NCERT Exemplar Class 12 Physics Solutions Chapter 7 Alternating Current

NCERT Exemplar Class 12 Physics Solutions Chapter 7 Alternating Current

Vishal kumarUpdated on 12 Apr 2025, 01:05 AM IST

NCERT Exemplar Class 12 Physics Solutions Chapter 7 Alternating Current helps the students to understand how AC circuits work in real life. This chapter explains concepts like AC voltage, different types of reactance, impedance, resonance, different types of circuits and power in AC circuits. You will also learn about devices like transformers and AC generators, which are widely used in electricity distribution. With detailed solutions, students can easily grasp important formulas and problem-solving techniques.

This Story also Contains

  1. NCERT Exemplar Class 12 Physics Solutions Chapter 7: MCQI
  2. NCERT Exemplar Class 12 Physics Solutions Chapter 7 MCQII
  3. NCERT Exemplar Class 12 Physics Solutions Chapter 7: Very Short Answer
  4. NCERT Exemplar Class 12 Physics Solutions Chapter 7: Short Answer
  5. NCERT Exemplar Class 12 Physics Solutions Chapter 7: Long Answer
  6. NCERT Exemplar Class 12 Physics Solutions Chapter 7 Main subtopics
  7. What will the students learn with NCERT Exemplar Class 12 Physics Solutions Chapter 7?
  8. NCERT Exemplar Class 12 Physics Chapter 7 Alternating Current: Important Concepts & Formulas
  9. NCERT Exemplar Class 12 Physics Solutions Chapter-Wise
  10. NCERT Exemplar Class 12 Solutions
NCERT Exemplar Class 12 Physics Solutions Chapter 7 Alternating Current
NCERT Exemplar Class 12 Physics Solutions Chapter 7 Alternating Current

NCERT Exemplar Class 12 Physics Solutions Chapter 7 Alternating Current explains how AC voltage drives current and its importance in electrical devices. It covers concepts like power in ac circuit, nature of the circuit and the phasor diagram. These solutions help the students to learn the concept of complex problems and improve their understanding. Mastering this chapter is very important for scoring well in board exams and competitive tests.

Also Read

Also see - NCERT Solutions for Class 12 Other Subjects

NCERT Exemplar Class 12 Physics Solutions Chapter 7: MCQI

Question:1

If the RMS current in a 50 Hz ac circuit is 5 A, the value of the current 1/300 seconds after its value becomes zero is
a. $5\sqrt{2}A$
b.$5\sqrt{\frac{3}{2A}}$

c.$\frac{5}{6A}$

d.$\frac{5}{\sqrt{2}A}$

Answer:

The answer is the option (b)
$I=I_0 \sin \omega t\\It \:is \:given\: that\: I_0=\sqrt{2} *5=5\sqrt{2}A,\nu=50Hz, t=\frac{1}{300s}\\ I=5\sqrt{2}\sin 2 \pi \nu t\\ =5\sqrt{2}sin2 \pi *50*\frac{1}{300}$
$\\=5\sqrt{2}sin\left ( \frac{ \pi }{3} \right )= 5\sqrt{2}\sin60^{\circ}\\ =5\sqrt{2}*\frac{\sqrt{3}}{2}=5\sqrt{\frac{3}{2}}A$

Question:2

An alternating current generator has an internal resistance Rg and an internal reactance Xg. It is used to supply power to a passive load consisting of a resistance Rg and a reactance XL. For maximum power to be delivered from the generator to the load, the value of XL is equal to
A. zero.
B. Xg.
C. – Xg.
D. Rg

Answer:

For maximum power to be delivered from the generator to the load
$X_L + X_g = 0$(the total reactance should be equal to zero)
$\Rightarrow X_L=-X_g$

Question:3

When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220V. This means
A. input voltage cannot be AC voltage, but a DC voltage.
B. maximum input voltage is 220V.
C. the meter reads not v but <v2> and is calibrated to read $\sqrt{< v^{2}> }$
D. the pointer of the meter is stuck by some mechanical defect.

