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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.

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 Wise Links
  9. Some Applications of Alternating Current

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}

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.

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_Cwill 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)?

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.

Answer:


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.

Answer:


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:


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.

(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.

(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.

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

(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

  • 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

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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 Wise Links

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.

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Frequently Asked Question (FAQs)

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 ?
I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

Hello aspirant,

The purpose of graphic design extends beyond the brand's look. Nevertheless, by conveying what the brand stands for, it significantly aids in the development of a sense of understanding between a company and its audience. The future in the field of graphic designing is very promising.

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Read Complete Answer
Sajal Trivedi 29 January,2024
i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Read Complete Answer
Upasana Barman 24 August,2022
i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


Believe in Yourself! You can make anything happen


All the very best.

Read Complete Answer
Mayankjha.student2007 23 August,2022
if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

Read Complete Answer
Nuthalapati Vyshnavi 21 August,2022
I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

Read Complete Answer
Muskan Dhalwal 17 August,2022
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Popular CBSE Class 12th Questions

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)

0.34\; J

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In the reaction,

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

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6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

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0.02

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more than 3 but less than 6
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more than 6 but less than 9

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