Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# NCERT Solutions for Class 12 Physics Chapter 7 - Alternating Current

Last updated date: 09th Aug 2024
Total views: 810.3k
Views today: 24.10k

## NCERT Solutions for Class 12 Physics Chapter 7 - Alternating Current

NCERT Solutions for CBSE Class 12 Physics Chapter 7 - Alternating Current are available in Vedantu. These NCERT Solutions are created as per the latest Syllabus of NCERT Physics for Class 12. This PDF covers solutions for all questions covered in the CBSE Class 12 Physics textbook in Chapter 7. All the solutions are explained in a step-by-step manner. Students can refer to these solutions to learn the important questions and prepare for their board exams. These NCERT Solutions for CBSE Class 12 Physics Chapter 7 Alternating Current are available in a PDF format and can be downloaded for free.

 Class: NCERT Solutions for Class 12 Subject: Class 12 Physics Chapter Name: Chapter 7 - Alternating Current Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

## Topics Covered in Alternating Current Solutions are as follows:

• C Voltage Applied to a Resistor.

• Representation of AC Current and Voltage by Rotating Vectors.

• AC Voltage Applied to an Inductor.

• AC Voltage Applied to a Capacitor.

• AC Voltage Applied to a Series LCR Circuit.

• Power in AC Circuit: The Power Factor.

• LC Oscillations.

• Transformers.

Would you like to view a summarized version of this chapter? Check out the 'Chapter at a glance' section below the PDF of NCERT Solutions.

Competitive Exams after 12th Science
More Free Study Material for Alternating Current
Revision notes
Important questions
Ncert books

## AC & EM Waves Chapter at a Glance - Class 12 NCERT Solutions

• An alternating voltage υ = υm sin ωt applied to a resistor R drives a current i = im sin ωt in the resistor, $i_m=\frac{\nu _m}{R}.$ The current is in phase with the applied voltage.

• For an alternating current i = im sin ωt passing through a resistor R, the average power loss P (averaged over a cycle) due to joule heating is $\frac{1}{2}(i_m)^2 R.$ To express it in the same form as the dc power (P = I2R), a special value of current is used. It is called root mean square (rms) current and is denoted by I:

$I=\frac{i_m}{\sqrt{2}}=0.707i_m$

Similarly, the rms voltage is defined by

$V=\frac{\nu_m}{\sqrt{2}}=7.07\nu_m$

We have P = IV = I2 R

• An ac voltage υ = υm sin ωt applied to a pure inductor L, drives a current in the inductor i = im sin (ωt – π/2), where im = υm/XL. XL = ωL is called inductive reactance. The current in the inductor lags the voltage by π/2. The average power supplied to an inductor over one complete cycle is zero.

• An ac voltage υ = υm sin ωt applied to a capacitor drives a current in the capacitor: i = im sin (ωt + π/2). Here,

$i_m=\frac{\nu_m}{X_C},\;X_C=\frac{1}{\omega C}$ is called capacitive reactance.

The current through the capacitor is π/2 ahead of the applied voltage. As in the case of inductor, the average power supplied to a capacitor over one complete cycle is zero.

• For a series RLC circuit driven by voltage υ = υm sin ωt, the current is given by i = im sin (ωt + φ)

where $i_m=\frac{\nu_m}{\sqrt{R^2 +(X_L)^2}}$.

and $\phi=\tan ^{-1}\frac{X_C-X_L}{R}$

$Z=\sqrt{R^2 + (X_C-X_L)^2}$ is called the impedance of the circuit.

• In a purely inductive or capacitive circuit, cosφ = 0 and no power is dissipated even though a current is flowing in the circuit. In such cases, current is referred to as a wattless current.

• An interesting characteristic of a series RLC circuit is the phenomenon of resonance. The circuit exhibits resonance, i.e., the amplitude of the current is maximum at the resonant frequency, $\omega_0=\frac{1}{\sqrt{LC}}.$

The quality factor Q defined by $Q=\frac{\omega_0 L}{R}=\frac{1}{\omega_0 CR}$  is an indicator of the sharpness of the resonance, the higher value of Q indicating sharper peak in the current.

• A circuit containing an inductor L and a capacitor C (initially charged) with no ac source and no resistors exhibits free oscillations. The charge q of the capacitor satisfies the equation of simple harmonic motion:

$\frac{d^2 q}{dt^2}+\frac{1}{LC}q=0$

and therefore, the frequency ω of free oscillation is $\omega_0=\frac{1}{\sqrt{LC}}.$ The energy in the system oscillates between the capacitor and the inductor but their sum or the total energy is constant in time.

• A transformer consists of an iron core on which are bound a primary coil of Np turns and a secondary coil of Ns turns. If the primary coil is connected to an ac source, the primary and secondary voltages and the currents are related by

$V_s=\left ( \frac{N_s}{N_p} \right )V_p$  and $I_s=\left ( \frac{N_p}{N_s} \right )I_p$

If the secondary coil has a greater number of turns than the primary, the voltage is stepped-up (Vs > Vp). This type of arrangement is called a step- up transformer. If the secondary coil has turns less than the primary, we have a step-down transformer.

• When a value is given for ac voltage or current, it is ordinarily the rms value. The voltage across the terminals of an outlet in your room is normally 240 V. This refers to the rms value of the voltage. The amplitude of this voltage is $\upsilon _m=\sqrt{2}=\sqrt{2}(240)=340V$

• The power rating of an element used in ac circuits refers to its average power rating.

• The power consumed in an ac circuit is never negative.

• In an ac circuit, while adding voltages across different elements, one should take care of their phases properly. For example, if VR and VC are voltages across R and C, respectively in an RC circuit, then the total voltage across RC combination is $V_{RC}=\sqrt{V^2_R + V^2_C}$  and not VR + VC since Vc is π/2 out of phase of VR.

• In a RLC circuit, resonance phenomenon occur when XL = XC or $\omega_0=\frac{1}{\sqrt{LC}}.$ For resonance to occur, the presence of both L and C elements in the circuit is a must. With only one of these (L or C) elements, there is no possibility of voltage cancellation and hence, no resonance is possible.

• The power factor in a RLC circuit is a measure of how close the circuit is to expending the maximum power.

• Displacement current is that current which comes into play in the region in which the electric field and the electric flux is changing with time. $I_D=\in _0= \frac{d\varphi_E}{dt}$

Maxwell modified Ampere’s circuital law in order to make the same logically consistent.

$\int \vec{B}\cdot d\vec{l}=\mu 0(I+I_{0})=\mu _{0}\left ( I+\varepsilon _{0}\dfrac{d\varphi _{E} }{dt} \right )$

• The orderly distribution of electromagnetic radiations according to their wavelength or frequency is called the electromagnetic spectrum.

• Radio waves are the electromagnetic wave of frequency range from 5 × 105 Hz to 109 Hz. These waves are produced by oscillating electric circuits having an inductor and capacitor.

• Microwaves are the electromagnetic waves of frequency range 1 GHz to 300 GHz. They are produced by special vacuum tubes. namely; klystrons, magnetrons and Gunn diodes etc.

• Infrared waves were discovered by Herschell. These are the electromagnetic waves of frequency range 3 × 1011 Hz to 4 × 1014 Hz. Infrared waves are sometimes called heat waves. Infrared waves are produced by hot bodies and molecules. These waves are not detected by the human eye but snakes can detect them.

• It is the narrow region of the electromagnetic spectrum, which is detected by the human eye. Its frequency ranges from 4×1014 Hz to 8×1014 Hz. It is produced due to atomic excitation.

• The ultraviolet rays were discovered by Ritter in 1801. The frequency range of ultraviolet rays is 8 × 1014 Hz to 5 × 1016 Hz. The ultraviolet rays are produced by sun, special lamps and very hot bodies.

