NCERT Solutions for Class 12 Physics Chapter 7: Alternating Current

These Class 12 Physics Chapter 7 solutions cover Alternating Current from the NCERT textbook (session 2026–27). Every NCERT Exercise question is reproduced verbatim and solved step by step, with each numerical worked out fully and verified against the official answer key — rms values, reactance, impedance, resonance and power. The chapter explains how ac voltage behaves across resistors, inductors and capacitors, the series LCR circuit, resonance, power factor and the transformer.

Class: 12 Subject: Physics Chapter: 7 Name: Alternating Current Exercises: 7.1 – 7.8 Session: 2026–27
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Class 12 Physics Chapter 7 Solutions – Overview

Chapter 7, Alternating Current, studies voltages and currents that vary sinusoidally with time, like the mains supply in our homes. It begins with ac voltage applied to a pure resistor, where voltage and current stay in phase, and introduces the root-mean-square (rms) value, the equivalent dc value that produces the same average heating. The chapter then analyses a pure inductor (current lags voltage by π/2, reactance XL = ωL) and a pure capacitor (current leads voltage by π/2, reactance XC = 1/ωC). Combining all three gives the series LCR circuit, solved using phasors to obtain impedance Z and phase angle φ. At resonance (XL = XC), impedance is minimum and current is maximum — the basis of radio tuning. Finally, the chapter covers power in ac circuits (the power factor cosφ and wattless current) and the transformer, which steps voltage up or down using mutual induction.

Key Concepts & Definitions

Alternating voltage: a voltage that varies sinusoidally with time, v = vm sinωt, where vm is the peak (amplitude) and ω the angular frequency.

rms value: the effective value of an ac quantity; I = im/√2 = 0.707 im and V = vm/√2 = 0.707 vm. It is the dc current that gives the same average Joule heating.

Inductive reactance (XL): opposition of an inductor to ac, XL = ωL = 2πνL; it increases with frequency. Current lags voltage by π/2.

Capacitive reactance (XC): opposition of a capacitor to ac, XC = 1/ωC = 1/(2πνC); it decreases with frequency. Current leads voltage by π/2.

Impedance (Z): the total opposition of a series LCR circuit, Z = √[R² + (XL − XC)²]. SI unit: ohm (Ω).

Resonance: occurs when XL = XC; then Z = R (minimum) and current is maximum. Resonant frequency ω0 = 1/√(LC).

Power factor (cosφ): the cosine of the phase angle between voltage and current; average power P = VI cosφ. A pure L or C gives cosφ = 0 (wattless current); only R dissipates energy.

Transformer: changes ac voltage using mutual induction: Vs/Vp = Ns/Np = Ip/Is.

Important Formulas

rms current: I = im/√2 = 0.707 im  |  rms voltage: V = vm/√2 = 0.707 vm

Resistor: v and i in phase, I = V/R, P = I²R = VI

Inductor: XL = ωL = 2πνL, I = V/XL, current lags v by π/2, P = 0

Capacitor: XC = 1/ωC = 1/(2πνC), I = V/XC, current leads v by π/2, P = 0

Series LCR: Z = √[R² + (XL − XC)²],   tanφ = (XL − XC)/R,   I = V/Z

Resonance: ω0 = 1/√(LC),   νr = 1/(2π√(LC)),   Z = R,   im = vm/R

Average power: P = VI cosφ = I²R,   power factor = cosφ = R/Z

Transformer: Vs/Vp = Ns/Np = Ip/Is

NCERT Exercises 7.1–7.8 — Solutions

Questions reproduced verbatim from the NCERT textbook; all answers are original, worked step by step, and numerical results verified against the NCERT answer key (with units).

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?

ANSWER Given: R = 100 Ω, V = 220 V (rms), ν = 50 Hz. (a) For a pure resistor, I = V/R = 220 V / 100 Ω = 2.20 A. (b) Net power consumed P = VI = (220 V)(2.20 A) = 484 W. (Equivalently P = I²R = (2.20)² × 100 = 484 W.) Net power consumed over a full cycle = 484 W. A resistor dissipates power because Joule heating depends on i², which is always positive.

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?

