Finding AC circuit problems difficult to solve? Chapter 7, Alternating Current, includes several concepts and formulas that can be confusing when you are solving questions on your own. RMS and peak values, reactance, impedance, resonance, power and transformers all require you to apply the right formula and follow the correct steps.

These NCERT Solutions for Class 12 Physics Chapter 7 explain the chapter-end questions step by step. The solutions show the relevant formula, substitution and calculation, along with a simple explanation wherever needed. This makes it easier to check your answers, understand the steps involved and revise the chapter.

You can also download the NCERT Solutions for Class 12 Physics Chapter 7 PDF for revision and offline study.

Important Concepts in Alternating Current One Must Know Before Solving Numericals

Chapter 7, Alternating Current, builds on the ideas introduced in Chapter 6: Electromagnetic Induction. The concepts of changing magnetic fields, induced emf and inductance help you understand how alternating current is produced and how AC circuits work.

Chapter 7 also connects with Chapter 8: Electromagnetic Waves. Both chapters deal with quantities that change with time. Concepts such as phase and phase difference used in AC circuits are also useful when studying changing electric and magnetic fields in electromagnetic waves.

What is Alternating Voltage and Current

An alternating voltage changes its magnitude and direction periodically with time. A sinusoidal AC voltage can be written as:

v = Vm sin ωt

where Vm is the peak voltage and ω is the angular frequency.

The corresponding current is also periodic. NCERT introduces AC because electrical mains supplies use alternating voltage and because AC voltage can be changed from one value to another using transformers.

What are RMS and Peak Values

The RMS value is the effective value of an AC voltage or current.

Vrms = Vm/√2

Irms = Im/√2

Therefore,

Vm = √2 Vrms

Im = √2 Irms

For example, if the RMS voltage is 220 V, the peak voltage is approximately 311 V.

AC Through a Resistor

For a pure resistor, current and voltage are in phase.

I = V/R

The average power consumed is:

P = VI = I2R = V2/R

The RMS values of voltage and current are normally used in these calculations.

AC Through an Inductor

An inductor opposes changes in current through inductive reactance:

XL = ωL = 2πfL

For a pure inductor, current lags the voltage by 90°. Its average power consumption over a complete cycle is zero.

AC Through a Capacitor

The capacitive reactance is:

XC = 1/ωC

For a pure capacitor, current leads the voltage by 90°. Its average power consumption over a complete cycle is also zero.

Phasors

Phasors are rotating vectors used to represent the phase relationship between AC voltage and current. They make it easier to combine quantities that are not in phase.

Series LCR Circuit and Impedance

For a series LCR circuit:

Z = √[R2 + (XL − XC)2]

The current is:

I = V/Z

where V and I are RMS values when the supply voltage is given as an RMS value and Z is the impedance of the circuit.

Resonance

Resonance occurs in a series LCR circuit when:

XL = XC

Therefore,

ω0 = 1/√LC

At resonance, the impedance becomes minimum:

Z = R

and the current reaches its maximum value. Both the inductor and capacitor must be present for resonance to occur.

Power Factor and Wattless Current

The average power consumed by an AC circuit is:

P = VI cos φ

where cos φ is the power factor.

In a pure inductive or pure capacitive circuit, the phase difference is 90°. Therefore:

cos 90° = 0

and:

P = 0

The current flowing in such a circuit is called wattless current because it flows without average power being dissipated.

Transformer

A transformer works on the principle of mutual induction. It has primary and secondary coils wound around a soft-iron core. An alternating current in the primary produces changing magnetic flux, which induces an emf in the secondary.

For an ideal transformer:

Vs/Vp = Ns/Np

A transformer with more turns in the secondary is a step-up transformer, while one with fewer secondary turns is a step-down transformer.

Chapter 7 Weightage in CBSE Physics

For the 2026–27 CBSE Class 12 Physics syllabus, Chapter 7 is included in Unit IV: Electromagnetic Induction and Alternating Currents, which contains Chapters 6 and 7 and carries 18 marks in total. CBSE does not assign a separate fixed mark allocation to Chapter 7 alone.

NCERT Class 12 Physics Chapter 7: All Exercises

The exercises cover important numerical and conceptual questions based on RMS values, inductive and capacitive reactance, AC power, LC oscillations, series LCR resonance and transformers.

