NCERT Solutions Class 12 Physics Chapter 7 Alternating Current: Key Topics, Question with Solutions PDF

Q: 7.11 Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = 40 Ω. (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (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.

7.11 Inductance of the Inductor, L = 5.0 H Resistance of the resistor, R = 40 Ω Capacitance of the capacitor, C = 80 μF= 80 ×10-6 F Potential of the voltage source, V = 230 V Resonance angular frequency is given as ωr = 1LC = 15×80×10-6 = 50 rad/s Hence,thecircuitwillcomeinresonanceforasourcefrequencyof50rad/s The impedance of the circuit is given as Z = R2+(XL-XC)2 where XL = Inductive reactance and XC = Capacitive reactance At resonance, XL = XC so Z = R = 40Ω Amplitude of the current at the resonating frequency is given as Io = V0Z where V0 = Peak voltage = √2 V Io = √2VZ = √2×23040 = 8.13 A Hence, at resonance, the impedance of the circuit is 40 Ω and the amplitude of the current is 8.13 A. rms potential drop across the inductor is given by (Vl)rms = Irms ×ωrL Irms= Io√2 = 8.13√2 = 5.75 A Hence, (Vl)rms = 5.75 ×50×5 = 1437.5 V Potential drop across capacitor, (VC)rms = Irms ×1ωr×C Hence, (Vl)rms = 5.75 ×150×80×10-6 = 1437.5 V Potential drop across resistor, (VR)rms = Irms ×R = 5.75 ×40 = 230 V Potential drop across LC combination (VLC)rms = Irms (XL-XC) At resonance, XL = XC) , hence (VLC)rms=0 Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency.

Q: 7.6 Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?

7.6 Inductance, L = 2.0 H Capacitance, C = 32 μF = 32 ×10-6 F Resistance, R = 10 Ω Resonant frequency is given by the relation, ωr = 1LC = 12×32×10-6 = 125 rad /s Now Q value of the circuit is given as Q = 1RLC = 110232×10-6 = 25

Q: 7.26 Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V 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?

7.26 The rating of step-down transformer = 40000 V – 220 V Hence, the input voltage, V1 = 40000 V Output voltage, V2 = 220 V Total electric power required, P = 800 kW = 800 ×103 W Source potential, V = 220 V Voltage at which electric plant generates power, V’ = 440 V Distance between the town and power generating station, d = 15 km Resistance of the two wires lines carrying power = 0.5 Ω /km Total resistance of the wire, R = 2 ×15×0.5Ω = 15 Ω rms current in the wire lines is given as I = PV1 = 800×10340000 = 20 A Line power loss = I2R = 202× 15 = 6 ×103 W = 6 kW Since the leakage power loss is negligible, the total power to be supplied by plant = 800 + 6 = 806 kW Voltage drop in the transmission line = IR = 20 ×15 = 300 V. Hence, the total voltage transmitted from plant = 300 + 40000 = 40300 V Since the power generated is 440V, the transformer rating at the plant should be 440 V – 40300 V Hence, power loss during transmission = 6806 ×100 = 0.744 % In the previous exercise, power loss during transmission = 6001400 ×100 = 42.86 % Since the power loss is less for a high voltage transmission, high voltage transmission is preferred.

Q: 7.9 A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V 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?

7.9 Given values: Resistance R = 20 Ω Inductance, L = 1.5 H Capacitance, C = 35 μF = 35 ×10-6 F AC power supply, V = 200 V Impedance of the circuit is given by the relation, Z = R2+(XL-XC)2 where XL = Inductive reactance and XC = Capacitive reactance At resonance XL = XC Hence Z = R = 20 Ω Current in the circuit, I = VZ = 20020 = 10 A Hence, the average power transferred to the circuit in one complete cycle = VI = 200 ×10 = 2000 W

Q: 7.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.