ANSWER (a) rms voltage V = vm/√2 = 300 V / 1.414 = 212.1 V. (b) Peak current im = √2 × I = 1.414 × 10 A = 14.1 A (i.e. 10√2 A).

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.

ANSWER Given: L = 44 mH = 44 × 10−3 H, V = 220 V, ν = 50 Hz. Inductive reactance: XL = 2πνL = 2 × 3.14 × 50 × 44 × 10−3 = 13.82 Ω. rms current: I = V/XL = 220 V / 13.82 Ω = 15.9 A.

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.

ANSWER Given: C = 60 µF = 60 × 10−6 F, V = 110 V, ν = 60 Hz. Capacitive reactance: XC = 1/(2πνC) = 1 / (2 × 3.14 × 60 × 60 × 10−6) = 1 / (0.02262) = 44.2 Ω. rms current: I = V/XC = 110 V / 44.2 Ω = 2.49 A.

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.

ANSWER Net power absorbed = zero in each case. In the inductive circuit (7.3) the current lags the voltage by π/2, and in the capacitive circuit (7.4) the current leads the voltage by π/2. In both cases the phase difference φ = π/2, so the power factor cosφ = cos(90°) = 0. Therefore the average power P = VI cosφ = 0. Over a complete cycle a pure inductor (or capacitor) stores energy in its magnetic (or electric) field during one quarter cycle and returns it to the source during the next, so no net energy is dissipated. The current is “wattless.”

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

ANSWER Given: C = 30 µF = 30 × 10−6 F, L = 27 mH = 27 × 10−3 H. Angular frequency of free LC oscillations: ω = 1/√(LC). LC = (27 × 10−3)(30 × 10−6) = 810 × 10−9 = 8.1 × 10−7 s². √(LC) = √(8.1 × 10−7) = 9.0 × 10−4 s. ω = 1 / (9.0 × 10−4) = 1.1 × 103 rad/s (s−1).

7.7 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?

ANSWER Given: R = 20 Ω, L = 1.5 H, C = 35 µF, V = 200 V. When the source frequency equals the natural (resonant) frequency, XL = XC, so the impedance is minimum: Z = R, and the phase angle φ = 0 (power factor cosφ = 1). At resonance the current is I = V/R = 200/20 = 10 A. Average power: P = VI cosφ = (200)(10)(1) = 2000 W. Equivalently P = V²/R = (200)²/20 = 2000 W.

7.8 Figure 7.17 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.

ANSWER Given: V = 230 V (rms), L = 5.0 H, C = 80 µF = 80 × 10−6 F, R = 40 Ω. (a) Resonant angular frequency ω0 = 1/√(LC). LC = 5.0 × 80 × 10−6 = 4.0 × 10−4; √(LC) = 2.0 × 10−2. So ω0 = 1/(2.0 × 10−2) = 50 rad/s. (Frequency νr = ω0/2π = 50/6.28 ≈ 7.96 Hz.) (b) At resonance XL = XC, so impedance Z = R = 40 Ω. rms current I = V/Z = 230/40 = 5.75 A. Amplitude of current im = √2 × I = 1.414 × 5.75 = 8.1 A. (c) First find the reactances at ω0 = 50 rad/s: XL = ω0L = 50 × 5.0 = 250 Ω; XC = 1/(ω0C) = 1/(50 × 80 × 10−6) = 1/(4.0 × 10−3) = 250 Ω. Potential drop across R: VR = I·R = 5.75 × 40 = 230 V. Potential drop across L: VL = I·XL = 5.75 × 250 = 1437.5 V. Potential drop across C: VC = I·XC = 5.75 × 250 = 1437.5 V. Potential drop across the LC combination: Since VL and VC are equal in magnitude but exactly π out of phase (VL leads I by π/2 while VC lags I by π/2), VLC = I(ω0L − 1/ω0C) = 5.75 × (250 − 250) = 0. Thus the entire source voltage appears across R, confirming resonance.

Extra Practice Questions

Short Answer Type Questions

Q1. Why is it advantageous to transmit electrical power at high voltage and low current?

ANSWERPower loss in transmission lines is I²R. Stepping up the voltage (and so reducing the current for the same power P = VI) drastically reduces I²R losses. Transformers make this voltage conversion easy, which is the main reason ac is preferred for transmission.