• The X–rays were discovered by German Physicist W. Roentgen. Their frequency range is 1016 Hz to 3 × 1021 Hz. These are produced when high energy electrons are stopped suddenly on a metal of high atomic number. X–rays have high penetrating power.

• γ–rays are the electromagnetic waves of frequency range 3 × 1018 Hz to 5 × 1022 Hz. γ–rays have nuclear origin. These rays are highly energetic and are produced by the nucleus of the radioactive substances.

• Intensity of an electromagnetic wave at a point is defined as the energy crossing per second per unit area normally around that point during the propagation of electromagnetic waves.

$I=\frac{1}{2}\frac{B^2_0}{\mu_0}c = \frac{1}{\mu_0}B^2 \text{rms}^C$

## Alternating Current Class 12 NCERT Solutions PDF

1. A $100\Omega$ resistor is connected to a $220V$, $50Hz$ ac supply.

a) What is the rms value of current in the circuit?

Ans: It is given that,

Resistance, $R=100\Omega$

Voltage, $V=220V$

Frequency, $f=50Hz$

It is known that,

${{I}_{rms}}=\frac{{{V}_{rms}}}{R}$

$\Rightarrow {{I}_{rms}}=\frac{220}{100}=2.2A$

Therefore, the rms value of current in the circuit is${{I}_{rms}}=2.2A$.

b) What is the net power consumed over a full cycle?

Ans: It is known that,

$Power=V\times I$

$\Rightarrow Power=220\times 2.2$

$\Rightarrow Power=484W$

Therefore, the net power consumed over a full cycle is$484W$.

2.

a) The peak voltage of an ac supply is $300V$. What is the rms voltage?

Ans: It is given that,

Peak voltage of the ac supply, ${{V}_{0}}=300V$

It is known that,

${{V}_{rms}}=\frac{{{V}_{0}}}{\sqrt{2}}$

$\Rightarrow {{V}_{rms}}=\frac{300}{\sqrt{2}}$

$\Rightarrow {{V}_{rms}}=212.1V$

Therefore, the rms voltage is $212.1V$.

c) The rms value of current in an ac circuit is $10A$. What is the peak current?

Ans: It is given that,

Rms value of current in an ac circuit, ${{I}_{rms}}=10A$

It is known that,

${{I}_{0}}=\sqrt{2}\times {{I}_{rms}}$

$\Rightarrow {{I}_{0}}=1.414\times 10$

$\Rightarrow {{I}_{0}}=14.14A$

Therefore, the peak current is $14.14A$.

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

Ans: It is known that,

Inductance, $L=44mH=44\times {{10}^{-3}}H$

Voltage,$V=220V$

Frequency, ${{f}_{L}}=50Hz$

Angular frequency, ${{\omega }_{L}}=2\pi {{f}_{L}}$

It is known that,

Inductive reactance, ${{X}_{L}}={{\omega }_{L}}L=2\pi {{f}_{L}}L$

$\Rightarrow {{X}_{L}}=2\times 3.14\times 50\times 44\times {{10}^{-3}}\Omega$

$\Rightarrow {{X}_{L}}=13.8\Omega$

${{I}_{rms}}=\frac{V}{{{X}_{L}}}$

$\Rightarrow {{I}_{rms}}=\frac{220}{13.82}$

$\Rightarrow {{I}_{rms}}=15.92A$

Therefore, the rms value of the current in the circuit is $15.92A$.

4. A $60\mu F$ capacitor is connected to a $110V$,$60Hz$ ac supply. Determine the rms value of the current in the circuit.

Ans: It is given that,

Capacitance, $C=60\mu F=60\times {{10}^{-6}}F$

Voltage, $V=110V$

Frequency, ${{f}_{C}}=60Hz$

It is known that,

${{I}_{rms}}=\frac{V}{{{X}_{C}}}$

${{X}_{C}}=\frac{1}{{{\omega }_{C}}C}=\frac{1}{2\pi {{f}_{C}}C}$

$\Rightarrow {{X}_{C}}=\frac{1}{2\times 3.14\times 60\times 60\times {{10}^{-6}}}$

$\Rightarrow {{X}_{C}}=44.248\Omega$

$\Rightarrow {{I}_{rms}}=\frac{110}{44.28}$

$\Rightarrow {{I}_{rms}}=2.488A$

Therefore, the rms value of the current in the circuit is $2.488A$.

5. In exercises $4$ and $5$ What is the net power absorbed by each circuit over a complete cycle? Explain your answer.

Ans: From the inductive circuit,

Rms value of current, ${{I}_{rms}}=15.92A$

Rms value of voltage, ${{V}_{rms}}=220V$

It is known that,

Net power absorbed, $P={{V}_{rms}}\times {{I}_{rms}}\cos \phi$

Where,

$\phi$ is the phase difference between voltage and current

For a pure inductive circuit, the phase difference between alternating voltage and current is ${{90}^{0}}$i.e., $\phi ={{90}^{0}}$

$\Rightarrow P=220\times 15.92\cos {{90}^{0}}=0$

Therefore, net power absorbed is zero in a pure inductive circuit.

In a capacitive circuit,

Rms value of current, ${{I}_{rms}}=2.49A$

Rms value of voltage, ${{V}_{rms}}=110V$

It is known that,

Net power absorbed, $P={{V}_{rms}}\times {{I}_{rms}}\cos \phi$

Where,

$\phi$ is the phase difference between voltage and current

For a pure capacitive circuit, the phase difference between alternating voltage and current is ${{90}^{0}}$i.e., $\phi ={{90}^{0}}$

$\Rightarrow P=110\times 2.49\cos {{90}^{0}}=0$

Therefore, net power absorbed is zero in a pure capacitive circuit.

6. Obtain the resonant frequency ${{\omega }_{r}}$ of a series LCR circuit with $L=2.0H$, $C=32\mu F$ and $R=10\Omega$ . What is the Q-value of this current?

Ans: It is given that,

Inductance, $L=2H$

Capacitance, $C=32\mu F=32\times {{10}^{-6}}F$

$R=10\Omega$

It is known that,

Resonant frequency, ${{\omega }_{r}}=\frac{1}{\sqrt{LC}}$

$\Rightarrow {{\omega }_{r}}=\frac{1}{\sqrt{2\times 32\times {{10}^{-6}}}}$

$\Rightarrow {{\omega }_{r}}=\frac{1}{8\times {{10}^{-3}}}$

$\Rightarrow {{\omega }_{r}}=125{rad}/{s}\;$

$Q-value=\frac{{{\omega }_{r}}L}{R}$

$\Rightarrow Q=\frac{1}{R}\sqrt{\frac{L}{C}}$

$\Rightarrow Q=\frac{1}{10}\sqrt{\frac{2}{32\times {{10}^{-6}}}}$

$\Rightarrow Q=\frac{1}{10\times 4\times {{10}^{-3}}}$

$\Rightarrow Q=25$

Therefore, the resonant frequency is $125rad/s$ and Q-value is $25$.

7. A charged $30\mu F$ capacitor is connected to a $27mH$ inductor. What is the angular frequency of free oscillations of the circuit?

Ans: It is given that,

Capacitance, $C=30\mu F=30\times {{10}^{-6}}F$

Inductance, $L=27mH=27\times {{10}^{-3}}H$

It is known that,

Angular frequency of free oscillations, ${{\omega }_{r}}=\frac{1}{\sqrt{LC}}$

$\Rightarrow {{\omega }_{r}}=\frac{1}{\sqrt{27\times {{10}^{-3}}\times 30\times {{10}^{-6}}}}$

$\Rightarrow {{\omega }_{r}}=\frac{1}{9\times {{10}^{-4}}}$

$\Rightarrow {{\omega }_{r}}=1.11\times {{10}^{3}}rad/s$

Therefore, the angular frequency of free oscillations of the circuit is $1.11\times {{10}^{3}}rad/s$.

