PhysicsCBSE Class 12 · Chapter 7

Alternating Current Class 12: Notes, Phasors and Solved Numericals

Updated 14 October 2026 · Fact-checked

Alternating Current is Chapter 7 of CBSE Class 12 Physics. It covers peak and rms values, the behaviour of a resistor, inductor and capacitor in an AC circuit, reactance and impedance, the series LCR circuit using phasors, resonance, power and power factor, wattless current, the AC generator and the transformer.

What Alternating Current is about

Most electrical power is supplied as alternating current, because AC voltages can be stepped up and down easily with transformers. In this chapter a sinusoidal voltage $v = v_{m} sin ω t$ is applied in turn to a resistor, an inductor and a capacitor. In a resistor, current and voltage are in phase. In an inductor the current lags the voltage by $\frac{π}{2}$ , and in a capacitor it leads by $\frac{π}{2}$ ; each opposes the current through a reactance ( $X_{L} = ω L$ , $X_{C} = \frac{1}{ω C}$ ) that depends on frequency.

Rotating arrows called phasors let you combine these in a series LCR circuit to get the impedance $Z = R^{2} + (X_{C} - X_{L})^{2}$ and the phase angle. When $X_{L} = X_{C}$ the circuit is at resonance: Z is smallest and the current largest, which is how a radio is tuned. The average power is $P = V_{r m s} I_{r m s} cos ϕ$ , where $cos ϕ$ is the power factor; a current component at $9 0^{\circ}$ to the voltage does no work (wattless current). The chapter ends with the transformer; the syllabus also lists the AC generator here (NCERT explains it in Chapter 6).

For 2026-27, the series LCR circuit is done with phasors only; the analytical solution, LC oscillations and the sharpness of resonance are not in the syllabus. In our question bank this chapter appears in all eight board papers from 2023 to 2026, with 12 of its 29 questions being 5-mark long answers, mostly on LCR resonance and transformers.

Weightage: This chapter is part of Units III and IV, Magnetic Effects of Current and Magnetism; Electromagnetic Induction and Alternating Currents, which carry 17 of the 70 theory marks together with Moving Charges and Magnetism, Magnetism and Matter and Electromagnetic Induction.Source: CBSE Physics syllabus 2026-27, Units III and IV Magnetic Effects of Current and Magnetism; Electromagnetic Induction and Alternating Currents

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 4.4 marks per paper. See the questions

Key concepts

AC through a resistor; peak and rms values

NCERT section 7.2 and 7.3

Apply $v = v_{m} sin ω t$ to a resistor R. The current is $i = \frac{v _{m}}{R} sin ω t = i_{m} sin ω t$ : in phase with the voltage. The power $i^{2} R$ varies, but its average over a cycle is $\frac{1}{2} i_{m}^{2} R$ because the average of $sin^{2} ω t$ is $\frac{1}{2}$ .

The root mean square (rms) or effective current is the steady current that would give the same average heating: $I = \frac{i _{m}}{2} \approx 0.707 i_{m}$ . Similarly $V = \frac{v _{m}}{2}$ . Then $P = I^{2} R = V I$ as for DC.

AC meters and the ratings of supplies and appliances give rms values: a 220 V supply has a peak of $2202 \approx 311 V$ . The average of a sinusoidal current over a full cycle is zero; over a half cycle it is $\frac{2 i _{m}}{π}$ .

Phasors: a voltage or current is drawn as a vector of length equal to its peak value, rotating anticlockwise at ω; its projection on the vertical axis gives the instantaneous value. The angle between two phasors is their phase difference.

Worked example

A lamp is rated 60 W, 240 V. Find (a) its resistance, (b) the rms and peak currents through it, (c) the peak voltage of the supply. [2 marks]

Try it first, then show the solutionSolution

(a) $R = \frac{V ^{2}}{P} = \frac{24 0 ^{2}}{60} = 960 Ω$
(b) $I = \frac{P}{V} = \frac{60}{240} = 0.25 A$ (rms); $i_{m} = 2 \times 0.25 \approx 0.354 A$
(c) $v_{m} = 2 \times 240 \approx 339 V$

Answer: (a) 960 Ω (b) 0.25 A rms, 0.354 A peak (c) about 339 V

AC through an inductor: inductive reactance

NCERT section 7.4

For a pure inductor, $v_{m} sin ω t = L \frac{d i}{d t}$ . Integrating: $i = - \frac{v _{m}}{ω L} cos ω t = i_{m} sin (ω t - \frac{π}{2})$ , with $i_{m} = \frac{v _{m}}{ω L}$ .

The current lags the voltage by $\frac{π}{2}$ (a quarter cycle).
The opposition is the inductive reactance $X_{L} = ω L = 2 π ν L$ (ohm). It is zero for DC and grows in proportion to frequency.
The average power over a cycle is zero: energy is stored in the magnetic field for a quarter cycle and returned for the next.

Worked example

A pure inductor of 0.1 H is connected to a 220 V, 50 Hz supply. Find the inductive reactance and the rms current. What power does it consume? [2 marks]

Try it first, then show the solutionSolution

$X_{L} = 2 π ν L = 2 π \times 50 \times 0.1 = 10 π \approx 31.4 Ω$
$I = \frac{V}{X _{L}} = \frac{220}{31.4} \approx 7.0 A$
Average power is zero for a pure inductor.