Q2. Define rms value of alternating current and give its relation with the peak value.

ANSWERThe rms (root-mean-square) value is the value of steady dc current that produces the same average heating effect as the ac in the same resistance over a cycle. It is related to the peak value by I = im/√2 = 0.707 im.

Q3. What is a wattless current?

ANSWERIt is the current in a purely inductive or purely capacitive circuit where the phase difference between voltage and current is π/2, so cosφ = 0 and the average power consumed is zero. Although current flows, no net power is dissipated — hence “wattless.”

Q4. Why does a capacitor block dc but allow ac to pass?

ANSWERFor dc the frequency is zero, so XC = 1/(2πνC) becomes infinite and no steady current flows once the capacitor is charged. For ac, ν is finite so XC is finite; the capacitor charges and discharges every half cycle, allowing ac to flow.

Q5. State two energy losses in a real transformer and how each is reduced.

ANSWER(i) Eddy-current loss — reduced by using a laminated core. (ii) Hysteresis loss — reduced by using a soft magnetic material (e.g. soft iron) with low hysteresis. (Other losses: flux leakage, reduced by good winding; copper/I²R loss, reduced by thick wire.)

Long Answer Type Questions

Q1. Derive the expression for impedance and phase angle of a series LCR circuit using the phasor method.

ANSWERIn a series LCR circuit the same current i = im sin(ωt + φ) flows through all elements. Drawing phasors: VR = imR is in phase with I; VL = imXL leads I by π/2; VC = imXC lags I by π/2. Since VL and VC are opposite, their resultant is im(XL − XC). The source voltage is the hypotenuse of a right triangle with sides VR and (VL − VC): vm² = (imR)² + im²(XL − XC)². Hence im = vm/√[R² + (XL − XC)²] = vm/Z, where impedance Z = √[R² + (XL − XC)²]. The phase angle is tanφ = (XL − XC)/R. If XL > XC the circuit is inductive (current lags); if XC > XL it is capacitive (current leads).

Q2. Explain resonance in a series LCR circuit. Derive the resonant frequency and discuss its application in radio tuning.

ANSWERAs the source frequency ω is varied, XL = ωL increases while XC = 1/ωC decreases. At a particular frequency ω0 they become equal, XL = XC, so the net reactance is zero, impedance is minimum (Z = R), and the current is maximum, im = vm/R. Setting ω0L = 1/ω0C gives ω0² = 1/LC, so the resonant frequency ω0 = 1/√(LC). Resonance is exhibited only when both L and C are present, since only then do their voltages cancel. Application: in a radio, the antenna receives many station signals; the tuning circuit’s capacitance is varied so its resonant frequency matches the desired station’s frequency. At resonance the current for that station is maximum, so it is selected and amplified while others are rejected.

Q3. Describe the construction and working of a transformer, and obtain the relation between primary and secondary voltages and currents.

ANSWERA transformer has two insulated coils — a primary (Np turns) and a secondary (Ns turns) — wound on a common soft-iron laminated core. It works on mutual induction. When ac is applied to the primary, it produces an alternating flux φ that links the secondary and induces an emf in it. For an ideal transformer, εs = −Ns(dφ/dt) and εp = −Np(dφ/dt), giving Vs/Vp = Ns/Np. Assuming no power loss, ipvp = isvs, hence Ip/Is = Ns/Np. If Ns > Np it is a step-up transformer (voltage increases, current decreases); if Ns < Np it is a step-down transformer. Transformers make long-distance power transmission economical by stepping voltage up (reducing I²R loss) and then stepping it down for use.