Answer:

The answer is the option (c)
This means that the voltage measuring device connected to AC mains is calibrated to read
the rms value $\sqrt{< v^{2}> }$

Question:4

To reduce the resonant frequency in an LCR series circuit with a generator
A. the generator frequency should be reduced.
B. another capacitor should be added in parallel to the first.
C. the iron core of the inductor should be removed.
D. dielectric in the capacitor should be removed.

Answer:

The answer is the option (b)
$v_0=\frac{1}{2\pi\sqrt{LC}}$
According to the formula to reduce v0 we have to increase either of the L or C.
And for increasing the capacitance or C, another capacitor should be in parallel connection to the first

Question:5

Which of the following combinations should be selected for better tuning of an LCR circuit used for communication?
A. R = 20 Ω, L = 1.5 H, C = 35μF.
B. R = 25 Ω, L = 2.5 H, C = 45μF.
C. R = 15 Ω, L = 3.5 H, C = 30μF.
D. R = 25 Ω, L = 1.5 H, C = 45μF.

Answer:

The quality factor
$Q=\frac{1}{R}\sqrt{\frac{L}{C}}$
where R is resistance, L is the inductance and C is the capacitance of the circuit
For a higher Q, L must be larger, and R and C must be smaller in value.

Question:6

An inductor of reactance 1Ω and a resistor of 2 Ω are connected in series to the terminals of a 6 V (rms) a.c. source. The power dissipated in the circuit is
A. 8 W.
B. 12 W.
C. 14.4 W.
D. 18 W.

Answer:

$X_L=1\Omega, R=2 \Omega \: \: E_{rms}=6V$
The average power dissipated in the circuit
$\\P_{av}=E_{rms}I_{rms}\cos\Phi \\ I_{rms}=\frac{E_{rms}}{Z}\\\\ Z=\sqrt{R^2+X_{L}^{2}}=\sqrt{4+1}=\sqrt{5}\\ \\ I_{rms}=\frac{6}{\sqrt{5}}A\\ \\ \cos \Phi \frac{R}{Z}=\frac{2}{\sqrt{5}}$
$\\P_{av}=6*\frac{6}{\sqrt{5}}*\frac{2}{\sqrt{5}}= \frac{72}{5}=14.4\: W$

Question:7

The output of a step-down transformer is measured to be 24 V when connected to a 12-watt light bulb. The value of the peak current is
$\\A.\: 1/\sqrt{2}A\\ B.\: \sqrt{2}A\\ C. \: 2A\\ D.\:2\sqrt{2}A$

Answer:

$\\ V_s=24 V P_s=12 W\\I_s V_s=12\\I_s=\frac{12}{V_s} =\frac{12}{24}=0.5 \: ampere\\ I_0=I_s \sqrt{2}=0.5\sqrt{2}=\frac{1}{\sqrt{2} }\: ampere$

NCERT Exemplar Class 12 Physics Solutions Chapter 7 MCQII

Question:8

As the frequency of an ac circuit increases, the current first increases and then decreases. What combination of circuit elements is most likely to comprise the circuit?
A. Inductor and capacitor.
B. Resistor and inductor.
C. Resistor and capacitor.
D. Resistor, inductor and capacitor.

Answer:

For I to be maximum in a circuit, the reactance should be minimum. In the LCR circuit, if the frequency is increased, the inductance will increase while the capacitance decreases.
$Z=\sqrt{R^2+\left ( X_L-X_C \right )^2}=\sqrt{R^2+\left ( 2\pi \nu L-\frac{1}{2\pi \nu C} \right )^2}$
On increasing v, there will come a stage where we will reach resonance
$\left ( X_L=X_C \right )$

Question:9

In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?
A. Only resistor.
B. Resistor and an inductor.
C. Resistor and a capacitor.
D. Only a capacitor.

Answer:

The correct answer is the options (c, d)
In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of the supply. When the current increases, the capacitive reactance of circuit decreases, and resistance does not depend on frequency.

Question:10

Electrical energy is transmitted over large distances at high alternating voltages. Which of the following statements is (are) correct?
A. For a given power level, there is a lower current.
B. Lower current implies less power loss.
C. Transmission lines can be made thinner.
D. It is easy to reduce the voltage at the receiving end using step-down transformers.

