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NCERT Solutions for Class 12 Physics Chapter 7 - Alternating Current

NCERT Solutions for Class 12 Physics Chapter 7 - Alternating Current

Edited By Vishal kumar | Updated on Apr 30, 2025 01:42 PM IST

Imagine a day without electricity...

No lights, no fans, no internet—even your phone battery would be useless! All of this is possible because of the Alternating Current (AC), which powers almost everything around us. In this chapter, you’ll learn how AC works, why it changes direction, and how it helps in sending electricity over long distances. It’s a super important and scoring Chapter that connects Physics to real life!

This Story also Contains
  1. NCERT Solutions for Class 12 Physics Chapter 7: Download PDF
  2. Class 12 Physics Alternating Current Chapter 7 - Exercise Solutions
  3. Additional Questions
  4. Class 12 physics NCERT Chapter 7: Higher Order Thinking Skills (HOTS) Questions
  5. Approach to Solve Questions of Alternating Current
  6. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  7. NCERT Solutions for Class 12 Physics: Chapter-Wise
NCERT Solutions for Class 12 Physics Chapter 7 - Alternating Current
NCERT Solutions for Class 12 Physics Chapter 7 - Alternating Current

Alternating Current is an important and scoring chapter in the Class 12 Physics syllabus. This NCERT solutions page gives you clear, step-by-step answers prepared by Physics experts from Careers360.

The electricity we use at home is alternating current (AC). But do you know the voltage and frequency of the current that comes to our homes? You’ll find answers to such questions in this chapter. The NCERT questions covered in Class 12 Physics Chapter 7 are based on single-phase AC, and these solutions will help you understand the concepts easily.

Background wave

NCERT Solutions for Class 12 Alternating Current are super helpful for understanding the basics needed for your CBSE board exam. By going through these solutions, you’ll get clear answers to all your questions about alternating current. In Chapter 7 of Class 12 Physics, you’ll also learn about phasor diagrams and how to solve problems related to them simply and easily.

Download the NCERT Class 12 Physics Chapter 7 (Alternating Current) Exercise Solutions PDF for free.

Perfect for quick revision and scoring well in the CBSE board exam! These expert-written solutions make tough concepts easy and help you practice important questions with confidence.

NCERT Solutions for Class 12 Physics Chapter 7: Download PDF

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Class 12 Physics Alternating Current Chapter 7 - Exercise Solutions

Q7.1 (a) A 100Ω resistor is connected to a 220V , 50Hz ac supply.

a)What is the RMS value of current?

Answer:

Given,

RMS voltage in the circuit Vrms=220V

Resistance in the circuit R=100Ω

Now,

RMS current in the circuit:

Irms=VrmsR=220100=2.2A

Hence, the RMS value of the current is 2.2A.

What is the net power consumed over a full cycle?

Answer:

Given,

Supplied RMS Voltage Vrms=220V

Supplied RMS Current Irms=2.2A

The net power consumed over a full cycle:

P=VrmsIrms=2202.2=484W

Hence net power consumed is 484W.

Q7.2 (a) The peak voltage of an ac supply is 300V . What is the RMS voltage?

Answer:

Given

Peak Value of ac supply:

Vpeak=300V

Now as we know in any sinusoidal function

RMSvalue=peakvalue2

Since our ac voltage supply is also sinusoidal

Vrms=Vpeak2=3002=212.13V

Hence RMS value of voltage os 212.13V.

Q7.2 (b) The RMS value of current in an ac circuit is 10A . What is the peak current?

Answer:

Given,

RMS value of current Irms=10A

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

Irms=Ipeak2

Ipeak=2Irms=210=14.1A

Hence the peak value of current is 14.1A.

Q7.3 A 44mH inductor is connected to 220V , 50Hz ac supply. Determine the RMS value of the current in the circuit.

Answer:

Given

Supply Voltage V=220V

Supply Frequency f=50Hz

The inductance of the inductor connected L=44mH=44103H

Now

Inductive Reactance

XL=ωL=2πfL=2π5044103

RMS Value of the current :

Irms=VrmsXL=2202π5044103=15.92A

Hence the RMS Value of current is 15.92A.

