NCERT Exemplar Class 12 Physics Solutions Chapter 7 Alternating Current

NCERT Exemplar Class 12 Physics Solutions Chapter 7 Alternating Current

Edited By Safeer PP | Updated on Sep 14, 2022 12:57 PM IST | #CBSE Class 12th

NCERT Exemplar Class 12 Physics solutions chapter 7 defines AC Current as the current driven by a voltage that varies sinusoidally with time (alternating voltage; AC voltage). It highlights the importance of AC voltage in most of the electrical devices that we use and how the electrical energy sold by power companies is distributed and transmitted as alternating current. NCERT Exemplar Class 12 Physics chapter 7 solutions would help you graphically deduce answers to complex problems and also deal with difficult concepts to make them comprehensively sound in your head.

A careful study of Class 12 Physics NCERT Exemplar solutions chapter 7 would help one score better in 12 boards and competitive exams. Students can make use of NCERT Exemplar Class 12 Physics solutions chapter 7 PDF download prepared by experts, for further use. The topics covered are as follows:

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

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

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.

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.

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.

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

$X_L=1\Omega, R=2 \Omega \: \: E_{rms}=6V$
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
$\\A.\: 1/\sqrt{2}A\\ B.\: \sqrt{2}A\\ C. \: 2A\\ D.\:2\sqrt{2}A$

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

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.

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.

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.

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.

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.

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 $\left | \phi \right |< \frac{\pi}{2}$

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?

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

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.

(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)?

$I_{rms} \: in (a)=\frac{V_{rms}}{R}$
$\\I_{rms} in (b)=\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C )}}\\\\ \\(a)Now (I_{rms} )_a=(I_{rms} )_b\\\\ \frac{V_{rms}}{R}=\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C )^2}}\\\\ \\R=\sqrt{R^2+(X_L-X_C )}$
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 )}
$(X_L-X_C )^2<0$
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?

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

$I=\frac{E_0}{\sqrt{2}}= \frac{1}{\sqrt{2}}= 0.707\: Amp$
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

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

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.

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
For resonance

$X_L

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.

(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’.

(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.

(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?

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.

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

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.

$P_{s}=60\; W$
$I_{s}=0.54\; A$
$V_{s}=\frac{P_{s}}{I_{s}}=\frac{60}{0.54}=110V$
$V_{P}=220V$
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.

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.

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.

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.
$(\rho _{cu}=1.7\times10^{-8} SI \; unit)$

$L=20\; km=20000\; m$
$R=\rho _{cu}\frac{L}{A}=\frac{1.7*10^{-8}*2*10^{4}}{\pi (0.5*10^{-2})^{2}}=4\Omega$
$I=\frac{10^{6}}{220}=0.45*10^{4}A$
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

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.

$V=V_{m}\sin \omega t$
$L\frac{di}{dt}+iR+\frac{q}{C}=V_{m}\sin \omega t$
$Li\frac{di}{dt}+i^{2}R+\frac{q}{C}i=iV_{m}\sin \omega t$
$Li\frac{di}{dt}+i^{2}R+\frac{q}{C}i=iV$
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 $v=v_{m}\sin \omega t$

(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.

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

• Introduction
• AC Voltage applied to a resistor
• Representation of AC Current and Voltage by Rotating Vectors-Phasors
• AC Voltage applied to an Inductor
• AC Voltage applied to a Capacitor
• AC Voltage applied to a Series LCR Circuit
• Phasor Diagram solution
• Analytical solution
• Resonance
• Sharpness of Resonance
• Power in AC Circuit: The Power Factor
• Resistive circuit
• Purely inductive or capacitive circuit
• LCR Series circuit
• Power dissipated at resonance in LCR circuit
• LC Oscillations
• Transformers
• Step-up transformers
• Step-down transformers

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 Wise Links

 Chapter 1 Electric Charges and Fields Chapter 2 Electrostatic Potential and Capacitance Chapter 3 Current Electricity Chapter 4 Moving Charges and Magnetism Chapter 5 Magnetism and Matter Chapter 6 Electromagnetic Induction Chapter 8 Electromagnetic Waves Chapter 9 Ray Optics and Optical Instruments Chapter 10 Wave Optics Chapter 11 Dual Nature of Radiation and Matter Chapter 12 Atoms Chapter 13 Nuclei Chapter 14 Semiconductor Electronics Chapter 15 Communication Systems

Some Applications of Alternating Current

· NCERT Exemplar Solutions For Class 12 Physics chapter 7 covers different aspects of Alternating Current. An alternating current (AC) is the type of electric current generated by a large number of power plants and used by a majority of power providers. Alternating current is cheap, efficient, and has fewer energy losses than the transmission of direct current. Some problems related to applications of alternating current are discussed in NCERT Exemplar Class 12 Physics solutions chapter 7

· Devices in massive factories that are attached to the grid utilise alternating current; electrical outlets at homes and industrial areas use AC current as well. Many devices make use of an AC adapter which helps in an efficient conversion of AC into DC current.

· AC is popularised and widely used because it provides voltage stepping by utilising transformers. As explained, this allows smooth and energy-efficient electrical transmission via power-lines. Questions related to transformers are solved in NCERT Exemplar Class 12 Physics chapter 7 solutions.

NCERT Exemplar Class 12 Solutions

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1. Why is alternating current an important chapter?

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.

2. How to use these solutions for preparation?

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.

3. What all topics is there in the chapter?

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

4. Who has solved these questions?

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

Have a question related to CBSE Class 12th ?

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Yuvan 01 September,2024

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9