Answer:

Since the power is to be transmitted over the large distances at high alternating voltages, the current flowing through the wires will below.
The power loss due to transmission lines with resistance R and current
$I_{rms}$ in the circuit is given by $I_{rms}^{2}R$
$P=E_{rms}I_{rms}(I_{rms}is \: low\: when\: E_{rms}\: is\: high)$
$Power\:loss=I_{rms}^{2}R=low$
The voltage at the receiving end is reduced by step-down transformers.

Question:11

For an LCR circuit, the power transferred from the driving source to the driven oscillator is $P = I^{}2Z \cos\phi$

A. Here, the power factor cos$\phi \geq 0, P \geq 0.$
B. The driving force can give no energy to the oscillator (P = 0) in some cases.
C. The driving force cannot syphon out (P < 0) the energy out of oscillator.
D. The driving force can take away energy out of the oscillator.

Answer:

The answers are the options (a, b, c)
According to the question
$P = I^{}2Z \cos\phi$
Power factor
$\cos \phi =\frac{R}{Z}\: \: As\: R>0\:and\:Z>0$
$Thus\:\cos \phi =\frac{R}{Z}\: \: Is \:positive\\and \:where\: \phi =\frac{\pi}{2},P=0$

Question:12

When an AC voltage of 220 V is applied to the capacitor C
A. the maximum voltage between plates is 220 V.
B. the current is in phase with the applied voltage.
C. the charge on the plates is in phase with the applied voltage.
D. power delivered to the capacitor is zero.

Answer:

The correct answer are the options (c, d) On connecting the capacitor to ac supply the plate of the capacitor connected to the +ve terminal is at higher potential compared to the plate connected to negative terminal.
The power applied to the circuit is given by
$P_{av}=V_{rms}I_{rms}\cos \phi\\ \phi =90^{\circ}\: f\! or\: pure \: capacitor \: circuit\\ P_{av}=V_{rms}I_{rms}\cos 90^{\circ} =0$

Question:13

The line that draws power supply to your house from street has
A. zero average current.
B. 220 V average voltage.
C. voltage and current out of phase by 90°.
D. voltage and current possibly differing in phase $\phi$ such that $\left | \phi \right |< \frac{\pi}{2}$

Answer:

The correct answer are the options (a,d)
It is a known fact that AC supply is used in houses.
The average current over a cycle of AC is zero. L and C are connected in household circuit so R and Z cannot be equal.
$So \: Power\: factor\: = \cos \phi=\frac{R}{Z}\neq 0=\phi\neq \frac{\pi}{2}$
$\phi< \frac{\pi}{2} \:i.\:e.phase \:angle\: between\: voltage\: and\: current\: lies\: between\: 0\: and\: \frac{\pi}{2}$

NCERT Exemplar Class 12 Physics Solutions Chapter 7: Very Short Answer

Question:14

If a LC circuit is considered analogous to a harmonically oscillating spring block system, which energy of the LC circuit would be analogous to potential energy and which one analogous to kinetic energy?

Answer:

If a L-C circuit is considered analogous to a harmonically oscillating spring block system, energy due to motion of charge particle i.e., magnetic energy is analogous to kinetic energy and electrostatic energy due to charging of capacitor is analogous to potential energy.

Question:15

Draw the effective equivalent circuit of the circuit shown in Fig 7.1, at extremely high frequencies and find the effective impedance.
RLC circuit

Answer:

Inductive reactance
$X_L=\omega L=2\pi fL$
At very high frequencies, $X_L$ will be high or L can be considered an open circuit for the high frequency of ac circuit.
Capacitive reactance
$X_C=\frac{1}{2\pi \nu C}$
At high frequency, $X_C$will below.
Therefore reactance of capacitance can be considered negligible and capacitor can be considered short-circuited.

Question:16

Study the circuits (a) and (b) shown in Fig 7.2 and answer the following questions.
R and RLC circuit
(a) Under which conditions would the rms currents in the two circuits be the same?
(b) Can the rms current in circuit (b) be larger than that in (a)?