Q7.4 A 60μF capacitor is connected to a 100V , 60Hz ac supply. Determine the rms value of the current in the circuit.

Answer:

Given,

Supply Voltage V=110V

Supply Frequency f=60Hz

The capacitance of the connected capacitor C=60μF=60106F

Now,

Capacitive Reactance

XC=1ωC=12πfC=12π6060106

RMS Value of current

Irms=VrmsXC=VωC=V2πfC=1102π6060106=2.49A

Hence the RMS Value of current is 2.49A.

Q7.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.

Answer:

As we know,

Power absorbed P=VIcosϕ

Where ϕ is the phase difference between voltage and current.

ϕ for the inductive circuit is -90 degree and ϕ 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.

Answer:

Given

Capacitance C=30μF=30106

Inductance L=27mH=27103H

Now,

Angular Frequency

ωr=1LC=13010627103=1.11103rad/sec

Hence Angular Frequency is 1.11103rad/sec

Answer:

Given,

Resistance R=20Ω

Inductance L=1.5H

Capacitance C=35μF=35106F

Voltage supply V=200V

At resonance, the 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 cancel each other. in other words,

Z=(ωL1ωC)2+R2=02+R2=R=20Ω

As

ωL=1ωC

Now,

Current in the circuit

I=VZ=20020=10A

Average Power transferred in the circuit :

P=VI=20010=2000W

Hence average power transferred is 2000W.

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

1594271673048

Answer:

Given,

Variable frequency supply voltage V = 230V

Inductance L=5.0H

Capacitance C=80μF=80106F

Resistance R=40Ω

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

wresonance=1LC=1580106=100020=50rad/sec

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

Q7.8 (b) Figure shows a series LCR circuit connected to a variable frequency 230V source.

L=5.0H , C=80μF , R=40Ω .

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

1594271689346

Answer:

Given,

Variable frequency supply voltage V = 230V

Inductance L=5.0H

Capacitance C=80μF=80106F

Resistance R=40Ω

Now,

The impedance of the circuit is

Z=(wL1wC)2+R2

At Resonance Condition

wL=1wC

Z=R=40Ω

Hence, the Impedance at resonance is 40 Ω .

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

Current in the Resonance circuit is Given by

Iresonance=VpeakZ=223040=8.13A

Hence amplitude of the current at resonance is 8.13A.

Q7.8 (c)Figure shows a series LCR circuit connected to a variable frequency 230V source.

L=5.0H , C=80μF , R=40Ω .

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.

1594271702160

Answer:

Potential difference across any element = Irms(impedance)

Irms=Ipeak2=8.132=5.85A

Now

The potential difference across the capacitor:

Vcapacitor=Irms(1wresonanceC)=5.85(15080106)=1437.5V

The potential difference across the inductor

Vinductor=Irms(wresonanceL)=5.85505=1437.5V

The potential difference across Resistor

=40 Irms=230V

The potential difference across LC combination

VLC=Irms(wL1wC)=5.850=0

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

Additional Questions

Q1. (a) An LC circuit contains a 20mH inductor and a 50μF capacitor with an initial charge of 10mC . The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0 .

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

Answer :

Given,

The inductance of the inductor:

L=20mH=20103H

The capacitance of the capacitor :

C=50μF=50106F

The initial charge on the capacitor:

Q=10mC=10103C

Total energy present at the initial moment:

Einitial=Q22C=(10103)2250106=1J

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

Q1. (b) An LC circuit contains a 20mH inductor and a 50μF capacitor with an initial charge of 10mC . The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0 .

What is the natural frequency of the circuit?

Answer:

Given,

The inductance of the inductor:

L=20mH=20103H

The capacitance of the capacitor :

C=50μF=50106F

The initial charge on the capacitor:

Q=10mC=10103C

The natural angular frequency of the circuit:

wnatural=1LC=1(2010350106)=103rad/sec

Hence the natural angular frequency of the circuit is 103rad/sec .

The natural frequency of the circuit:

fnatural=wnatural2π=1032π=159Hz

Hence the natural frequency of the circuit is 159Hz.

Q1. (c-i) An LC circuit contains a 20mH inductor and a 50μF capacitor with an initial charge of 10mC . The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0 . At what time is the energy stored completely electrical (i.e., stored in the capacitor)?