Note: As per the current rationalized NCERT syllabus, the earlier "Additional Exercises" set for this chapter has been removed.

Key Formulas for Chapter 7

NCERT Solutions For Alternating Current with Step-by-Step Solutions

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?

Solution

(a) RMS current

Given: R = 100 Ω, V = 220 V

For a resistor:

I = V/R

Substitute the values:

I = 220/100 = 2.2 A

(b) Net power consumed

For a pure resistor:

P = VI

P = 220 × 2.2 = 484 W

The resistor consumes electrical energy as heat, so its average power consumption is not zero.

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?

Solution

(a) RMS voltage

Given: Vm = 300 V

The relation between peak and RMS voltage is:

V = Vm/√2

V = 300/√2 ≈ 212.1 V

(b) Peak current

Given: I = 10 A

Im = √2 I

Im = √2 × 10 ≈ 14.14 A

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

Solution

Given: L = 44 mH = 44 × 10−3 H, V = 220 V, f = 50 Hz

Step 1: Find inductive reactance

XL = 2πfL

XL = 2π(50)(44 × 10−3) ≈ 13.82 Ω

Step 2: Find RMS current

For a pure inductor:

I = V/XL

I = 220/13.82 ≈ 15.9 A

The inductor limits the AC current through its inductive reactance rather than ordinary resistance.

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.

Solution

Given: C = 60 μF = 60 × 10−6 F, V = 110 V, f = 60 Hz

Step 1: Find capacitive reactance

XC = 1/(2πfC)

XC = 1/(2π(60)(60 × 10−6)) ≈ 44.21 Ω

Step 2: Find RMS current

For a pure capacitor:

I = V/XC

I = 110/44.21 ≈ 2.49 A

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

Solution

Exercise 7.3 contains a pure inductor, while Exercise 7.4 contains a pure capacitor.

For the inductor

In a pure inductive circuit, current lags voltage by 90°. Therefore φ = 90°. The average power is P = VI cos φ, so P = VI cos 90° = 0.

For the capacitor

In a pure capacitive circuit, current leads voltage by 90°. Again, P = VI cos 90° = 0.

Why?

The inductor and capacitor can exchange energy with the AC source during a cycle, but they do not dissipate net energy over a complete cycle. NCERT therefore describes the current in such circuits as wattless current.

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

Solution

Given: C = 30 μF = 30 × 10−6 F, L = 27 mH = 27 × 10−3 H

For an LC circuit:

ω0 = 1/√LC

Step 1: Calculate LC

LC = (27 × 10−3)(30 × 10−6) = 810 × 10−9 = 8.10 × 10−7

Step 2: Calculate angular frequency

ω0 = 1/√(8.10 × 10−7) ≈ 1.11 × 103 rad s−1

The capacitor and inductor exchange energy between electric and magnetic forms, producing free oscillations.

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?

Solution

Given: R = 20 Ω, L = 1.5 H, C = 35 μF, V = 200 V

The supply frequency is equal to the natural frequency, so the circuit is at resonance.

Step 1: Use the resonance condition

At resonance:

XL = XC

Therefore, the reactive components cancel each other. Hence:

Z = R = 20 Ω

Step 2: Find RMS current

I = V/Z = 200/20 = 10 A

Step 3: Find average power

At resonance, φ = 0. Therefore:

P = VI cos φ = (200)(10) cos 0° = 2000 W

Alternatively: P = I2R = (10)2(20) = 2000 W

At resonance, the impedance is minimum and the power factor is 1, so the power transferred to the resistive part of the circuit is maximum.

8. Figure 7.17 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 μF, 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.

Solution

Given: L = 5.0 H, C = 80 μF = 80 × 10−6 F, R = 40 Ω, V = 230 V

(a) Resonant frequency

The resonant angular frequency is:

ω0 = 1/√LC

ω0 = 1/√[(5.0)(80 × 10−6)] = 50 rad s−1

Convert angular frequency to frequency:

f0 = ω0/2π = 50/2π ≈ 7.96 Hz

(b) Impedance and amplitude of current

At resonance:

XL = XC ⇒ Z = R = 40 Ω

The given 230 V is an RMS value. First find the RMS current:

I = V/Z = 230/40 = 5.75 A

The question asks for the amplitude of current, so:

Im = √2 I = √2(5.75) ≈ 8.13 A

(c) RMS potential drops

First calculate the reactances at resonance:

XL = ω0L = (50)(5) = 250 Ω XC = XL = 250 Ω

VR = IR = (5.75)(40) = 230 V

VL = IXL = (5.75)(250) = 1437.5 V

VC = IXC = (5.75)(250) = 1437.5 V

At resonance, VL and VC are equal and opposite — they cancel, leaving VR equal to the supply voltage

Why is the LC potential drop zero?