Answer:

$I_{rms} \: in (a)=\frac{V_{rms}}{R}$
$\\I_{rms} in (b)=\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C )}}\\\\ \\(a)Now (I_{rms} )_a=(I_{rms} )_b\\\\ \frac{V_{rms}}{R}=\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C )^2}}\\\\ \\R=\sqrt{R^2+(X_L-X_C )}$
Squaring both sides
$\\R^2=R^2+\left ( X_L-X_C \right )^{2}\\\\ Or \left ( X_L-X_C \right )^{2}=0\\ \\X_L=X_C$
So, $I_{rms}$ in circuits a and b will be equal if $\\ \\X_L=X_C$
$(b)\: \: For (I_{rms} )_b>(I_{rms })_a$

$\\\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C )}}>\frac{V_{rms}}{R}\\ \\ As V_{rms=V} \\\\So\: \: \sqrt{R^2+(X_L-X_C )}<R \\ \\ Squaring\: \: both \: \: sides R^2+(X_L-X_C )^2<R^2$
$(X_L-X_C )^2<0$
Square of a number can'tbe negative.
Therefore, the rms current in circuit (b) has to be larger than that in (a).

Question:17

Can the instantaneous power output of an ac source ever be negative? Can the average power output be negative?

Answer:

The instantaneous power output of an AC source can be negative because there are chances for power to get absorbed at some point of time. However, the average output over a cycle cannot be negative at any instant. It can only be either positive or zero as in any LC circuit.
$\\Let \: the\: applied \: e.m.f=E=E_0 \sin \omega t\\ I=I_0 \: \sin \left ( \omega t-\phi \right )\\Instantaneous \: power \: output \: o\! f \: ac \: source P=EI\\E_0 \sin \omega t\: I_0 \: \sin \left ( \omega t-\phi \right )= E_0 I_0 \sin \omega t\left [ \sin \omega t \cos\phi - \cos \omega t\sin \phi\right ]\\=E_0 I_0\left [ \frac{1-\cos 2\: \omega t}{2}\cos \phi-\frac{1}{2}\sin 2 \omega t \sin \phi \right ]$
$\\=\frac{E_0 I_0}{2}\left [ \cos\phi -\cos2 \omega t \cos \phi -\sin 2 \omega t \sin \phi \right ]\\\\=\frac{E_0 I_0}{2}\left [ \cos\phi -\left ( \cos2 \omega t \cos \phi +\sin 2 \omega t \sin \phi \right )\right ]\\P=\frac{E_0 I_0}{2}\left [ \cos\phi-\cos\left ( 2 \omega t-\phi \right ) \right ]$
The phase angle taken φ, ±ve

Question:18

In series LCR circuit, the plot of $I_{max} v s \: \omega$ is shown in Fig 7.3. Find the bandwidth and mark in the figure.
Peak current vs angular frequency graph

Answer:

Peak current vs angular frequency graph
$I=\frac{E_0}{\sqrt{2}}= \frac{1}{\sqrt{2}}= 0.707\: Amp$
From graph we get
$\omega _1=0.8\frac{rad}{s}\: and\:\omega _2=1.2\frac{rad}{s}$
$Bandwidth =1.2-0.8=0.4\: rad/sec$

Question:19

The alternating current in a circuit is described by the graph shown in Fig 7.4. Show RMS current in this graph.
Current - time graph

Answer:

current time graph
$I_{rms}=\sqrt{\frac{\left ( 1^{2}+2^{2} \right )}{2}}=\sqrt{\frac{5}{2}}\cong 1.6A$

Question:20

How does the sign of the phase angle $\theta$, by which the supply voltage leads the current in an LCR series circuit, change as the supply frequency is gradually increased from very low to very high values.

Answer:

RLC circuit and phasor diagram

Phase angle $\theta$ by which Voltage leads current is given by,

$\tan \theta= \frac{X_L-X_C}{R}=\frac{\omega L-\frac{1}{\omega C}}{R}$
For low frequencies, $X_L<X_C\Rightarrow \tan\theta <0$
For resonance

$X_L<X_C\Rightarrow \tan\theta =0\left (\nu =\nu _0=\frac{1}{2\pi\sqrt{2C}} \right )$

NCERT Exemplar Class 12 Physics Solutions Chapter 7: Short Answer

Question:21

A device ‘X’ is connected to an a .c source. The variation of voltage, current and power in one complete cycle is shown in Fig 7.5.
Variation of voltage current and power
(a) Which curve shows power consumption over a full cycle?
(b) What is the average power consumption over a cycle?
(c) Identify the device ‘X’.