Answer:

at any instant, the charge on the capacitor is:

Q=Q0cos(wnaturalt)=Q0cos(2πfnaturalt)=Q0cos(2πtT)

Where time period :

T=1fnatural=1159=6.28ms

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

Q=Q0

Q0=Q0cos(2πT)

cos(2πtT)=1

this is possible when

t=0,T2,T,3T2....

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

t=0,T2,T,3T2.... .

Q1.(c-ii)An LC circuit contains a 20mH inductor and a 50μF capacitor with an initial charge of 10mC . The resistance of the circuit is negligible.
Let the instant the circuit is closed be t=0 .
At what time is the energy stored completely magnetic (i.e., stored in the inductor)?

Answer:

The stored energy will be 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

t=T4,3T4,5T4..

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

Q1. (d) An LC circuit contains a 20mH inductor and a 50μF capacitor with an initial charge of 10mC . The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0 . 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.

Q22C=12Q022C

From here, we got

Q=Q02

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

Q02=Q0cos(2πtT)

12=cos(2πtT)

From here,

t=T8,3T8,5T8..

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

Q1. (e) An LC circuit contains a 20mH inductor and a 50μF capacitor with an initial charge of 10mC . The resistance of the circuit is negligible. Let the instant the circuit is closed be t=0 . 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.

Q2. (a) A coil of inductance 0.50H and resistance 100Ω is connected to a 240V , 50Hz ac supply. What is the maximum current in the coil?

Answer:

Given,

The inductance of the coil L=0.50H

The resistance of the coil R=100Ω

Supply voltage V=240V

Supply voltage frequency f=50Hz

Now, as we know peak voltage = 2 (RMS Voltage)

Peak voltage

Vpeak=2240=339.4V

The impedance of the circuit :

Z=R2+(wL)2=1002+(2π(.5)50)2

Now peak current in the circuit :

Ipeak=VpeakZ=3391002+(2π(.5)50)2=1.82A

Hence peak current is 1.82A in the circuit.

Q2. (b) A coil of inductance 0.50H and resistance 100Ω is connected to a 200V , 50Hz ac supply. What is the time lag between the voltage maximum and the current maximum?

Answer:

Let the voltage in the circuit be

V=V0coswt and

Current in the circuit be

I=I0cos(wtϕ)

Where ϕ is the phase difference between voltage and current.

V is maximum At

t = 0

I is maximum At

t=wϕ

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

wϕ .

For phase difference ϕ we have

tanϕ=wLR=2π500.5100=1.57

ϕ=57.50

t=ϕw=57.5π1802π50=3.2ms

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

Q3 Obtain the answers (a) to (b) ,if the circuit is connected to a high-frequency supply (240V,10kHz) . 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:

Given,

The inductance of the coil L=50H

The resistance of the coil R=100Ω

Supply voltage V=240V

Supply voltage frequency f=10kHz

a)

Now, as we know peak voltage = 2 (RMS Voltage)

Peak voltage Vpeak=2240=339.4V

Now,

The impedance of the circuit :

Z=R2+(wL)2=1002+(2π1010350)2

Now peak current in the circuit :

Ipeak=VpeakZ=3391002+(2π1010350)2=1.1102A

Hence peak current is 1.1102A 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 ϕ we have

tanϕ=wLR=2π101030.5100=100π

ϕ=89.820

Now

t=ϕw=89.82π1802π103=25μs

Hence time lag between the maximum voltage and the maximum current is 25μs.

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

Q4. (a) A 100μF capacitor in series with a 40Ω resistance is connected to a 110V , 60Hz supply.
What is the maximum current in the circuit?

Answer:

Given,

The capacitance of the capacitor C=100μF

The resistance of the circuit R=40Ω

Voltage supply V=100V

Frequency of voltage supply f=60Hz

The maximum current in the circuit

Imax=VmaxZ=2VR2+(1wC)2=2110402+(12π60100106)2=3.24A

Hence maximum current in the circuit is 3.24A.

Q4. (b) A 100μF capacitor in series with a 40Ω resistance is connected to a 110V , 60Hz supply. What is the time lag between the current maximum and the voltage maximum?