8. Suppose the initial charge on the capacitor in exercise $7$ is $6mC$ . What is the total energy stored in the circuit initially? What is the total energy at a later time?

Ans: It is known that,

Capacitance of the capacitor, $C=30\mu F=30\times {{10}^{-6}}F$

Inductance of the capacitor, $L=27mH=27\times {{10}^{-3}}H$

Charge on the capacitor, $Q=6mC=6\times {{10}^{-3}}C$

It is known that,

Energy,$E=\frac{1}{2}\frac{{{Q}^{2}}}{C}$

$\Rightarrow E=\frac{1}{2}\frac{{{(6\times {{10}^{-3}})}^{2}}}{30\times {{10}^{-6}}}$

$\Rightarrow E=\frac{6}{10}=0.6J$

Therefore, the energy stored in the circuit initially is $E=0.6J$.

Total energy at later time will remain same as the initially stored i.e., $0.6J$ because energy is shared between the capacitor and the inductor.

9. A series LCR circuit with $R=20\Omega$, $L=1.5H$ and $C=35\mu F$ is connected to a variable frequency $200V$ ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Ans: It is known that,

Resistance, $R=20\Omega$

Inductance, $L=1.5H$

Capacitance, $C=35\mu F=35\times {{10}^{-6}}F$

Voltage, $V=200V$

It is known that,

Impedance,$Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

At resonance, ${{X}_{L}}={{X}_{C}}$

$\Rightarrow Z=R=20\Omega$

$I=\frac{V}{Z}=\frac{200}{20}$

$\Rightarrow I=10A$

Average power, $P={{I}^{2}}R$

$\Rightarrow P={{10}^{2}}\times 20$

$\Rightarrow P=2000W$

Therefore, the average power transferred is $2000W$.

10. A radio can tune over the frequency range of a portion of $MW$ broadcast band: ($800kHz$ to $1200kHz$). If its LC circuit has an effective inductance of $200\mu H$, what must be the range of its variable capacitor?

(Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radio wave.)

Ans: It is given that,

The range of frequency$(f)$ of a radio is $800kHz$ to $1200kHz$.

Effective inductance of the circuit, $L=200\mu H=200\times {{10}^{-6}}H$

It is known that,

Capacitance of variable capacitor for ${{f}_{1}}$ is ${{C}_{1}}=\frac{1}{{{\omega }_{1}}^{2}L}$

Where,

${{\omega }_{1}}$ is the angular frequency for capacitor for ${{f}_{1}}=2\pi {{f}_{1}}$

$\Rightarrow {{\omega }_{1}}=2\times 3.14\times 800\times {{10}^{3}}rad/s$

$\Rightarrow {{C}_{1}}=\frac{1}{{{(2\times 3.14\times 800\times {{10}^{3}})}^{2}}\times 200\times {{10}^{-6}}}$

$\Rightarrow {{C}_{1}}=1.9809\times {{10}^{-10}}F$

$\Rightarrow {{C}_{1}}=198.1pF$

${{C}_{2}}=\frac{1}{{{\omega }_{2}}^{2}L}$

$\Rightarrow {{C}_{2}}=\frac{1}{{{(2\times 3.14\times 1200\times {{10}^{3}})}^{2}}\times 200\times {{10}^{-6}}}$

$\Rightarrow {{C}_{2}}=0.8804\times {{10}^{-10}}F$

$\Rightarrow {{C}_{2}}=88.04pF$

Therefore, the range of the variable capacitor is from $88.04pF$ to $198.1pF$.

11. Figure shows a series LCR circuit connected to a variable frequency $230V$ source. $L=5.0H$, $C=80\mu F$, $R=40\Omega$ .

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

Ans: It is given that,

Voltage, $V=230V$

Inductance, $L=5.0H$

Capacitance, $C=80\mu F=80\times {{10}^{-6}}F$

Resistance, $R=40\Omega$

It is known that,

Source frequency at resonance$=\frac{1}{\sqrt{LC}}$

$\Rightarrow \frac{1}{\sqrt{5\times 80\times {{10}^{-6}}}}=50rad/s$

Therefore, the source frequency of the circuit in resonance is $50rad/s$.

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

Ans: It is known that,

At resonance, Impedance,$Z=$ Resistance,$R$

$\Rightarrow Z=R=40\Omega$

$I=\frac{V}{Z}$

$\Rightarrow I=\frac{230}{40}=5.75A$

Amplitude, ${{I}_{0}}=1.414\times I$

$\Rightarrow {{I}_{0}}=1.414\times 5.75$

$\Rightarrow {{I}_{0}}=8.13A$

Therefore, the impedance of the circuit is $40\Omega$ and the amplitude of current at resonating frequency is $8.13A$.

c) 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.

Ans: It is known that,

Potential drop, $V=IR$

Across resistor, ${{V}_{R}}=IR$

$\Rightarrow {{V}_{R}}=5.75\times 40=230V$

Across capacitor, ${{V}_{C}}=I{{X}_{C}}=\frac{I}{\omega C}$

$\Rightarrow {{V}_{C}}=5.75\times \frac{1}{50\times 80\times {{10}^{-6}}}$

$\Rightarrow {{V}_{C}}=1437.5V$

Across Inductor, ${{V}_{L}}=I{{X}_{L}}=I\omega L$

$\Rightarrow {{V}_{L}}=5.75\times 50\times 5$

$\Rightarrow {{V}_{L}}=1437.5V$

Across LC combination, ${{V}_{LC}}=I({{X}_{L}}-{{X}_{C}})$

At resonance, ${{X}_{L}}={{X}_{C}}$

$\Rightarrow {{V}_{LC}}=0$

Therefore, the rms potential drop across Resistor is $230V$, Capacitor is $1437.5V$, Inductor is $1437.5V$ and the potential drop across LC combination is zero at resonating frequency.

12. An LC circuit contains a $20mH$ inductor and a $50\mu F$ capacitor with an initial charge of $10mC$. The resistance of the circuit is negligible. Let the instant the circuit is closed be $t=0$.

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

Ans: It is given that,

Inductance of the inductor, $L=20mH=20\times {{10}^{-3}}H$

Capacitance of the capacitor, $C=50\mu F=50\times {{10}^{-6}}F$

Initial charge on the capacitor, $Q=10mC=10\times {{10}^{-3}}C$

It is known that,

Total energy stored initially in the circuit, $E=\frac{1}{2}\frac{{{Q}^{2}}}{C}$

$\Rightarrow E=\frac{{{(10\times {{10}^{-3}})}^{2}}}{2\times 50\times {{10}^{-6}}}=1J$

Therefore, the total energy stored in the LC circuit will be conserved because there is no resistor $(R=0)$ connected in the circuit.

b) What is the natural frequency of the circuit?

Ans: It is known that,

Natural frequency of the circuit, $\nu =\frac{1}{2\pi \sqrt{LC}}$

$\Rightarrow \nu =\frac{1}{2\pi \sqrt{20\times {{10}^{-3}}\times 50\times {{10}^{-6}}}}$

$\Rightarrow \nu =\frac{{{10}^{3}}}{2\pi }=159.24Hz$

Natural angular frequency, ${{\omega }_{r}}=\frac{1}{\sqrt{LC}}$

$\Rightarrow {{\omega }_{r}}=\frac{1}{\sqrt{20\times {{10}^{-3}}\times 50\times {{10}^{-6}}}}$

$\Rightarrow {{\omega }_{r}}=\frac{1}{\sqrt{{{10}^{-6}}}}={{10}^{3}}rad/s$

Therefore, the natural frequency is $159.24Hz$ and the natural angular frequency is ${{10}^{3}}rad/s$.

c) At what time is the energy stored $(i)$ completely electrical (i.e., stored in the capacitor)? $(ii)$ completely magnetic (i.e., stored in the inductor)?