The voltages across the inductor and capacitor are equal in magnitude but opposite in phase at resonance. Therefore:

VLC = VL − VC = 1437.5 − 1437.5

This does not mean that the individual voltages across the inductor and capacitor are zero. They are both large, but they cancel each other because they are 180° apart in their phasor contribution. NCERT specifically notes that the voltages across L and C cancel at resonance.

Additional Practice Questions

The following two questions are not from NCERT. They are included to bring the total to 10 questions and cover important Chapter 7 concepts that are not tested directly by the eight current chapter-end exercises.

9. Additional Practice Question: A step-down transformer has 500 turns in its primary coil and 100 turns in its secondary coil. The primary is connected to a 250 V AC supply. The secondary supplies 2 A at 50 V. If the transformer is 80% efficient, determine the input current.

Solution

A step-down transformer: fewer secondary turns give a lower secondary voltage

Step 1: Identify the transformer

Given: Np = 500, Ns = 100

Since Ns < Np, the transformer is a step-down transformer.

Step 2: Find the secondary voltage

For an ideal transformer:

Vs/Vp = Ns/Np

Vs = Vp(Ns/Np) = 250 × (100/500) = 50 V

So the given secondary voltage is consistent with the turns ratio.

Step 3: Calculate output power

Pout = VsIs = 50 × 2 = 100 W

Step 4: Use transformer efficiency

Given: η = 80%

η = (Pout/Pin) × 100

0.80 = 100/Pin ⇒ Pin = 100/0.80 = 125 W

Step 5: Find input current

Pin = VpIp ⇒ Ip = Pin/Vp = 125/250 = 0.5 A

The secondary has fewer turns, so the transformer reduces voltage. In an actual transformer, some energy is lost through effects such as flux leakage, winding resistance, eddy currents and hysteresis.

10. Additional Practice Question: A pure inductive AC circuit is connected to a 220 V supply and carries an rms current of 5 A. Find the average power consumed by the circuit over one complete cycle. Explain why the current is called wattless current.

Solution

Given: V = 220 V, I = 5 A

For a pure inductor, current lags voltage by 90°. Therefore φ = 90°.

In a pure inductor, current and voltage are 90° out of phase

Step 1: Use the AC power formula

P = VI cos φ

P = (220)(5) cos 90°

Since cos 90° = 0, we get:

P = 0 W

Why is it called wattless current?

Current is still flowing through the inductor, but the average power consumed over a complete cycle is zero. The current is therefore called wattless current. NCERT defines the current in a purely inductive or capacitive circuit this way because the phase difference is 90° and the power factor is zero.

How to Approach Alternating Current Numericals

Before substituting values, identify which component or circuit is involved.

For a resistor

Use: I = V/R and P = VI

For an inductor

First calculate: XL = 2πfL. Then: I = V/XL

For a capacitor

First calculate: XC = 1/(2πfC). Then: I = V/XC

For a series LCR circuit

Calculate XL = 2πfL and XC = 1/(2πfC). Then Z = √[R2 + (XL − XC)2] and I = V/Z.

If the circuit is at resonance

Immediately use XL = XC, therefore Z = R. This can considerably shorten the calculation.

Key Features of These NCERT Solutions

• Step-by-step solutions for the questions in the chapter
• Clear explanations of formulas and calculations used in each solution
• Solved problems on AC circuits, reactance, resonance, power and transformers
• NCERT-based practice with additional questions on important chapter topics
• Exam-ready answers that show the required formulas, calculations and final answers
• Free access to the solutions and related study resources

Why Use Infinity Learn for Class 12 Physics NCERT Solutions?

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Students can use the solutions alongside the NCERT textbook: first attempt the question independently, then compare the method, calculation and final answer with the solution.

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