Answer:

(a) $\text {Power (P)}=\text {VI}$
Power consumption will have the maximum amplitude between the three curves.
(b) Average Power consumption is zero covering a positive and a similar negative area in a complete cycle.
Variation of voltage current and power
(c) For the given circuit, the phase difference between V and I is $\frac{\pi }{2}$. The device ‘X’ can be an inductor or capacitor or a series combination of inductors and capacitors.

Question:22

Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current?

Answer:

In a DC circuit, 1 Ampere is defined as 1 Coulomb per second, but for AC circuit, the direction of current keeps changing. Instead, Joule’s heating is used to define the rms value of AC current. This is done because Joule’s heating is independent of the direction of the current. RMS value of AC is equal to the value of DC required to generate the same amount of heat through a given resistor in a given time.

Question:23

A coil of 0.01 henry inductance and 1-ohm resistance is connected to 200 volts, 50 Hz ac supply. Find the impedance of the circuit and time lag between max. alternating voltage and current.

Answer:

$Inductance (L)=0.01 \; H$
$Resistance (R)=1\Omega$
$Voltage (V)=200 V$
$Frequency (f)=50 Hz$
$X_{L}=2\pi fL=3.14\Omega$
$z=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{10.86}=3.3\Omega$
$\tan \phi =\frac{\omega L}{R}=\frac{2\pi fL}{R}=3.14$
$\text {Phase different}(\phi )=\tan ^{-1}3.14=72^{o}=\frac{72}{180}*\pi$
$\text {Time lag}(\Delta t )=\frac{\phi }{\omega }=\frac{\frac{72\pi }{180}}{2\pi *50}=0.004\; s$

Question:24

A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil? Comment on the type of transformer being used.

Answer:

$P_{s}=60\; W$
$I_{s}=0.54\; A$
$V_{s}=\frac{P_{s}}{I_{s}}=\frac{60}{0.54}=110V$
$V_{P}=220V$
As, $V_{P} >V_{s},$ it is a step down transformer
$r=\frac{V_{s}}{V_{p}}=\frac{I_{P}}{I_{S}}$
$I_{P}=0.54*\frac{110}{220}=0.27 A$

Question:25

Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency.

Answer:

Under the effect of AC current, capacitor plates charge and discharge alternately. Current is a direct result of this charging and discharging. On increasing the frequency of current, the capacitor will pass more current through it.

Question:26

Explain why the reactance offered by an inductor increases with increasing frequency of an alternating voltage.

Answer:

Under the flow of current, Inductor develops a back emf. On decreasing the current, the induced emf will be such that it tries to maintain the existing flow of current. Induced emf is directly proportional to the rate of change of current.

NCERT Exemplar Class 12 Physics Solutions Chapter 7: Long Answer

Question:27

An electrical device draws 2kW power from AC mains $(Voltage\; 223V(rms)=\sqrt{50,000}v)$ The current differs (lags) in phase by $\phi \left ( \tan \phi =\frac{-3}{4} \right )$ as compared to voltage. Find (i) R, (ii) XC – XL, and (iii) IM.

Answer:

Impedance,
$Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}=\sqrt{R^{2}+\left ( \omega L-\frac{1}{\omega C} \right )^{2}}$
$P=2000\; W$
$\tan \phi =-\frac{3}{4}$
$V_{rms}=223 \; V$
$P=\frac{V^{2}}{Z}$
$Z=25 \Omega$
$\frac{X_{c}-X_{L}}{R}=-\frac{3}{4}$
$X_{L}-X_{C}=-\frac{3}{4}R$
$R^{2}+(X_{L}-X_{C})^{2}=Z^{2}$
$R^{2}+\frac{9}{16}R^{2}=25^{2}$
$R^{2}=625*\frac{16}{25}=400$
$R=20 \Omega$
$X_{C}-X_{L}=-\frac{3}{4}*20=-15\Omega$
$I_{M}=\sqrt{2}I=\sqrt{2}\frac{V}{Z}=12.6 \; A$