7.5 In the inductive circuit: rms value of current, I = 15.92 A rms value of voltage, V = 220 V The net power absorbed can be obtained by P = VI cos ∅ , where ∅ is the phase difference between alternating voltage and current = 90 ° So P = VI cos 90° = 0 In the capacitive circuit: rms value of current, I = 2.49 A rms value of voltage, V = 110 V The net power absorbed can be obtained by P = VI cos ∅ , where ∅ is the phase difference between alternating voltage and current = 90 ° So P = VI cos 90° = 0

Updated on Jul 3, 2025 16:27 IST

By Pallavi Pathak, Assistant Manager Content

Class 12 Physics Ch 7 NCERT Solutions Alternating Current explores the behavior of electrical currents that vary with time. It is the foundation of AC electricity used in industries, homes, and across global power grids.
The NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current provides clear, step-by-step answers to all textbook questions. The following are the main points covered:

The working of AC circuits containing capacitors, resistors, and inductors.
How alternating current (AC) differs from direct current (DC).
In AC systems, how LC circuits, transformers and resonance operate.
The concepts of power factor, phase difference and impedance.

For a comprehensive study material on Class 11 and Class 12 Chemistry, Physics and Maths, do check out - NCERT Solutions Class 11 and 12.

Table of content

Insight into Chapter 7 Alternating Current
Class 12 Alternating Current: Key Topics, and Weightage
Class 12 Alternating Current Chapter NCERT Solution PDF: Download PDF for Free
Chapter 7 Alternating Current Important Formulas & Concepts
Alternating Current Solutions
Benefits of Using NCERT Solutions for Class 12 Physics Chapter 7
NCERT Physics Chapter 7 Alternating Current – FAQs

Insight into Chapter 7 Alternating Current

Here is a snapshot of the Chapter 7 Alternating Current:

The students will learn why alternating current is preferred for power distribution and how it differs from direct current.
In AC circuits, how do voltage and current vary sinusoidally?
To study phase relationships in circuits, learn to interpret phasor diagrams.
Students will get an insight into circuit elements and their behavior. They will understand how inductors, resistors, and capacitors affect current flow in AC circuits.
Grasp concepts like capacitive reactance, inductive reactance, and impedance. Also, how they influence the amplitude and phase of the current.
Understand how power is conserved and consumed in different AC circuits, learn about rms (root mean square) values of voltage and current and why rms is used. They will learn about the significance of the power factor and how it affects the efficiency of electrical systems.
The chapter will provide the concept of resonance in series LCR circuits and its importance in tuning TVs and radios. They will know why resonance leads to maximum power transfer and how current amplitude changes with frequency.
In electric power transmission and distribution, understand the vital role of transformers. Find out how the principle of mutual induction is used in the transformers.
The chapter will provide enhanced analytical thinking by analyzing time-varying signals and understanding vector (phasor) representations. Students will learn how to handle phase differences in complex circuits and combine multiple elements.
Students will be able to connect the theoretical knowledge to practical uses like radios, chargers, electric power grids, medical equipment, metal detectors, and motors.

Class 12 Alternating Current: Key Topics, and Weightage

Class 12 Physics Chapter Alternating Current (AC) is an important topic that clarifies how AC circuits work, AC circuits properties, and their real-life applications in electrical systems. Unlike direct current (DC), which flows in one direction, AC changes direction periodically and is used in household electricity, transformers, and power transmission.

These topics are covered in Class 12 Physics Ch 7 NCERT Solutions:

Exercise	Topics Covered
7.1	Introduction To Alternating Current
7.2	AC Voltage Applied To A Resistor
7.3	Representation Of AC Current And Voltage By Rotating Vectors — Phasors
7.4	AC Voltage Applied To An Inductor
7.5	AC Voltage Applied To A Capacitor
7.6	AC Voltage Applied To A Series LCR Circuit
7.7	Power In AC Circuit: The Power Factor
7.8	Transformers

Key Topics in Alternating Current

Here are the ket topics included in the chapter Alternating Current class 12.