Answer:

In the case of a capacitor, we have

tanϕ=1wCR=1wCR

So,

tanϕ=1wCR=12π6010010640=0.6635

ϕ=33.560

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

t=ϕw=33.56π1802π60=1.55ms

Q5. Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110V , 10kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.

Answer:

Given,

The capacitance of the capacitor C=100μF

The resistance of the circuit R=40Ω

Voltage supply V=100V

Frequency of voltage supply f=12kHz

The maximum current in the circuit

Imax=VmaxZ=2VR2+(1wC)2=2110402+(12π12103100106)2=3.9A

Hence maximum current in the circuit is 3.9A.

b)

In the case of capacitor, we have

tanϕ=1wCR=1wCR

So,

tanϕ=1wCR=12π1010310010640=196π

ϕ=0.20

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

t=ϕw=0.2π1802π60=0.04μs

At high frequencies, ϕ 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.

Q6. Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L , C and R 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:

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

1Z=1R2+(wC1wL)2

I=VZ=V1R2+(wC1wL)2

The current will be minimal when

wC=1wL

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

Irms=VrmsR=23040=5.75A

RMS value in Inductor

Iinductor=VrmswL=230550=0.92A

RMS value in capacitor

Icapacitor=Vrms1/wL=2305080106=0.92A

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

Q7. (a) A circuit containing a 80mH inductor and a 60μF capacitor in series is connected to a 230V , 50Hz supply. The resistance of the circuit is negligible. Obtain the current amplitude and rms values.

Answer:

The inductance of the inductor L=80mH=80103H

The capacitance of the capacitor C=60μF

Voltage supply V=230V

Frequency of voltage supply f=50Hz .

Here, we have

V=Vmaxsinwt=Vmaxsin2πft

Impedance

Z=R2+(wL1wC)2

Z=02+(2π508010312π5060106)2=8π10006π

Now,

Current in the circuit will be

I=VZ=VmaxsinwtZϕ=Imaxsin(wtϕ)

where,

Imax=VmaxZ=22308π10006π=11.63A

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

I=11.63sin(wtϕ)

here

tanϕ=wL1wCR

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

ϕ=900

Hence

I=11.63sin(wtπ2)

Now,

RMS value of this current:

Irms=Imax2=11.632=8.22A .

Q7. (b) A circuit containing a 80mH inductor and a 60μF capacitor in series is connected to a 230V , 50Hz supply. The resistance of the circuit is negligible. Obtain the rms values of potential drops across each element.

Answer:

As we know,

RMS potential drop across an element with impedance Z:

Velement=IrmsZelement

SO,

RMS potential difference across inductor:

Vinductor=IrmswL=8.222π6080103=206.61V

RMS potential drop across capacitor

Vcapacitor=Irms1wC=8.2212π6060106=436.3V

Q7. (c) A circuit containing a 80mH inductor and a 60μF capacitor in series is connected to a 230V , 50Hz supply. The resistance of the circuit is negligible

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

Answer:

Since

I=Imaxsin(wtϕ)

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.

Q7. (d) A circuit containing a 80mH inductor and a 60μF capacitor in series is connected to a 230V , 50Hz supply. The resistance of the circuit is negligible . What is the average power transferred to the capacitor?

Answer:

As we know,

Average power P=VIcosθ where θ is the phase difference between voltage and current.

Since in the circuit, phase difference θ is π/2 , the average power is zero.

Q7. (e) A circuit containing a 80mH inductor and a 60μF capacitor in series is connected to a 230V , 50Hz supply. The resistance of the circuit is negligible . What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.]

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.

Q8. Suppose the circuit in Exercise 7.18 has a resistance of 15Ω . Obtain the average power transferred to each element of the circuit, and the total power absorbed.

Answer:

The inductance of the inductor L=80mH=80103H

The capacitance of the capacitor C=60μF

The resistance of a 15Ω resistor

Voltage supply V=230V

Frequency of voltage supply f=50Hz

As we know,

Impedance

Z=R2+(wL1wC)2

Z=152+(2π508010312π5060106)2=31.728

Current flowing in the circuit :

I=VZ=23031.72=7.25A

Now,

Average power transferred to the resistor:

Presistor=I2R=(7.25)215=788.44W

Average power transferred to the inductor = 0

Average power transferred to the capacitor = 0:

Total power absorbed by circuit :

Presistor+pinductor+Pcapacitor=788.44+0+0=788.44W

Hence circuit absorbs 788.44W.