Ans:

(i) Completely electrical:

It is known that,

Time period for LC oscillations, $T=\frac{1}{\nu }$

$\Rightarrow T=\frac{1}{159.24}=6.28ms$

Total charge on the capacitor at time $t$, $Q'=Q\cos \left( \frac{2\pi }{T}t \right)$

If energy stored is electrical, $Q'=\pm Q$

Therefore, it can be inferred that the energy stored in the capacitor is completely electrical at time, $t=0,\frac{T}{2},T,\frac{3T}{2},..............$ where, $T=6.3ms$.

(ii) Completely magnetic:

Magnetic energy is maximum, when electrical energy $Q'$ is equal to $0$.

Therefore, it can be inferred that the energy stored is completely magnetic at time, $t=\frac{T}{4},\frac{3T}{4},\frac{5T}{4},...........$ where, $T=6.3ms$.

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

Ans: Consider, $Q'$ be the charge on capacitor when total energy is equally shared between the capacitor and the inductor at time $t$.

When total energy is equally shared between the inductor and capacitor, the energy stored in the capacitor$={}^{1}/{}_{2}($maximum energy$)$.

$\Rightarrow \frac{1}{2}\frac{{{\left( Q' \right)}^{2}}}{C}=\frac{1}{2}\left( \frac{1}{2}\frac{{{Q}^{2}}}{C} \right)$

$\Rightarrow \frac{1}{2}\frac{{{\left( Q' \right)}^{2}}}{C}=\frac{1}{4}\frac{{{Q}^{2}}}{C}$

$\Rightarrow Q'=\frac{Q}{\sqrt{2}}$

It is known that, $Q'=Q\cos \frac{2\pi }{T}t$

$\Rightarrow \frac{Q}{\sqrt{2}}=Q\cos \frac{2\pi }{T}t$

$\Rightarrow \cos \frac{2\pi }{T}t=\frac{1}{\sqrt{2}}=\cos (2n+1)\frac{\pi }{4};$ $n=0,1,2,3.....$

$\Rightarrow t=(2n+1)\frac{T}{8}$

Therefore, the total energy is equally shared between the inductor and the capacitor at the time,$t=\frac{T}{8},\frac{3T}{8},\frac{5T}{8}...........$.

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

Ans: If a resistor is included in the circuit, then the total initial energy gets dissipated as heat energy in the circuit. The LC oscillation gets damped due to the resistance.

13. A coil of inductance $0.5H$ and resistance $100\Omega$ is connected to a $240V,50Hz$ ac supply.

a) What is the maximum current in the coil?

Ans: It is given that,

Inductance of the inductor, $L=0.5H$

Resistance of the resistor, $R=100\Omega$

Potential of the supply voltage, $V=240V$

Frequency of the supply, $\nu =50Hz$

It is known that,

Peak voltage, ${{V}_{0}}=\sqrt{2}V$

$\Rightarrow {{V}_{0}}=\sqrt{2}\times 240$

$\Rightarrow {{V}_{0}}=339.41V$

Angular frequency of the supply, $\omega =2\pi \nu$

$\Rightarrow \omega =2\pi \times 50=100\pi rad/s$

Maximum current in the circuit, ${{I}_{0}}=\frac{{{V}_{0}}}{\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}}$

$\Rightarrow {{I}_{0}}=\frac{339.41}{\sqrt{{{(100)}^{2}}+{{(100\pi )}^{2}}{{(0.50)}^{2}}}}=1.82A$

Therefore, the maximum current in the coil is $1.82A$.

b) What is the time lag between the voltage maximum and the current maximum?

Ans: It is known that,

Equation for voltage, $V={{V}_{0}}\cos \omega t$

Equation for current, $I={{I}_{0}}\cos (\omega t-\phi )$

Where,

$\phi$ is the phase difference between voltage and current.

At time $t=0$, $V={{V}_{0}}$ (voltage is maximum)

If $\omega t-\phi =0$ i.e., at $t=\frac{\phi }{\omega }$ , $I={{I}_{0}}$ (current is maximum)

Therefore, the time lag between maximum voltage and maximum current is $\frac{\phi }{\omega }$ .

$\Rightarrow \tan \phi =\frac{\omega L}{R}$

$\Rightarrow \tan \phi =\frac{2\pi \times 50\times 0.5}{100}=1.57$

$\Rightarrow \phi ={{\tan }^{-1}}(1.57)$

$\Rightarrow \phi ={{57.5}^{\circ }}=\frac{57.5\pi }{180}rad$

Time lag, $t=\frac{\phi }{\omega }$

$\Rightarrow t=\frac{57.5\pi }{180\times 2\pi \times 50}$

$\Rightarrow t=3.19\times {{10}^{-3}}s$

$\Rightarrow t=3.2ms$

Therefore, the time lag between the maximum voltage and maximum current is $3.2ms$.

14. Obtain the answers $(a)$ to $(b)$ in Exercise $13$ 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?

Ans: It is given that,

Inductance of the inductor, $L=0.5H$

Resistance of the resistor, $R=100\Omega$

Potential of the supply voltage, $V=240V$

Frequency of the supply, $\nu =10kHz={{10}^{4}}Hz$

Angular frequency, $\omega =2\pi \nu =2\pi \times {{10}^{4}}rad/s$

Peak Voltage, ${{V}_{0}}=V\sqrt{2}=110\sqrt{2}V$

1. Maximum current, ${{I}_{0}}=\frac{{{V}_{0}}}{\sqrt{{{R}^{2}}+{{\omega }^{2}}{{C}^{2}}}}$

$\Rightarrow {{I}_{0}}=\frac{240\sqrt{2}}{\sqrt{{{(100)}^{2}}+{{(2\pi \times {{10}^{4}})}^{2}}\times {{(0.5)}^{2}}}}=1.1\times {{10}^{-2}}A$

Therefore, the maximum current in the coil is $1.1\times {{10}^{-2}}A$.

1. The time lag between maximum voltage and maximum current is $\frac{\phi }{\omega }$ .

For phase difference $\phi$:$\tan \phi =\frac{\omega L}{R}$

$\Rightarrow \tan \phi =\frac{2\pi \times {{10}^{4}}\times 0.5}{100}=100\pi$

$\Rightarrow \phi ={{\tan }^{-1}}(100\pi )$

$\Rightarrow \phi ={{89.82}^{\circ }}=\frac{89.82\pi }{180}rad$

Time lag, $t=\frac{\phi }{\omega }$

$\Rightarrow t=\frac{89.82\pi }{180\times 2\pi \times {{10}^{4}}}$

$\Rightarrow t=25\times {{10}^{-6}}s$

$\Rightarrow t=25\mu s$

Therefore, the time lag between the maximum voltage and maximum current is $25\mu s$.

It can be observed that ${{I}_{0}}$ is very small in this case.

Thus, at high frequencies, the inductor amounts to an open circuit.

In a dc circuit, after a steady state is achieved, $\omega =0$. Thus, inductor $L$ behaves like a pure conducting object.

15. A $100\mu F$ capacitor in series with a $40\Omega$ resistance is connected to a $110V,60Hz$ supply.

a) What is the maximum current in the circuit?