Basics of Alternating Current
Peak and RMS Value of AC
Reactance & Impedance
AC Circuits with Resistance, Inductance, and Capacitance
Phasor Diagrams
Power in AC Circuits
Transformers
LC Oscillations and Resonance in AC Circuits

Also read:

Class 12 Alternating Current Chapter NCERT Solution PDF: Download PDF for Free

Class 12 Physics Chapter 7 NCERT solutions provide the link to the Alternating Current NCERT PDF. The students should download it and practice to score good marks in the CBSE Board exams and other competitive exams.

Class 12 Alternating Current Chapter NCERT Solution PDF: Download Free PDF

Chapter 7 Alternating Current Important Formulas & Concepts

Refer to the table below for the important formulas and concepts of Chapter 7, Alternating Current:

Category	Title	Formula / Concept Description
Concept	Alternating Current (AC)	A current that varies sinusoidally with time: $I(t) = I_0 \sin(\omega t)$
Concept	Angular Frequency	$\omega = 2\pi f$ , where $f$ is frequency in Hz
Formula	Mean Value of AC	$I_{\text{mean}} = \frac{2 I_0}{\pi}$ , over half cycle
Formula	RMS Value of AC	$I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$ , $V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$
Formula	Peak Current / Voltage	$I_0 = \sqrt{2} I_{\text{rms}}$ , $V_0 = \sqrt{2} V_{\text{rms}}$
Concept	AC Through Resistor (R)	$I = \frac{V_0}{R} \sin(\omega t)$ ; Voltage and current are in phase
Concept	AC Through Inductor (L)	$I = \frac{V_0}{\omega L} \sin(\omega t - \frac{\pi}{2})$ ; Current lags voltage by 90°
Concept	AC Through Capacitor (C)	$I = V_0 \omega C \sin(\omega t + \frac{\pi}{2})$ ; Current leads voltage by 90°
Formula	Inductive Reactance	$X_L = \omega L$
Formula	Capacitive Reactance	$X_C = \frac{1}{\omega C}$
Formula	Impedance in R-L Circuit	$Z = \sqrt{R^2 + (\omega L)^2}$
Formula	Impedance in R-C Circuit	$Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}$
Formula	Impedance in R-L-C Circuit	$Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$
Formula	Power in AC Circuit	$P = V_{\text{rms}} I_{\text{rms}} \cos\phi$ , where $\phi$ is the phase difference
Formula	Power Factor	$\cos \phi = \frac{R}{Z}$
Concept	Resonance in RLC Circuit	Occurs when $X_L = X_C$ , i.e., $\omega = \frac{1}{\sqrt{LC}}$ ; impedance is minimum
Formula	Resonant Frequency	$f_0 = \frac{1}{2\pi \sqrt{LC}}$
Formula	Quality Factor (Q)	$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$
Concept	LC Oscillations	Energy oscillates between L and C; ideal oscillation is undamped
Formula	Transformer Ratio	$\frac{V_s}{V_p} = \frac{N_s}{N_p}$ ; $V \propto N$

Alternating Current Solutions

Q.7.1 A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.

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

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

Ans.7.1 Resistance of the resistor, R = 100 Ω

Supply voltage, V = 220 V

Supply frequency, $ν$ = 50 Hz

The rms value of current is given as, I = $\frac{V}{R}$ = $\frac{220}{100}$ = 2.2 A

The net power consumed over a full cycle is given as P = VI = 220 $\times 2.2$ = 484 W

Q.7.2 (a) The peak voltage of an ac supply is 300 V. What is the rms voltage?

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

Ans.7.2 Peak voltage of the ac supply, $V_{o}$ = 300 V

RMS voltage is given by $V_{r m s}$ = $\frac{V_{o}}{\sqrt 2}$ = $\frac{300}{\sqrt 2}$ = 212.13 V

The rms value of current, I = 10 A

Peak current $I_{o}$ = $\sqrt[]{2} I$ = $\sqrt[]{2} \times 10 = 14.1 A$

Q.7.3 A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.