Q9. (a) A series LCR circuit with L=0.12H , C=480nF , R=23Ω is connected to a 230V variable frequency supply.

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

Answer:

The inductance of the inductor L=0.12H

The capacitance of the capacitor C=480nF

The resistance of the resistor R=23Ω

Voltage supply V=230V

Frequency of voltage supply f=50Hz

As we know,

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

wnatural=1LC=10.12480109)=4166.67rad/sec

Also, at this frequency,

Z=R=23

SO,

The maximum current in the circuit :

Imax=VmaxZ=VmaxR=223023=14.14A

Hence maximum current is 14.14A.

Q9. (b) A series LCR circuit with L=0.12H , C=480nF , R=23Ω is connected to a 230V variable frequency supply.

What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.

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

f=wr2π=4166.672π=663.48Hz

Maximum Power Absorbed

P=I2R=(14.14)223=2299.3W .

Q9. (c) A series LCR circuit with L=0.12H , C=480nF , R=23Ω is connected to a 230V variable frequency supply . What is the Q factor of the given circuit?

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

Q=wrLR=4166.670.1223=21.74

Hence Q-factor for the circuit is 21.74.

Q9. (d) A series LCR circuit with L=0.12H , C=480nF , R=23Ω is connected to a 230V variable frequency supply. 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?

Answer:

As

Power P=I2R

Power P will be half when the current I is 1/2 times the maximum current.

As,

I=ImaxSin(wtϕ)

At half powerpoint :

imax2=ImaxSin(wtϕ)

12=Sin(wtϕ)

wt=ϕ+π4

here,

ϕ=tan1(wL1wCR)

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

wt=ϕ+π4

And they are:

w1=678.75Hz

w2=648.22Hz

Also,

The current amplitude at these frequencies

Ihalfpowerpoint=Imax2=14.141.414=10A

Q10. Obtain the resonant frequency and ϱ -factor of a series LCR circuit with L=0.3H , C=27μF , and R=7.4Ω . 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 L=0.3H

The capacitance of the capacitor C=27μF

The resistance of the resistor R=7.4Ω

Now,

Resonant frequency

wr=1LC=10.327106=111.11rad/sec

Q-Factor of the circuit

Q=wrLR=111.110.37.4=45.0446

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

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

Rnew=R2=7.42=3.7Ω

Q11. (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?

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.

Q11.(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.

Q11. (c) Answer the following questions:

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 C and the ac signal across L .

Answer:

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

XL=wL=LargevalueAndXC=1wC=Verysmall

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:

XL=wL=VerysmallAndXC=1wC=Largevalue

Hence DC signal appears across Capacitor only.

Q11.(d) Answer the following questions:

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.

Answer:

For a steady state DC, the increasing inductance value by inserting iron core in the choke, has no effect on 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.

Q11. (e) Answer the following questions:

Why is a 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:

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.

Q12. A power transmission line feeds input power at 2300V to a stepdown transformer with its primary windings having 40000 turns. What should be the number of turns in the secondary in order to get output power at 230V ?

Answer:

Given,

Input voltage:

Vinput=2300V

Number of turns in the primary coil

Nprimary=4000

Output voltage:

Voutput=230V

Now,

Let the number of turns in secondary be

N=Nsecondary

Now as we know, in a transformer,

VprimaryVsecondary=NprimaryNsecondary

Nsecondary=VsecondaryVprimaryNprimary=23023004000=400

Hence the number of turns in secondary winding is 400.

Q13. At a hydroelectric power plant, the water pressure head is at a height of 300m and the water flow available is 100m3s1 . If the turbine generator efficiency is 600/0 , estimate the electric power availablefrom the plant (g=9.8ms2) .

Answer:

Given,

Height of the water pressure head

h=300m

The volume of the water flow per second

V=100m3s1

Turbine generator efficiency

η=0.6

Mass of water flowing per second

M=100103=105kg

The potential energy stored in the fall for 1 second

P=Mgh=1059.8300=294106J

Hence input power

Pinput=294106J/s

Now as we know,

η=PoutputPinput

Poutput=ηPinput=0.6294106=176.4106W

Hence output power is 176.4 MW.