MCQs & Assertion–Reason

1. The rms value of an alternating current of peak value 14.14 A is approximately:

(a) 7.07 A    (b) 10 A    (c) 14.14 A    (d) 20 A

2. In a purely inductive ac circuit, the current:

(a) leads the voltage by π/2    (b) lags the voltage by π/2    (c) is in phase with voltage    (d) lags by π

3. The inductive reactance of an inductor:

(a) decreases with frequency    (b) is independent of frequency    (c) increases with frequency    (d) is zero for ac

4. The SI unit of capacitive reactance is:

(a) farad    (b) henry    (c) ohm    (d) watt

5. At resonance in a series LCR circuit, the impedance equals:

(a) XL    (b) XC    (c) R    (d) zero

6. The resonant angular frequency of an LC circuit is:

(a) √(LC)    (b) 1/√(LC)    (c) LC    (d) 1/(LC)

7. The power factor of a purely capacitive circuit is:

(a) 1    (b) 0.5    (c) 0    (d) −1

8. In a step-up transformer, compared with the primary, the secondary has:

(a) more turns and higher current    (b) more turns and lower current    (c) fewer turns and higher voltage    (d) fewer turns and lower current

9. Eddy-current losses in a transformer core are reduced by:

(a) using a thicker solid core    (b) laminating the core    (c) using a copper core    (d) increasing the frequency

10. A current is called wattless when the phase angle between voltage and current is:

(a) 0    (b) π/4    (c) π/2    (d) π

Answer key: 1-(b), 2-(b), 3-(c), 4-(c), 5-(c), 6-(b), 7-(c), 8-(b), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: A pure inductor connected to an ac source consumes no average power.

Reason: The phase difference between voltage and current in a pure inductor is π/2, so cosφ = 0.

A-R 2. Assertion: At resonance, the current in a series LCR circuit is maximum.

Reason: At resonance the impedance is minimum and equals R because XL = XC.

A-R 3. Assertion: A capacitor allows ac to pass but blocks dc.

Reason: Capacitive reactance is directly proportional to the frequency of the source.

A-R 4. Assertion: In a series LCR circuit at resonance, the voltage across the LC combination is zero.

Reason: The voltages across L and C are equal in magnitude and opposite in phase at resonance.

A-R 5. Assertion: Electrical energy is transmitted over long distances at high voltage.

Reason: Transmitting at high voltage reduces the current and hence the I²R power loss in the lines.

Answer key: 1-(A), 2-(A), 3-(C), 4-(A), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Adding voltages across R, L and C arithmetically. They are out of phase — combine using V = √[VR² + (VL − VC)²].
  • Confusing XL = ωL (increases with ν) with XC = 1/ωC (decreases with ν).
  • Forgetting that the given ac voltage/current is the rms value, not the peak; convert with √2 when the question asks for amplitude.
  • Using degrees instead of radians for ωt, or dropping the factor 2π when converting frequency ν to angular frequency ω.
  • Thinking a pure inductor or capacitor dissipates power. Only the resistor dissipates energy; cosφ = 0 for pure L or C.
  • Mixing up the transformer turn ratios — voltage ratio follows Ns/Np, but current ratio is the inverse, Np/Ns.

Exam Tips

How to score full marks in this chapter

Always write the formula, substitute values with units, then compute — examiners award marks for each step. Keep π = 3.14 and √2 = 1.414 ready. For LCR numericals, first compute XL and XC, then Z, then I, then the required quantity. State clearly whether the circuit is inductive or capacitive from the sign of (XL − XC). Remember rms-to-peak conversions and that average power = VI cosφ. For resonance questions, immediately use Z = R, I = V/R and ω0 = 1/√(LC).

Frequently Asked Questions

What is Class 12 Physics Chapter 7 about?

Chapter 7, Alternating Current, deals with ac voltage applied to resistors, inductors and capacitors, the series LCR circuit, impedance, resonance, power in ac circuits (power factor and wattless current), and transformers. It explains rms values and why ac is used for power transmission.

How many exercise questions are there in Class 12 Physics Chapter 7?

The NCERT textbook gives 8 exercise questions, numbered 7.1 to 7.8 (several with sub-parts). All of them are reproduced verbatim and solved step by step on this page, with numerical answers verified against the NCERT answer key.

What is the resonant frequency formula in a series LCR circuit?

The resonant angular frequency is ω0 = 1/√(LC), and the resonant frequency is νr = 1/(2π√(LC)). At resonance XL = XC, impedance is minimum (Z = R) and the current is maximum.

Are these Class 12 Physics Chapter 7 solutions free?

Yes. All solutions are free and follow the official NCERT Physics textbook for session 2026–27.

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