Ans: It is given that,

Capacitance of the capacitor, $C=100\mu F=100\times {{10}^{-6}}F$

Resistance of the resistor, $R=40\Omega$

Supply voltage, $V=110V$

Frequency oscillations, $\nu =60Hz$

Angular frequency, $\omega =2\pi \nu =2\pi \times 60rad/s$

It is known that,

For a RC circuit, Impedance: $Z=\sqrt{{{R}^{2}}+\frac{1}{{{\omega }^{2}}{{C}^{2}}}}$

Peak Voltage, ${{V}_{0}}=V\sqrt{2}=110\sqrt{2}V$

Maximum current; ${{I}_{0}}=\frac{{{V}_{0}}}{Z}$

$\Rightarrow {{I}_{0}}=\frac{{{V}_{0}}}{\sqrt{{{R}^{2}}+\frac{1}{{{\omega }^{2}}{{C}^{2}}}}}$

$\Rightarrow {{I}_{0}}=\frac{110\sqrt{2}}{\sqrt{{{(40)}^{2}}+\frac{1}{{{(120\pi )}^{2}}{{({{10}^{-4}})}^{2}}}}}$

$\Rightarrow {{I}_{0}}=\frac{110\sqrt{2}}{\sqrt{1600+\frac{1}{{{(120\pi )}^{2}}{{({{10}^{-4}})}^{2}}}}}=3.24A$

Therefore, the maximum current in the circuit is $3.24A$.

b) What is the time lag between the current maximum and the voltage maximum?

Ans: It is known that,

In a capacitor circuit, the voltage lags behind the current by a phase angle of $\phi$.

$\tan \phi =\frac{\frac{1}{\omega C}}{R}=\frac{1}{\omega CR}$

$\Rightarrow \tan \phi =\frac{1}{120\pi \times {{10}^{-4}}\times 40}=0.6635$

$\Rightarrow \phi ={{\tan }^{-1}}(0.6635)$

$\Rightarrow \phi ={{33.56}^{\circ }}=\frac{33.56\pi }{180}rad$

It is known that,

Time lag, $t=\frac{\phi }{\omega }$

$\Rightarrow t=\frac{33.56\pi }{180\times 120\pi }$

$\Rightarrow t=1.55\times {{10}^{-3}}s$

$\Rightarrow t=1.55ms$

Therefore, the time lag between maximum current and maximum voltage is $1.55ms$.

16. Obtain the answers to $(a)$ and $(b)$ in Exercise $15$ if the circuit is connected to a $\mathbf{110V},\mathbf{12kHz}$ 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.

Ans: It is given that,

Capacitance of the capacitor, $C=100\mu F=100\times {{10}^{-6}}F$

Resistance of the resistor, $R=40\Omega$

Supply voltage, $V=110V$

Frequency oscillations, $\nu =12kHz=12\times {{10}^{3}}Hz$

Angular frequency, $\omega =2\pi \nu =2\pi \times 12\times {{10}^{3}}rad/s=24\pi \times {{10}^{3}}rad/s$

Peak Voltage, ${{V}_{0}}=V\sqrt{2}=110\sqrt{2}V$

1. It is known that,

For a RC circuit, Impedance: $Z=\sqrt{{{R}^{2}}+\frac{1}{{{\omega }^{2}}{{C}^{2}}}}$

Maximum current; ${{I}_{0}}=\frac{{{V}_{0}}}{Z}$

$\Rightarrow {{I}_{0}}=\frac{{{V}_{0}}}{\sqrt{{{R}^{2}}+\frac{1}{{{\omega }^{2}}{{C}^{2}}}}}$

$\Rightarrow {{I}_{0}}=\frac{110\sqrt{2}}{\sqrt{{{(40)}^{2}}+\frac{1}{{{(24\pi \times {{10}^{3}})}^{2}}{{({{10}^{-4}})}^{2}}}}}$

$\Rightarrow {{I}_{0}}=\frac{110\sqrt{2}}{\sqrt{1600+{{\left( \frac{10}{24\pi } \right)}^{2}}}}=3.9A$

Therefore, the maximum current in the circuit is $3.9A$.

1. It is known that,

In a capacitor circuit, the voltage lags behind the current by a phase angle of $\phi$.

$\tan \phi =\frac{\frac{1}{\omega C}}{R}=\frac{1}{\omega CR}$

$\Rightarrow \tan \phi =\frac{1}{24\pi \times {{10}^{3}}\times {{10}^{-4}}\times 40}=\frac{1}{96\pi }$

$\Rightarrow \phi ={{\tan }^{-1}}(\frac{1}{96\pi })$

$\Rightarrow \phi ={{0.2}^{\circ }}=\frac{0.2\pi }{180}rad$

It is known that,

Time lag, $t=\frac{\phi }{\omega }$

$\Rightarrow t=\frac{0.2\pi }{180\times 24\pi \times {{10}^{3}}}$

$\Rightarrow t=0.04\times {{10}^{-6}}s$

$\Rightarrow t=0.04\mu s$

Therefore, the time lag between maximum current and maximum voltage is $0.04\mu s$.

It can be concluded that $\phi$ tends to become zero at high frequencies. At a high frequency, capacitor $C$ acts as a conductor.

In a dc circuit, after the steady state is achieved, $\omega =0$. Therefore, capacitor $C$ amounts to an open circuit.

17. Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if 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 11 for this frequency.

Ans: It is given that,

An inductor $(L)$, a capacitor $(C)$ and a resistor $(R)$ is connected in parallel with each other in a circuit where,

Inductance, $L=5.0H$

Capacitance, $C=80\mu F=80\times {{10}^{-6}}F$

Resistance, $R=40\Omega$

Potential of the voltage source, $V=230V$

It is known that,

Impedance $(Z)$ of the given LCR circuit is given as:

$\frac{1}{Z}=\sqrt{\frac{1}{{{R}^{2}}}+{{\left( \frac{1}{\omega L}-\omega C \right)}^{2}}}$

Where,

$\omega$ is the angular frequency

At resonance: $\frac{1}{\omega L}-\omega C=0$

$\Rightarrow \omega =\frac{1}{\sqrt{LC}}$

$\Rightarrow \omega =\frac{1}{\sqrt{5\times 80\times {{10}^{-6}}}}=50rad/s$

Therefore, the magnitude of $Z$ is maximum at $50rad/s$ and the total current is minimum.

Rms current flowing through inductor $L$: ${{I}_{L}}=\frac{V}{\omega L}$

$\Rightarrow {{I}_{L}}=\frac{230}{50\times 5}=0.92A$

Rms current flowing through capacitor $C$: ${{I}_{C}}=\frac{V}{\frac{1}{\omega C}}=\omega CV$

$\Rightarrow {{I}_{C}}=50\times 80\times {{10}^{-6}}\times 230=0.92A$

Rms current flowing through resistor $R$: ${{I}_{R}}=\frac{V}{R}$

$\Rightarrow {{I}_{R}}=\frac{230}{40}=5.75A$

The current RMS value in the inductor is $0.92A$, in the capacitor is $0.92A$ and in the resistor is $5.75A$.

18. A circuit containing an $80mH$ inductor and a $60\mu F$ capacitor in series is connected to a $\mathbf{230V},\mathbf{50Hz}$ supply. The resistance of the circuit is negligible.

a) Obtain the current amplitude and rms values.

Ans: It is given that,

Inductance, $L=80mH=80\times {{10}^{-3}}H$

Capacitance, $C=60\mu F=60\times {{10}^{-6}}F$

Supply voltage, $V=230V$

Frequency, $\nu =50Hz$

Angular frequency, $\omega =2\pi \nu =100\pi rad/s$

Peak voltage, ${{V}_{0}}=V\sqrt{2}=230\sqrt{2}V$

It is known that,

Maximum current: ${{I}_{0}}=\frac{{{V}_{0}}}{(\omega L-\frac{1}{\omega C})}$

$\Rightarrow {{I}_{0}}=\frac{230\sqrt{3}}{(100\pi \times 80\times {{10}^{-3}}-\frac{1}{100\pi \times 60\times {{10}^{-6}}})}$

$\Rightarrow {{I}_{0}}=\frac{230\sqrt{3}}{(8\pi -\frac{1000}{6\pi })}=-11.63A$

The negative sign is because $\omega L<\frac{1}{\omega C}$

Amplitude of maximum current, $\left| {{I}_{0}} \right|=11.63A$

$\Rightarrow I=\frac{{{I}_{0}}}{\sqrt{2}}=\frac{-11.63}{\sqrt{2}}$

$\Rightarrow I=-8.22A$, which is the rms value of current.

b) Obtain the rms values of potential drops across each element.