Question:28

1MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if

(i) power is transmitted at 220V. Comment on the feasibility of doing this.
(ii) a step-up transformer is used to boost the voltage to 11000 V, power transmitted, then a step-down transformer is used to bring voltage to 220 V.
$(\rho _{cu}=1.7\times10^{-8} SI \; unit)$

Answer:

$L=20\; km=20000\; m$
$R=\rho _{cu}\frac{L}{A}=\frac{1.7*10^{-8}*2*10^{4}}{\pi (0.5*10^{-2})^{2}}=4\Omega$
$I=\frac{10^{6}}{220}=0.45*10^{4}A$
Power loss = $I^{2}R=4*0.45^{2}*10^{8}>10^{6}W$
ii) $P=10^{6}$
$V'=11000 \; V$
$I'=\frac{P}{V'}=\frac{1}{1.1}*10^{2}$
Power Loss =
$I'^{2}R=\frac{1}{1.21}*4*10^{4}=3.3*10^{4}W$
Fraction of power loss =
$\frac{3.3*10^{4}}{10^{6}}=3.3\; ^{o}/_{o}$

Question:29

Consider the LCR circuit shown in Fig 7.6. Find the net current i and the phase of i. Show that $i=\frac{v}{Z}$. Find the impedance Z for this circuit.
RLC circuit

Answer:

Taking Resistor andAlternating EMF circuit, $i=i_{1}+i_{2}$
$i_{2}R=V_{m}\sin \omega t\Rightarrow i_{2}=\frac{V_{m}\sin \omega t}{R}$
Taking Capacitor-Inductor and Alternating EMF circuit,
$V_{m}\sin \omega t=\frac{q_{1}}{C}+L\frac{d^{2}q}{dt^{2}}$
Assuming $q_{1}=q_{m}\; \sin (\omega t+\phi )$
$\frac{dq_{1}}{dt}=\omega q_{m}\cos (\omega t+\phi )$
$\frac{d^{2}q_{1}}{dt^{2}}=-\omega ^{2}q_{m}\sin (\omega +\phi )$
$V_{m}\sin \omega t=\frac{q_{m}\sin (\omega t+\phi )}{C}+L(-\omega ^{2}q_{m}\; \sin (\omega t+\phi ))$
$V_{m}\sin \omega t=q_{m}\left ( \frac{1}{C}-\omega ^{2}L \right )\sin (\omega t+\phi )$
$q_{m}=\frac{V_{m}}{\frac{1}{C}-\omega ^{2}L}$
$\sin \omega t=\sin (\omega t+\phi )$
$\phi =0$
$i_{1}=\frac{dq_{1}}{dt}=\omega \left ( \frac{V_{m}}{\frac{1}{C}-\omega ^{2}L} \right )\cos (\omega t)$
$i=i_{1}+i_{2}=\left ( \frac{V_{m}}{\frac{1}{\omega C}-\omega L} \right )\cos (\omega t)+\frac{V_{m}(\sin \omega t)}{R}$
$=A\; \sin (\omega t+\phi )$
$A=\sqrt{\left ( \left ( \frac{V_{m}}{\left ( \frac{1}{\omega C}-\omega L \right )} \right )^{2}+\left ( \frac{V_{m}}{R} \right )^{2} \right )}=V_{m}\sqrt{\left ( \left ( \frac{V_{m}}{\left ( \frac{1}{\omega C}-\omega L \right )} \right )^{2}+\left ( \frac{1}{R} \right )^{2} \right )}$
$\tan \phi =\frac{R}{\frac{1}{\omega C}-\omega L}$
$\phi =\tan ^{-1}\left ( \frac{R}{\frac{1}{\omega C}-\omega L} \right )$

Question:30

For an LCR circuit driven at frequency $\omega$, the equation reads
$L\frac{di}{dt}+Ri+\frac{q}{C}=v_{1}=v_{m}\; \sin \omega t$
(i) Multiply the equation by i and simplify where possible.
(ii) Interpret each term physically.
(iii) Cast the equation in the form of a conservation of energy statement.
(iv) Integrate the equation over one cycle to find that the phase difference between v and i must be acute.

