

** Answer: **

Given,

The inductance of the coil

the resistance of the coil

Supply voltage

Supply voltage frequency

a)

Now, as we know peak voltage = (RMS Voltage)

Peak voltage

Now,

The impedance of the circuit :

Now peak current in the circuit :

Hence peak current is in the circuit.

The current in the circuit is very small, which is one of the indications of inductor working as a nearly open circuit in the case of high frequency.

b)

For phase difference we have

Now

Hence time lag between the maximum voltage and the maximum current is .

In the DC circuit, after attaining the steady state, inductor behaves line short circuit as is Zero.

** Answer: **

Given,

The capacitance of the capacitor

The resistance of the circuit

Voltage supply

Frequency of voltage supply

The maximum current in the circuit

Hence maximum current in the circuit is 3.24A.

** Answer: **

In the case of a capacitor, we have

So,

So the time lag between max voltage and the max current is :

** Answer: **

Given,

The capacitance of the capacitor

The resistance of the circuit

Voltage supply

Frequency of voltage supply

The maximum current in the circuit

Hence maximum current in the circuit is 3.9A.

b)

In the case of capacitor, we have

So,

So the time lag between max voltage and max current is :

At high frequencies, tends to zero. which indicates capacitor acts as a conductor at high frequencies.

In the DC circuit, after a steady state is achieved, Capacitor acts like an open circuit.

** Answer: **

As we know, in the case of a parallel RLC circuit:

The current will be minimum when

Which is also the condition of natural frequency. Hence the total current is minimum when source frequency is equal to the natural frequency.

RMS value of current in R

RMS value in Inductor

RMS value in capacitor

Capacitor current and inductor current will cancel out each other so the current flowing in the circuit is 5.75A.

** Answer: **

The inductance of the inductor

The capacitance of the capacitor

Voltage supply

Frequency of voltage supply .

Here, we have

Impedance

Now,

Current in the circuit will be

where,

The negative sign is just a matter of the direction of current.so,

here

But, since the value of R is zero(since our circuit have only L and C)

Hence

Now,

RMS value of this current:

.

** Answer: **

As we know,

RMS potential drop across an element with impedance Z:

SO,

RMS potential difference across inductor:

RMS potential drop across capacitor

** Q7.18 (c) ** A circuit containing a inductor and a capacitor in series is connected to a , supply. The resistance of the circuit is negligible

(c) What is the average power transferred to the inductor?

** Answer: **

Since

Current flowing in the circuit is sinusoidal and hence average power will be zero as the average of sin function is zero.in other words, the inductor will store energy in the positive half cycle of the sin (0 degrees to 180 degrees) and will release that energy in the negative half cycle(180 degrees to 360 degrees), and hence average power is zero.

** Answer: **

As we know,

Average power where is the phase difference between voltage and current.

Since in the circuit, phase difference is , the average power is zero.

** Answer: **

Since the phase difference between voltage and current is 90 degree, even the total power absorbed by the circuit is zero. This is an ideal circuit, we can not have any circuit in practical that consumes no power, that is because practically resistance of any circuit is never zero. Here only inductor and capacitor are present and none of them consumes energy, they just store it and transfer it like they are doing it in this case.

** Q7.19 ** Suppose the circuit in Exercise 7.18 has a resistance of . Obtain the average power transferred to each element of the circuit, and the total power absorbed.

** Answer: **

The inductance of the inductor

The capacitance of the capacitor

The resistance of a resistor

Voltage supply

Frequency of voltage supply

As we know,

Impedance

Current flowing in the circuit :

Now,

Average power transferred to the resistor:

Average power transferred to the inductor = 0

Average power transferred to the capacitor = 0:

Total power absorbed by circuit :

Hence circuit absorbs 788.44W.

** Q7.20 (a) ** A series LCR circuit with , , is connected to a variable frequency supply.

What is the source frequency for which current amplitude is maximum. Obtain this maximum value.

** Answer: **

The inductance of the inductor

The capacitance of the capacitor

The resistance of the resistor

Voltage supply

Frequency of voltage supply

As we know,

the current amplitude is maximum at the natural frequency of oscillation, which is

Also, at this frequency,

SO,

The maximum current in the circuit :

Hence maximum current is 14.14A.

** Q7.20 (b) ** A series LCR circuit with , , is connected to a variable frequency supply.

** Answer: **

Since the resistor is the only element in the circuit which consumes the power, the maximum absorbed power by circuit will be maximum when power absorbed by the resistor will be maximum. power absorbed by the resistor will be maximum at when current is maximum which is the natural frequency case,

Hence when source frequency will be equal to the natural frequency, the power absorbed will be maximum.

Hence frequency

Maximum Power Absorbed

.

** Answer: **

The value of maximum angular frequency is calculated in the first part of the question and whose magnitude is 4166.67

Q-factor of any circuit is given by

Hence Q-factor for the circuit is 21.74.

** Answer: **

As

Power

Power will be half when the current is times the maximum current.