Ans: It is known that,

Potential difference across the inductor, ${{V}_{L}}=I\times \omega L$

$\Rightarrow {{V}_{L}}=8.22\times 100\pi \times 80\times {{10}^{-3}}$

$\Rightarrow {{V}_{L}}=206.61V$

Potential difference across the capacitor, ${{V}_{C}}=I\times \frac{1}{\omega C}$

$\Rightarrow {{V}_{C}}=8.22\times \frac{1}{100\pi \times 60\times {{10}^{-6}}}$

$\Rightarrow {{V}_{C}}=436.3V$, which is the rms value of potential drop.

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

Ans: Average power transferred to the inductor is zero as actual voltage leads the current by $\frac{\pi }{2}$.

d) What is the average power transferred to the capacitor?

Ans: Average power transferred to the capacitor is zero as actual voltage lags the current by $\frac{\pi }{2}$.

e) What is the total average power absorbed by the circuit? (‘Average’ implies ‘averaged over one cycle’.)

Ans: The total average power absorbed (averaged over one cycle) is zero.

19. Suppose the circuit in Exercise $18$  has a resistance of $15\Omega$ . Obtain the average power transferred to each element of the circuit, and the total power absorbed.

Ans:  It is given that,

Average power transferred to the resistor $=788.44W$

Average power transferred to the capacitor $=0W$

Total power absorbed by the circuit $=788.44W$

Inductance of inductor, $L=80mH=80\times {{10}^{-3}}H$

Capacitance of capacitor, $C=60\mu F=60\times {{10}^{-6}}F$

Resistance of resistor, $R=15\Omega$

Potential of voltage supply, $V=230V$

Frequency of signal, $\nu =50Hz$

Angular frequency of signal, $\omega =2\pi \nu =2\pi \times \left( 50 \right)=100\pi rad/s$

It is known that,

Impedance, $Z=\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}$

$\Rightarrow Z=\sqrt{{{(15)}^{2}}+{{\left( 100\pi (80\times {{10}^{-3}})-\frac{1}{\left( 100\pi \times 60\times {{10}^{-6}} \right)} \right)}^{2}}}$

$\Rightarrow Z=\sqrt{{{(15)}^{2}}+{{\left( 25.12-53.08 \right)}^{2}}}=31.728\Omega$

Now,

$I=\frac{V}{Z}$

$\Rightarrow I=\frac{230}{31.728}=7.25A$

The elements are connected in series to each other. Therefore, impedance of the circuit is given as current flowing in the circuit,

Average power transferred to resistance is given as: ${{P}_{R}}={{I}^{2}}R$

$\Rightarrow {{P}_{R}}={{\left( 7.25 \right)}^{2}}\times 15=788.44W$

Average power transferred to capacitor,${{P}_{C}}=$ Average power transferred to inductor, ${{P}_{L}}=0$

Total power absorbed by the circuit: ${{P}_{T}}={{P}_{R}}+{{P}_{C}}+{{P}_{L}}$

${{P}_{T}}=788.44+0+0=788.44W$

Therefore, the total power absorbed by the circuit is $788.44W$.

20. A series LCR circuit with $\mathbf{L}=\mathbf{0}.\mathbf{12H},\mathbf{C}=\mathbf{480nF},\mathbf{R}=\mathbf{23}\text{ }\mathbf{\Omega }$ is connected to a $230V$ variable frequency supply.

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

Ans: It is given that,

Inductance, $L=0.12H$

Capacitance, $C=480nF=480\times {{10}^{-9}}F$

Resistance, $R=23\Omega$

Supply voltage, $V=230V$

Peak voltage, ${{V}_{0}}=230\times \sqrt{2}=325.22V$

It is known that,

Current flowing in the circuit, ${{I}_{0}}=\frac{{{V}_{0}}}{\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}}$

Where,

${{I}_{0}}$ is maximum at resonance.

At resonance: ${{\omega }_{R}}L-\frac{1}{{{\omega }_{R}}C}=0$

Where,

${{\omega }_{R}}$is the resonance angular frequency

${{\omega }_{R}}=\frac{1}{\sqrt{LC}}$

$\Rightarrow {{\omega }_{R}}=\frac{1}{\sqrt{0.12\times 480\times {{10}^{-9}}}}$

$\Rightarrow {{\omega }_{R}}=4166.67rad/s$

Resonant frequency, ${{\nu }_{R}}=\frac{{{\omega }_{R}}}{2\pi }$

$\Rightarrow {{\nu }_{R}}=\frac{4166.67}{2\times 3.14}=663.48Hz$

Maximum current, ${{\left( {{I}_{0}} \right)}_{Max}}=\frac{{{V}_{0}}}{R}$

$\Rightarrow {{\left( {{I}_{0}} \right)}_{Max}}=\frac{325.22}{23}=14.14A$

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

Ans: It is known that,

Maximum average power absorbed by the circuit; ${{({{P}_{V}})}_{Max}}=\frac{1}{2}\left( {{I}_{0}} \right)_{Max}^{2}R$

$\Rightarrow {{({{P}_{V}})}_{Max}}=\frac{1}{2}\times {{(14.14)}^{2}}\times 23$

$\Rightarrow {{({{P}_{V}})}_{Max}}=2299.3W$

Therefore, the resonant frequency, ${{\nu }_{R}}=663.48Hz$

c) 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?

Ans: It is known that,

The power transferred to the circuit is half the power at resonant frequency.

Frequencies at which power transferred is half, $={{\omega }_{R}}\pm \Delta \omega =2\pi ({{\nu }_{R}}\pm \Delta \nu )$

Where,

$\Delta \omega =\frac{R}{2L}$

$\Rightarrow \Delta \omega =\frac{23}{2\times 0.12}=95.83rad/s$

Therefore, the change in frequency, $\Delta \nu =\frac{1}{2\pi }\Delta \omega$

$\Delta \nu =\frac{95.83}{2\pi }=15.26Hz$

${{\nu }_{R}}+\Delta \nu =663.48+15.26=678.74Hz$

${{\nu }_{R}}-\Delta \nu =663.48-15.26=648.22Hz$

Therefore, at $648.22Hz$ and $678.74Hz$ frequencies, the power transferred is half.

At these frequencies, current amplitude:$I'=\frac{1}{\sqrt{2}}\times {{\left( {{I}_{0}} \right)}_{Max}}$

$\Rightarrow I'=\frac{14.14}{\sqrt{2}}=10A$

Therefore, the current amplitude is $10A$.

d) What is the Q-factor of the given circuit?

Ans: It is known that,

Q-factor of the given circuit, $Q=\frac{{{\omega }_{r}}L}{R}$

$\Rightarrow Q=\frac{4166.67\times 0.12}{23}=21.74$

Therefore, the Q-factor of the given circuit is $21.74$.

21. Obtain the resonant frequency and Q-factor of a series LCR circuit with $L=3.0H$ ,$C=27\mu F$ and $R=7.4\Omega$. 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.

Ans: It is given that,

Inductance, $L=3.0H$

Capacitance, $C=27\mu F=27\times {{10}^{-6}}F$

Resistance,$R=7.4\Omega$

It is known that,

At resonance, angular frequency of the source for the given LCR series circuit is ${{\omega }_{r}}=\frac{1}{\sqrt{LC}}$

$\Rightarrow {{\omega }_{r}}=\frac{1}{\sqrt{3\times 27\times {{10}^{-6}}}}$

$\Rightarrow {{\omega }_{r}}=\frac{{{10}^{3}}}{9}=111.11rad/s$

Therefore, the resonant frequency is $111.11rad/s$.