Answer:

$V=V_{m}\sin \omega t$
$L\frac{di}{dt}+iR+\frac{q}{C}=V_{m}\sin \omega t$
$Li\frac{di}{dt}+i^{2}R+\frac{q}{C}i=iV_{m}\sin \omega t$
$Li\frac{di}{dt}+i^{2}R+\frac{q}{C}i=iV$
Power loss = $i^{2}R$
Rate of change of energy stored in Inductor =
$Li\frac{di}{dt}=\frac{d}{dt}\left ( \frac{1}{2}Li^{2} \right )$
Rate of change of energy stored in Capacitor =
$\frac{q}{C}i=\frac{d}{dt}\left ( \frac{q^{2}}{2C} \right )$
$\int_{0}^{T}\frac{d}{dt}\left ( \frac{1}{2}Li^{2}+\frac{q^{2}}{2C} \right )dt+\int_{0}^{T}i^{2}R dt=\int_{0}^{T}iV \; dt$
$0+(+ve)=\int_{0}^{T}iV\; dt>0$

Question:31

In the LCR circuit shown in Fig 7.7, the ac driving voltage is $v=v_{m}\sin \omega t$
RLC circuit
(i) Write down the equation of motion for q (t).
(ii) At $t=t_{0}$, the voltage source stops and R is short circuited. Now write down how much energy is stored in each of L and C.
(iii) Describe subsequent motion of charges.

Answer:

a) $V=V_{m}\sin \omega t$
$L\frac{di}{dt}+iR+\frac{q}{C}=V_{m}\sin \omega t$
$L\frac{d^{2}q}{dt^{2}}+\frac{dq}{dt}R+\frac{q}{C}=V_{m}\sin \omega t$
b) Let $q=q_{m}\sin \left ( \omega t+\phi \right )$
$\frac{dq}{dt}=\omega q_{m}\cos (\omega t+\phi )$
$i_{m}=\frac{V_{m}}{Z}=\frac{V_{m}}{\sqrt{(R)^{2}+(X_{c}-X_{L})^{2}}}$
$\phi =\tan ^{-1}\left ( \frac{X_{c}-X_{L}}{R} \right )$
At $t=t_{0},$ Resistance is short - circuited and the inductor and capacitor store energy
$U_{L}=\frac{1}{2}Li^{2}=\frac{1}{2}L\left [ \frac{V_{m}}{\sqrt{(R)^{2}+(X_{c}-X_{L})^{2}}} \right ]^{2}\sin ^{2} (\omega t_{0}+\phi )$
$U_{c}=\frac{q^{2}}{2C}=\frac{1}{2C\omega ^{2}}\left [ \frac{V_{m}}{\sqrt{(R)^{2}+(X_{C}-X_{L})^{2}}} \right ]^{2}\; \cos ^{2}(\omega t_{0}+\phi )$
(c) When R is short-circuited, the circuit becomes an L-C oscillator. When the capacitor discharges, all its energy goes from capacitor to inductor. Energy oscillates from electrostatic to magnetic and from magnetic to electrostatic.

NCERT Exemplar Class 12 Physics Solutions Chapter 7 Main subtopics

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What will the students learn with NCERT Exemplar Class 12 Physics Solutions Chapter 7?

AC voltages are preferred over DC because they can be efficiently converted to a voltage of choice through transformers, which can step up or step down the voltage. Class 12 Physics NCERT Exemplar solutions chapter 7 explains the graphical representation of AC current and voltage, principles behind an AC generator, defines angular frequency, time period, factors and phasors, dives deep into current and potential relations and explores Resistors, Capacitors, and Inductors, involves a study of Phasor algebra, circuit series and Power in an AC circuit.
Also, check NCERT Solutions for Class 12 Physics

NCERT Exemplar Class 12 Physics Chapter 7 Alternating Current: Important Concepts & Formulas

1. AC Voltage & Current Representation

  • AC voltage: $V=V_0 \sin (\omega t)$
  • AC current: $I=I_0 \sin (\omega t+\phi)$
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2. Impedance & Reactance

  • Resistance (R): $V=I R$
  • Inductive reactance: $X_L=\omega L=2 \pi f L$
  • Capacitive reactance: $X_C=\frac{1}{\Delta C}=\frac{1}{2 \pi f C}$
  • Impedance in an LCR circuit:

$
Z=\sqrt{R^2+\left(X_L-X_C\right)^2}
$

3. Phasor Representation

  • AC voltages and currents are represented using rotating vectors (phasors).
  • Phase difference: In a pure inductor, current lags voltage by $90^{\circ}$, while in a pure capacitor, current leads by $90^{\circ}$.