Q14. (a) A small town with a demand of 800kW of electric power at 220V is situated 15km away from an electric plant generating power at 440V . The resistance of the two wire line carrying power is 0.5Ω per km. The town gets power from the line through a 4000220V step-down transformer at a sub-station in the town. Estimate the line power loss in the form of heat.

Answer:

Power required

P=800kW=800103kW

The total resistance of the two-wire line

R=2150.5=15Ω

Input Voltage

Vinput=4000V

Output Voltage:

Voutput=220V

RMS Current in the wireline

I=PVinput=8001034000=200A

Now,

Power loss in the line

Ploss=I2R=200215=600103=600kW

Hence, power loss in line is 600kW.

Q14. (b) A small town with a demand of 800kW of electric power at 220V is situated 15km away from an electric plant generating power at 440V . The resistance of the two wire line carrying power is 0.5Ω per km. The town gets power from the line through a 4000220V step-down transformer at a sub-station in the town.

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

Answer:

Power required

P=800kW=800103kW

The total resistance of the two-wire line

R=2150.5=15Ω

Input voltage

Vinput=4000V

Output voltage:

Voutput=220V

RMS current in the wireline

I=PVinput=8001034000=200A

Now,

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

Q14.(c) A small town with a demand of 800kW of electric power at 220V is situated 15km away from an electric plant generating power at 440V . The resistance of the two wire line carrying power is 0.5Ω per km. The town gets power from the line through a 4000220V step-down transformer at a sub-station in the town. Characterise the step up transformer at the plant.

Answer:

Power required

P=800kW=800103kW

The total resistance of the two-wire line

R=2150.5=15Ω

Input Voltage

Vinput=4000V

Output Voltage:

Voutput=220V

RMS Current in the wireline

I=PVinput=8001034000=200A

Now,

Voltage drop in the power line = IR=20015=3000V

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

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

Q15. Do the same exercise as above with the replacement of the earlier transformer by a 40,000220V 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

P=800kW=800103kW

The total resistance of the two-wire line

R=2150.5=15Ω

Input Voltage

Vinput=40000V

Output Voltage:

Voutput=220V

RMS current in the wireline

I=PVinput=80010340000=20A

Now,

a) power loss in the line

Ploss=I2R=20215=6kW

b)

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

c)

Voltage drop in the power line = IR=2015=300V

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 less than low voltage transmission.

Class 12 physics NCERT Chapter 7: Higher Order Thinking Skills (HOTS) Questions

Q1:

The power factor of R-L circuit is 13 If the inductive reactance is 2Ω. The value of resistance is

Answer:

Power factor -

cosϕ=RZR resistance Z impedance 


And For RL circuit

Z=R2+(XL)2

where XL= inductive reactance
So
Power factor =cosϕ=RZ

13=RR2+22

on Solving we get

R=2Ω


Q2:

In the above circuit, C=32μF,R2=20Ω,L=310H and R1=10Ω. Current in L-R R1 path is I1 and in C-R 2 path it is I2. The voltage of A.C source is given by, V=2002sin(100t) volts. The phase difference between I1 and I2 is : (in degrees)

Answer:

Phase difference -

tanϕ=XLR=tan1(XLR)ϕ=tan1(ωLR)

- wherein

XL= inductive reactance 
R = resistance

Phase difference -

tanϕ=XcRϕ=tan1(XcR)=tan1(1ωcR)

For current I1

ϕ1=tan1ωLR1=tan1(10031010)=60 (lagging) 
For I2 current

ϕ2=tan1(1ωCR2)=tan1(106100×32×20)90 (leading)  Phase difference =90+60=150


Q3:

An inductance of (200π)mH is connected with an AC source 220 V,50 Hz. The inductive susceptance of the circuit is:

Answer:

As we learn
Inductive susceptance

SL=1XL=12πνL

XL (inductive reactance)

XL=2πfL=2π×50[200π×103]=20Ω

Inductive susceptance =1XL=120=0.05


Q4:

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

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 rms potential drop across R is given by