Q-factor of the series, $Q=\frac{{{\omega }_{r}}L}{R}$

$\Rightarrow Q=\frac{111.11\times 3}{7.4}=45.0446$

Therefore, the Q-factor is $45.0446$.

To improve the sharpness of the resonance by reducing ‘full width at half maximum’ by a factor of $2$without changing ${{\omega }_{r}}$ , reduce the resistance to half.

$\Rightarrow R=\frac{7.4}{2}=3.7\Omega$

Therefore, required resistance is $3.7\Omega$.

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?

Ans: Yes, in any ac circuit, the applied instantaneous voltage is equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit.

The same is not true for rms voltage because voltages across different elements may not be in phase.

b) A capacitor is used in the primary circuit of an induction coil.

Ans:  Yes, a capacitor is used in the primary circuit of an induction coil.

This is because, when the circuit is broken, a high induced voltage is used to charge the capacitor to avoid sparks.

c) 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.

Ans:  The dc signal will appear across capacitor $C$ because for dc signals, the impedance of an inductor $L$ is negligible while the impedance of a capacitor $C$ is very high (almost infinite).

Therefore, a dc signal appears across$C$.

For an ac signal of high frequency, the impedance of $L$is high and that of $C$ is very low.

Thus, an ac signal of high frequency appears across $L$.

d) 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.

Ans: When an iron core is inserted in the choke coil (which is in series with a lamp connected to an ac line), the lamp will glow dimly.

This is because the choke coil and the iron core increase the impedance of the circuit.

e) Why is 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?

Ans: As the choke coil reduces the voltage across the tube without wasting much power, it is used in the fluorescent tubes with ac mains. An ordinary resistor cannot be used instead of choke coil because it wastes power in the form of heat.

23. A power transmission line feeds input power at $2300V$ to a step-down transformer with its primary windings having $4000$ turns. What should be the number of turns in the secondary in order to get output power at $230V$?

Ans: It is given that,

Input voltage, ${{V}_{1}}=2300V$

Number of turns in primary coil, ${{n}_{1}}=4000$

Output voltage, ${{V}_{2}}=230V$

Number of turns in secondary coil ,${{n}_{2}}=?$

It is known that,

Voltage is related to number of terms: $\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{n}_{1}}}{{{n}_{2}}}$

$\Rightarrow \frac{2300}{230}=\frac{4000}{{{n}_{2}}}$

$\Rightarrow {{n}_{2}}=\frac{4000\times 230}{2300}=400$

Therefore, the number of turns in the second winding is $400$.

24. At a hydroelectric power plant, the water pressure head is at a height of $300m$ and the water flow available is $100{{m}^{3}}/s$ . If the turbine generator efficiency is $60%$ , estimate the electric power available from the plant ($g=9.8m/{{s}^{2}}$ ).

Ans: It is known that,

Height of water pressure head, $\mathbf{h}=\mathbf{300m}$

Volume of water flow per second, $V=100{{m}^{3}}/s$

Efficiency of turbine generator, $\mathbf{n}=\mathbf{60}%=\mathbf{0}.\mathbf{6}$

Acceleration due to gravity, $g=9.8m/{{s}^{2}}$

Density of water, $\mathbf{\rho }=\mathbf{1}{{\mathbf{0}}^{3}}\mathbf{kg}/{{\mathbf{m}}^{3}}$

It is known that,

Electric power available from the plant$=\eta \times h\rho gV$

$\Rightarrow P=0.6\times 300\times {{10}^{3}}\times 9.8\times 100$

$\Rightarrow P=176.4\times {{10}^{6}}W$

$\Rightarrow P=176.4MW$

Therefore, the estimated electric power available from the plant is $176.4MW$.

25. 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 $0.5\Omega per km$. The town gets power from the line through a $400-220V$ step-down transformer at a substation in the town.

a) Estimate the line power loss in the form of heat.

Ans: It is given that,

Total electric power required, $\mathbf{P}=\mathbf{800kW}=\mathbf{800}\times \mathbf{1}{{\mathbf{0}}^{3}}\mathbf{W}$

Supply voltage, $\mathbf{V}=\mathbf{220V}$

Voltage at which electric plant is generating power, $V'=440V$

Distance between the town and power generating station, $d=15km$

Resistance of the two wire lines carrying power$=0.5\Omega /km$

Total resistance of the wires, $\mathbf{R}=\left( \mathbf{15}+\mathbf{15} \right)\mathbf{0}.\mathbf{5}=\mathbf{15}\text{ }\mathbf{\Omega }$

A step-down transformer of rating $\mathbf{4000}-\mathbf{220V}$ is used in the sub-station.

Input voltage, ${{V}_{1}}=4000V$

Output voltage, ${{V}_{2}}=220V$

It is known that,

Rms current in the wire lines: $I=\frac{P}{{{V}_{1}}}$

$\Rightarrow I=\frac{800\times {{10}^{3}}}{4000}=200A$

Line power loss$={{I}^{2}}R$

$\Rightarrow {{(200)}^{2}}\times 15$

$\Rightarrow 600\times {{10}^{3}}W=600kW$

Therefore, the line power loss is $600kW$.

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

Ans: Assuming that there is negligible power loss due to leakage of the current:

Total power supplied by the plant$=800kW+600kW=1400kW$

Therefore, the plant must supply $1400kW$ of power.

c) Characterise the step up transformer at the plant.

Ans: It is known that,

Voltage drop in the power line$=IR$

$\Rightarrow V=200\times 15=3000V$

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

The power generated is $440V$.

Therefore, the rating of the step-up transformer situated at the power plant is $440V-7000V$.

26.Do the same exercise as above with the replacement of the earlier transformer by a $40,000-220V$  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?

Ans: It is given that,

Total electric power required, $\mathbf{P}=\mathbf{800kW}=\mathbf{800}\times \mathbf{1}{{\mathbf{0}}^{3}}\mathbf{W}$

Supply voltage, $\mathbf{V}=\mathbf{220V}$

Voltage at which electric plant is generating power, $V'=440V$

Distance between the town and power generating station, $d=15km$

Resistance of the two wire lines carrying power$=0.5\Omega /km$

Total resistance of the wires, $\mathbf{R}=\left( \mathbf{15}+\mathbf{15} \right)\mathbf{0}.\mathbf{5}=\mathbf{15}\text{ }\mathbf{\Omega }$

The rating of a step-down transformer is $40000V-220V$.

Input voltage, ${{V}_{1}}=40000V$

Output voltage, ${{V}_{2}}=220V$

1. It is known that,

Rms current in the wire lines: $I=\frac{P}{{{V}_{1}}}$

$\Rightarrow I=\frac{800\times {{10}^{3}}}{40000}=20A$

Line power loss$={{I}^{2}}R$

$\Rightarrow {{(20)}^{2}}\times 15$

$\Rightarrow 6\times {{10}^{3}}W=6kW$

Therefore, the line power loss is $6kW$.

1. Assume that there is negligible power loss due to leakage of the current:

Total power supplied by the plant$=800kW+6kW=806kW$

Therefore, the plant must supply $806kW$ of power.

1. It is known that,

Voltage drop in the power line$=IR$

$\Rightarrow V=20\times 15=300V$

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

The power generated in the plant is generated at $440V$.

Therefore, the rating of the step-up transformer situated at the power plant is $440V-40300V$.

Power loss during transmission$=\frac{600}{1400}\times 100=42.8%$

In previous exercise the power loss due to the same reason is $=\frac{6}{806}\times 100=0.744%$

As the power loss is less for a high voltage transmission, High voltage transmissions are preferred for this purpose.

### Alternating Current Class 12 NCERT Solutions PDF

The alternating current class 12 CBSE NCERT Solutions are available in PDF. Moreover, they are free and easily accessible for students who want to score good grades in board exams. The answers give an in-depth explanation of the topic with examples.