4. Power in AC Circuits

  • Instantaneous power: $P=V I$
  • Average power: $P_{\mathrm{avg}}=V_{\text {rma }} I_{\text {rms }} \cos \phi$
  • Power factor: $\cos \phi=\frac{R}{Z}$

5. Resonance in LCR Circuit

  • Resonant frequency:

$
f_r=\frac{1}{2 \pi \sqrt{L C}}
$

  • At resonance, $X_L=X_C$ and impedance is minimum ( $Z=R$ ), leading to maximum current.

6. LC Oscillations
* The energy oscillates between inductor and capacitor with frequency:

$
\omega=\frac{1}{\sqrt{L C}}
$

7. Transformers

  • Turns ratio: $\frac{N_1}{N_p}=\frac{V_p}{V_p}$
  • Step-up transformer: $V_s>V_p$ (increases voltage)
  • Step-down transformer: $V_s<V_p$ (decreases voltage)
  • Power transmission efficiency: $P=V I$ (ideally, $P_{\text {in }}=P_{\text {out }}$ )

NCERT Exemplar Class 12 Solutions

Check Class 12 Physics Chapter-wise Solutions

Also, Read NCERT Solution subject-wise

Must read NCERT Notes subject-wise

Also, Check NCERT Books and NCERT Syllabus

Frequently Asked Questions (FAQs)

Q: Why is alternating current an important chapter?
A:

For Class 12 students, AC is a crucial chapter from boards, entrance exam and higher education point of view. This is one of the major chapters in electrical engineering.

Q: How to use these solutions for preparation?
A:

 One can use these NCERT exemplar Class 12 Physics solutions chapter 7 for better understanding of the topics and how the questions will be formed in the board exam.

Q: What all topics is there in the chapter?
A:

Chapter covers topics like AC circuit, capacitor, resonance, LC oscillations, transformers, LCR circuit, etc.

Q: Who has solved these questions?
A:

Our esteemed physics teachers and team have solved each and every question as per the CBSE pattern and marking scheme in NCERT exemplar Class 12 Physics chapter 7 solutions

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Questions related to CBSE Class 12th

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Yes, if you’re not satisfied with your marks even after the improvement exam, many education boards allow you to reappear as a private candidate next year to improve your scores. This means you can register independently, study at your own pace, and take the exams without attending regular classes. It’s a good option to improve your results and open up more opportunities for higher studies or careers. Just make sure to check the specific rules and deadlines of your education board so you don’t miss the registration window. Keep your focus, and you will do better next time.

Hello Aspirant,

Yes, in the case that you appeared for the 2025 improvement exam and your roll number is different from what was on the previous year’s marksheet, the board will usually release a new migration certificate. This is because the migration certificate will reflect the most recent exam details, roll number and passing year. You can apply to get it from your board using the process prescribed by them either online or through your school/college.

Yes, if you miss the 1st CBSE exam due to valid reasons, then you can appear for the 2nd CBSE compartment exam.

From the academic year 2026, the board will conduct the CBSE 10th exam twice a year, while the CBSE 12th exam will be held once, as per usual. For class 10th, the second phase exam will act as the supplementary exam. Check out information on w hen the CBSE first exam 2026 will be conducted and changes in 2026 CBSE Board exam by clicking on the link .

If you want to change your stream to humanities after getting a compartment in one subject in the CBSE 12th Board Exam , you actually have limited options to qualify for your board exams. You can prepare effectively and appear in the compartment examination for mathematics again. If you do not wish to continue with the current stream, you can take readmission in the Humanities stream and start from Class 11th again, and continue studying for two more years to qualify for the 12th examination.

The GUJCET Merit List is prepared based on the Class 12th marks and GUJCET marks received by the students. CBSE students who are not from the Gujarat board can definitely compete with GSEB students, as their eligibility is decided based on the combined marks scored by them in GUJCET and the 12th board. The weightage of the GUJCET score is 40% and the weightage of the class 12 scores is 60%.