Vrms=IrmsR=5.75×40=230 V

The rms potential drop across L is given by

VLrms=IrmsXL=Irms(ω0L)=5.75×50×5=1437.5 V

The rms potential drop across C is given by

Vrms=IrmsXC=Irms×1ω0C=5.75×150×80×106=1437.5 V

The rms potential drop across L-C is given by

VLC=VLVC=1437.51437.5=0


Q5:

In the LCR circuit current resonant frequency is 600Hz, and half power points are at 650 and 550 Hz. The quality factor is-

Answer:

In a series LCR circuit, the quality factor is a measure of the sharpness of the resonance and is defined as the ratio of the resonant frequency (f0) to the bandwidth (Δf). The bandwidth is the difference between the half-power frequencies ( f1 and f2 ), which are the frequencies at which the power drops to half of its maximum value. Mathematically, the Q-factor is given by:

Q=f0Δf=f0f2f1

Given the following values:
Resonant frequency (f0)=600 Hz
Lower half-power frequency (f1)=550 Hz
Upper half-power frequency (f2)=650 Hz

The bandwidth ( Δf ) is:

Δf=f2f1=650 Hz550 Hz=100 Hz

Now, calculating the Q-factor:

Q=f0Δf=600 Hz100 Hz=6

Therefore, the Q-factor of the circuit is 6.


Approach to Solve Questions of Alternating Current

  • Understand basic terminology –

  1. Alternating current (AC): Current that reverses direction periodically

  2. Peak value, RMS value, frequency, and time period

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  • Get basic equations –

  1. Instantaneous current: I=I0sin(ωt)

  2. RMS current: Irms =I02

  3. RMS voltage: Vrms=V02

  • Know AC vs DC differences – Direction, value over time, frequency, and use

  • Understand impedance in AC circuits –

Z=R2+(XLXC)2XL=ωL,XC=1ωC

  • Apply power formulas in AC –

  1. Average power: P=VrmsIrmscosϕ
  2. Power factor: cosϕ
  • Phase difference information –
  1. Resistance → Current and voltage in phase
  2. Inductor → Current lags voltage
  3. Capacitor → Current leads voltage
  • Practice LCR circuit problems – Apply phasor diagrams and Ohm's law for AC: V=IZ
  • Understand resonance in AC –  Resonance in LCR: XL=XC, and Z=R (minimum impedance, maximum current) 
  • Remember unit conversions – Frequency: Hz, Time: seconds, Voltage/Current in volts/amperes
  • Solve numerals step-by-step – Put down given values clearly, choose suitable formula, and use units correctly.

What Extra Should Students Study Beyond NCERT for JEE/NEET?

NCERT Solutions for Class 12 Physics: Chapter-Wise

Here are the exercise-wise solutions of the NCERT Class 12 physics book:

Also, check NCERT Books and NCERT Syllabus here:

NCERT Solutions Subject-Wise

NCERT Exemplar Class 12 Solutions

Frequently Asked Questions (FAQs)

1. What type of questions are asked in board exams from the chapter alternating current?

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, NCERT exemplar questions and previous year papers.

2. What is the weightage of chapter for JEE Main?

One question for JEE main can be expected from the Class 12 chapter Alternating Current.

3. How important is the chapter for NEET?

One or two questions may be asked from NCERT chapter Alternating Current for NEET exam.

4. What is the frequency of AC in our homes?

In India, the AC supply has a frequency of 50 Hz, which means it changes direction 50 times in one second.

5. What is a phasor diagram?

A phasor diagram is a visual way to show the phase difference between voltage and current in an AC circuit using rotating vectors.

6. Why is AC preferred over DC for power transmission?

AC is easier to transmit over long distances and can be easily stepped up or down using transformers.

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0.34\; J

Option 2)

0.16\; J

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1.00\; J

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0.67\; J

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2.45×10−3 kg

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 6.45×10−3 kg

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 9.89×10−3 kg

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12.89×10−3 kg

 

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2,000 \; J - 5,000\; J

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200 \, \, J - 500 \, \, J

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2\times 10^{5}J-3\times 10^{5}J

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20,000 \, \, J - 50,000 \, \, J

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K/2\,

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\; K\;

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zero\;

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2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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0.02

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3.125 × 10-2

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