### Alternating Current Class 12 NCERT PDF

The solutions contain 11 questions from ch 7 physics class 12. Take a look.

Question 1: NCERT solutions of alternating current class 12.

This question asks students to determine the RMS value of the current and the net power consumed, where Resistance is 100 ohms, velocity is 220 and frequency is 50 Hz.

Question 2: NCERT solutions class 12 Physics alternating current.

In this question, the students have to find RMS voltage when the peak voltage of AC is 300 v. You are also required to calculate the peak current when the RMS value of an AC circuit is 10 A.

Question 3: NCERT solutions for class 12 Physics chapter 7 alternating current.

This third question asks you to determine the RMS value of current in a circuit when a 44 mh inductor is connected to 50 Hz ac supply and 220 V.

Question 4: Alternating current class 12 NCERT PDF.

The fourth question of alternating current class 12 NCERT solutions asks you to calculate root mean square value where 60 microfarads, frequency is 60Hz and velocity is 110 volts.

Question 5: NCERT solutions for class 12 Physics Chapter 7 PDF.

In question number 5, students have to find the net power absorbed by each circuit over a complete cycle, where RMS value of I or current is 15.92 A and RMS value of voltage is 220 volts. You have to use the formula p= VI cos Ø (phase value between p and I is 90°).

Question 6: NCERT solutions of chapter 7 physics class 12.

Question six of alternating current class 12 NCERT solutions asks to determine the Q value or the amount of energy absorbed of an LCR circuit.

Question 7: Class 12 Physics ac NCERT solutions.

This 7th question asks a student to calculate the angular frequency of free oscillation in the circuit where a 30µF capacitor is connected to a 27mH inductor.

Question 8: Class 12 Physics ch 7 NCERT solutions.

The eighth question asks a student to calculate the total energy stored in the circuit, initially which held 6mC in a capacitor.

Question 9: NCERT solutions for class 12 Physics Chapter 7.

The ninth question asks you to determine the frequency of supply equal to the natural frequency of the circuit, what the average power transferred to a circuit in one cycle where R is 30Ω, L is 1.5H and C is 35µF.

Question 10:  Alternating current class 12 NCERT PDF.

This question needs determining the range of variable capacitors of a radio over the frequency range of a portion of an MW broadcast band. Its LC circuit has an inductance of 200µH.

Question 11: class 12 Physics chapter 7 NCERT solutions

This question asks students to calculate the frequency in the image that drives the circuit in resonance. They have to find the impedance of the circuit and the amplitude of the current resonating frequency. Moreover, the rms potential drop across an LC combination is zero at resonating frequency is needed to be shown. Here the LCR circuit has V- 230, L-5.0H, C-80µF and R-40Ω.

Apart from these eleven main questions in alternating current class 12 solutions, fourteen additional exercises deal with voltage transformers, power loss in the form of heat, the voltage in power plants, turns in a secondary winding, behaviour of inductor in a DC circuit after steady-state, along with similar theoretical questions.

## Important Questions from Alternating Current (Short, Long & Practice)

1. Weber is the unit of which physical quantity? Hence define it.

2. Two identical loops, one of copper and another of aluminium are rotated with the same speed in the same magnetic field. In which case, the induced

a. emf.

b. current will be more and why?

3. Why the transformer cannot be used to step up d.c. voltage?

1. A power transmission line feeds input power at 2300V to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230V?

2. Show that in the free oscillations of an LC circuit, the sum of the energies stored in the capacitor and the inductor is constant in time.

3. Define mutual inductance. What is its S.I. unit? Write the expression for the mutual inductance between a pair of circular coils of radius r and R (R>r).

### Practice Questions

1. Explain with the help of a labelled diagram, the principle, construction and working of a transformer.

2. 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.

3. 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?

### Marks Weightage of ch 7 Physics Class 12

 Alternating Current Knowledge Application Very short answer 1(1) - Short answer II 2(1) 2(1) Short answer I - 3(1) Long answer - -

This chapter carries a total of 8 marks. Therefore, thorough practice and understanding the AC class 12 NCERT solutions will help secure good grades.

### Benefits of Class 12 Physics Ch 7 Solutions

Practising the Chapter 7 physics class 12 is beneficial for a student in the following ways:

• Solving the questions help students develop logical and reasoning skills.

• Derivations from this chapter will guide the young children to solve problems by applying the correct formulas.

• Students will have an improved understanding of the concepts which will be helpful in JEE or competitive exams.

To avail the free solution, check the Vedantu site today for a hassle-free download. Keep practising alternating current class 12 NCERT solutions to get good marks in board exams.

## Key Features of NCERT Solutions for CBSE Class 12 Physics Chapter 7

• Solutions are curated in the right manner to help students in quickly finding solutions.

• Step-by-step explanation for all questions from NCERT textbook.

• All solutions are easy to grasp and learn as they are prepared by subject experts to match the curriculum.

• NCERT solutions for CBSE Class 12 Physics Chapter 7 help develop strong conceptual basics for students, which is important in the final stages of preparation for board and competitive exams.

## Conclusion

NCERT Solutions for Class 12 Physics Chapter 7 - Alternating Current provide an indispensable resource for students delving into the world of alternating currents. This chapter explores fundamental concepts like AC voltage, impedance, resonance, and power in AC circuits. The solutions, available in various formats, including PDFs, offer clear explanations and step-by-step problem-solving techniques, aiding students in grasping complex concepts. Understanding alternating current is not only essential for academic excellence but also for comprehending the operation of numerous electrical devices in our daily lives. These NCERT solutions play a pivotal role in facilitating a deeper understanding of AC circuits, making them an invaluable asset for students striving for success in their physics studies.

## FAQs on NCERT Solutions for Class 12 Physics Chapter 7 - Alternating Current

Q1. What is Alternating Current?

Ans. The alternating current is an electric current that changes direction, in contrast to Direct Current (DC). If in an alternator, a loop of wire is spun rapidly inside of a magnetic field which produces an electric current along the wire, the current enters a different magnetic polarity, current alternates on the wire and the voltage.

If an AC circuit is hooked to an oscilloscope and its voltage is plotted, various waveforms are visible, such as square, sine, and triangle. You can check Vedantu’s AC solution class 12 for a detailed answer to the question.

Q2. Why does Alternating Current Alter its Direction?

Ans. Initially, generating stations produce electricity in 3 phase, 1 neutral wire systems where each phase wire gets a positive charge for ten milliseconds and then negative charge in a sine wave manner. All 3 phases are placed inside a generator at equal space, and as one phase has a positive charge, the other two may be having a negative control.

At any given moment, the sum of charge of all 3 phases is equal to zero. The Neutral point where one end of all 3 phases is connected, serving as a reference point of measurement, is kept at zero voltage.

Therefore every phase wire becomes plus and minus 50 times in a second, which is the reason why the current direction reverses. This way, all 3 phase wires mutually give and take current to each one and return current. This makes your appliances work.

Q3. What are the Ways to Score Well in Class 12?

Ans. A class 12 student needs to understand and practice the concepts to secure good grades in the exam. Students can maintain a separate notebook of formulas, theories or highlight the essential subtopics for easier access during revision. Moreover, sites like Vedantu provide solutions for different subjects in PDF format, which is free to download.

Students must make a time table and assign time for every specific topic to cover. They can also practice from online portals, practice papers, books to strengthen the base of the subject. Apart from devoting time for reading, aspirants must also maintain some time for relaxation and exercise to be exam ready.

Q4. How to measure the Alternating Current?

Ans: Measuring Alternating Current is really simple. It can be measured by using moving iron type ammeters or using tong testers or clamp meters. In a simple circuit, the current can be calculated by dividing the voltage by resistance. An alternating current is found by dividing the peak voltage by the resistance.