Ans.7.3 Inductance of the Inductor, L = 44 mH = 44 $\times 10^{- 3}$ H

Supply voltage, V = 220 V

Supply frequency, $ν$ = 50 Hz

Angular frequency, $ω$ = 2 $π ν$

Inductive reactance, $X_{L}$ = $ω L$ = 2 $π ν L$

rms value of current is given as

I = $\frac{V}{X_{L}}$ = $\frac{220}{2 π \times 50 \times 44 \times 10^{- 3}}$ = 15.92 A

Hence, the rms value of the current in the circuit is 15.92 A

Q.7.4 A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.

Ans.7.4 Capacitance of the capacitor, C = 60 $μ F$ = 60 $\times 10^{- 6}$ F

Supply voltage, V = 110 V

Supply frequency, $ν$ = 60 Hz

Angular frequency, $ω$ = 2 $π ν$

Capacitive reactance, $X_{c}$ = $\frac{1}{ω C}$ = $\frac{1}{2 π ν C}$

rms value of current is given by I = $\frac{V}{X_{c}}$ = 110 $\times 2 \times π \times 60 \times$ 60 $\times 10^{- 6}$ = 2.49 A

Hence, the rms value of current is 2.49 A.

Commonly asked questions

7.11 Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = 40 Ω.

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

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

(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.

7.6 Obtain the resonant frequency $ω_{r}$ of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?

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

7.9 A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V 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?

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

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

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

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

(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?

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

7.7 A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

7.21 Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 μF, and R = 7.4 Ω. 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.

7.12 An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. 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?

(b) What is the natural frequency of the circuit?

(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)?

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

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

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

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

7.22 Answer the following questions:

(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?

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

(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.

(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.

(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?

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

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

7.16 Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behavior with that of a capacitor in a dc circuit after the steady state.

7.4 A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.

7.8 Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?

7.10 A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, 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.]

7.1 A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.

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

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

7.13 A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply.

(a) What is the maximum current in the coil?

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

7.14 Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). 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?

7.15 A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.

(a) What is the maximum current in the circuit?

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

7.17 Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.

7.18 A circuit containing a 80 mH inductor and a 60 $μ$ F capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.

(a) Obtain the current amplitude and rms values.

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

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

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

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

7.23 A power transmission line feeds input power at 2300 V 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 230 V?

7.24 At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m³s^–1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms^–2 ).

7.25 A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.

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

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

(c) Characterize the step up transformer at the plant.

Benefits of Using NCERT Solutions for Class 12 Physics Chapter 7

By preparing from the NCERT Solutions for Class 12 Physics Chapter 7, students can easily grasp the complex topics such as impedance, reactance, AC circuits, and resonance. It offers all NCERT textbook solutions for concept clarity. The students will understand the key concepts given in the chapter. The step-by-step approach improves students' problem-solving skills as these solutions break down the calculations and derivations.
The NCERT solutions are aligned with the CBSE curriculum, which ensures that topics such as LC oscillations, transformers, and power in AC circuits are thoroughly covered as per the exam requirements. It can boost the students' exam preparation as it offers the solutions to all questions of the NCERT books and the questions that appear in the CBSE board exams are based on the NCERT syllabus only. Also, most of the prominent entrance examinations for the medical and engineering fields include questions based on the NCERT concepts. It also helps in students' time management as they get the solutions to all the numerical and theoretical questions here and they can quickly prepare and revise these questions. It also helps in improving accuracy and speed during the exams. It is one of the best tools for self-study and clarity of doubts, as these are given in an easy-to-understand language, which can be easily understood by the students without needing the help of external coaching centers.

NCERT Physics Chapter 7 Alternating Current – FAQs

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Insight into Chapter 7 Alternating Current

Class 12 Alternating Current: Key Topics, and Weightage

Key Topics in Alternating Current

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Class 12 Alternating Current Chapter NCERT Solution PDF: Download PDF for Free

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Alternating Current Solutions

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