As,

At half powerpoint :

here,

On putting values, we get, two values of for which

And they are:

Also,

The current amplitude at these frequencies

** Q7.21 ** Obtain the resonant frequency and -factor of a series circuit with , , and . 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: **

The inductance of the inductor

The capacitance of the capacitor

The resistance of the resistor

Now,

Resonant frequency

Q-Factor of the circuit

Now, to improve the sharpness of resonance by reducing its full width at half maximum, by a factor of 2 without changing ,

we have to change the resistance of the resistor to half of its value, that is

** Answer: **

Yes, at any instant the applied voltage will be distributed among all element and the sum of the instantaneous voltage of all elements will be equal to the applied. But this is not the case in RMS because all elements are varying differently and they may not be in the phase.

** 7.22 (b) ** Answer the following questions: A capacitor is used in the primary circuit of an induction coil.

** Answer: **

Yes, we use capacitors in the primary circuit of an induction coil to avoid sparking. when the circuit breaks, a large emf is induced and the capacitor gets charged from this avoiding the case of sparking and short circuit.

** Q 7.22 (c) ** Answer the following questions:

** Answer: **

For a high frequency, the inductive reactance and capacitive reactance:

Hence the capacitor does not offer resistance to a higher frequency, so the ac voltage appears across L.

Similarly

For DC, the inductive reactance and capacitive reactance:

Hence DC signal appears across Capacitor only.

** Q 7.22 (d) ** Answer the following questions:

** Answer: **

For a steady state DC, the increasing inductance value by inserting iron core in the choke, have no effect in the brightness of the connected lamp, whereas, for ac when the iron core is inserted, the light of the lamp will shine less brightly because of increase in inductive impedance.

** Q7.22 (e) ** Answer the following questions:

** Answer: **

We need choke coil in the use of fluorescent tubes with ac mains to reduce the voltage across the tube without wasting much power. If we use simply resistor for this purpose, there will be more power loss, hence we do not prefer it.

** Answer: **

Given,

Input voltage:

Number of turns in the primary coil

Output voltage:

Now,

Let the number of turns in secondary be

Now as we know, in a transformer,

Hence the number of turns in secondary winding id 400.

** Q7.24 ** At a hydroelectric power plant, the water pressure head is at a height of and the water flow available is . If the turbine generator efficiency is , estimate the electric power availablefrom the plant .

** Answer: **

Given,

Height of the water pressure head

The volume of the water flow per second

Turbine generator efficiency

Mass of water flowing per second

The potential energy stored in the fall for 1 second

Hence input power

Now as we know,

Hence output power is 176.4 MW.

** Answer: **

Power required

The total resistance of the two-wire line

Input Voltage

Output Voltage:

RMS Current in the wireline

Now,

Power loss in the line

Hence, power loss in line is 600kW.

How much power must the plant supply, assuming there is negligible power loss due to leakage?

** Answer: **

Power required

The total resistance of the two-wire line

Input voltage

Output voltage:

RMS current in the wireline

Now,

Total power delivered by plant = line power loss + required electric power = 800 + 600 = 1400kW.

** Answer: **

Power required

The total resistance of the two-wire line

Input Voltage

Output Voltage:

RMS Current in the wireline

Now,

Voltage drop in the power line =

Total voltage transmitted from the plant = 3000+4000=7000

as power is generated at 440V, The rating of the power plant is 440V-7000V.

** Q7.26 ** Do the same exercise as above with the replacement of the earlier transformer by a 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: **

Power required

The total resistance of the two-wire line

Input Voltage

Output Voltage:

RMS current in the wireline

Now,

a) power loss in the line

b)

Power supplied by plant = 800 kW + 6 kW = 806kW.

c)

Voltage drop in the power line =

Total voltage transmitted from the plant = 300+40000=40300

as power is generated at 440V, The rating of the power plant is 440V-40300V.

We prefer high voltage transmission because power loss is a lot lesser than low voltage transmission.

## ** NCERT solutions for class 12 physics chapter wise **

** NCERT solutions subject wise **

## ** Significance of ****NCERT solutions for class 12 physics chapter 7 alternating current in board exam ** :

From the unit electromagnetic induction and alternating current around 10% questions are asked in the CBSE board exam. If all the solutions of NCERT class 12 physics chapter 7 alternating current are covered then it is easy to answer questions asked in board exam related to the chapter. For JEE Main, NEET and other state board and competitive exams also this chapter is important. Prepare the chapter well with the help of CBSE NCERT solutions for class 12 physics chapter 7 alternating current.

## Frequently Asked Question (FAQs) - NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current

**Question: **Is the supply coming to our home alternating?

**Answer: **

Yes, the supply received in our home is alternating. Usually, domestic supplies are single-phase and supply to industries, factories etc are three-phase.

**Question: **What type of questions are asked in board exam from the chapter alternating current.

**Answer: **

In the previous year question papers, questions related to identifying the circuit elements are repeatedly asked. Along with this simple numerical questions are also asked from alternating current. For practising numerical solve NCERT exercise questions and previous year papers.

**Question: **What is the weightage of chapter for JEE Main?

**Answer: **

One question for JEE main can be expected from the class 12 chapter alternating current

**Question: **How important is the chapter for NEET?

**Answer: **

one or two questions may be asked from NCERT chapter alternating current for NEET exam