Answer: $X_{L} \approx 31.4 Ω$ ; $I \approx 7.0 A$ ; zero power

AC through a capacitor: capacitive reactance

NCERT section 7.5

For a capacitor, $q = C v = C v_{m} sin ω t$ , so $i = \frac{d q}{d t} = ω C v_{m} cos ω t = i_{m} sin (ω t + \frac{π}{2})$ , with $i_{m} = \frac{v _{m}}{1/ ω C}$ .

The current leads the voltage by $\frac{π}{2}$ .
The opposition is the capacitive reactance $X_{C} = \frac{1}{ω C} = \frac{1}{2 π ν C}$ . It is infinite for DC (a capacitor blocks steady current) and falls as frequency rises.
The average power is again zero: energy is stored in the electric field and returned.

A capacitor in series with a lamp: on DC the lamp stays off (after a brief charging pulse); on AC it glows, more brightly at higher frequency or larger C.

Worked example

A $50 μ F$ capacitor is connected to a 220 V, 50 Hz supply. Find its reactance and the rms current. [2 marks]

Try it first, then show the solutionSolution

$X_{C} = \frac{1}{2 π ν C} = \frac{1}{2 π \times 50 \times 50 \times 1 0 ^{- 6}} = \frac{1}{0.0157} \approx 63.7 Ω$
$I = \frac{220}{63.7} \approx 3.46 A$

Answer: $X_{C} \approx 63.7 Ω$ ; $I \approx 3.46 A$

Series LCR circuit by phasors: impedance and phase angle

NCERT section 7.6.1

In series, the same current $i = i_{m} sin (ω t + ϕ)$ flows through R, L and C, so draw the current phasor first. Then:

$V_{R}$ (= $i_{m} R$ ) is along $I$ ;
$V_{L}$ (= $i_{m} X_{L}$ ) is $9 0^{\circ}$ ahead of $I$ ;
$V_{C}$ (= $i_{m} X_{C}$ ) is $9 0^{\circ}$ behind $I$ .

$V_{L}$ and $V_{C}$ point in opposite directions, so they combine to $V_{C} - V_{L}$ (or $V_{L} - V_{C}$ ). This is at right angles to $V_{R}$ , so by Pythagoras

$v_{m}^{2} = (i_{m} R)^{2} + (i_{m} X_{C} - i_{m} X_{L})^{2} \Rightarrow i_{m} = \frac{v _{m}}{R ^{2} + ( X _{C} - X _{L} ) ^{2}}$

The impedance is $Z = R^{2} + (X_{C} - X_{L})^{2}$ and the phase angle between voltage and current is given by $tan ϕ = \frac{X _{C} - X _{L}}{R}$ . If $X_{L} > X_{C}$ the circuit is inductive and the current lags; if $X_{C} > X_{L}$ it is capacitive and the current leads. Note: the separate voltages across L and C can each be larger than the source voltage, because they partly cancel.

Worked example

A series circuit has $R = 30 Ω$ , $X_{L} = 80 Ω$ and $X_{C} = 40 Ω$ and is connected to a 200 V (rms) supply. Find the impedance, the current, the phase angle, and say whether the current leads or lags. [3 marks]

Try it first, then show the solutionSolution

$Z = 3 0^{2} + (80 - 40)^{2} = 900 + 1600 = 50 Ω$
$I = \frac{200}{50} = 4 A$
$tan ϕ = \frac{X _{L} - X _{C}}{R} = \frac{40}{30}$ , so $ϕ \approx 5 3^{\circ}$
$X_{L} > X_{C}$ : the circuit is inductive and the current lags the voltage by about $5 3^{\circ}$

Answer: Z = 50 Ω; I = 4 A; φ ≈ 53°, current lagging

Resonance

NCERT section 7.6.2

As the frequency of the source changes, $X_{L}$ rises and $X_{C}$ falls. At one frequency they are equal, and then $Z = R$ (its minimum) and the current is maximum, $i_{m} = \frac{v _{m}}{R}$ . This is resonance. From $ω_{0} L = \frac{1}{ω _{0} C}$ :

$ω_{0} = \frac{1}{L C}, ν_{0} = \frac{1}{2 π L C}$

At resonance the voltage and current are in phase (φ = 0) and the voltages across L and C are equal and opposite (they can each be much larger than the source voltage). Resonance needs both L and C; an RL or RC circuit has no resonance.

Use: the tuning circuit of a radio. Turning the knob changes C, shifting $ω_{0}$ ; the station whose frequency matches $ω_{0}$ produces a large current and is heard. (The sharpness of resonance, or Q factor, is not in the 2026-27 syllabus.)

Worked example

A series LCR circuit has L = 0.1 H, $C = 10 μ F$ and R = 20 Ω. Find its resonant angular frequency and frequency, and the rms current at resonance with a 220 V source. [2 marks]

Try it first, then show the solutionSolution

$ω_{0} = \frac{1}{L C} = \frac{1}{0.1 \times 1 0 ^{- 5}} = \frac{1}{1 0 ^{- 3}} = 1000 r a d s^{- 1}$
$ν_{0} = \frac{1000}{2 π} \approx 159 H z$
At resonance Z = R: $I = \frac{220}{20} = 11 A$

Answer: $ω_{0} = 1000 r a d s^{- 1}$ ; $ν_{0} \approx 159 H z$ ; I = 11 A

Power in an AC circuit: power factor and wattless current

NCERT section 7.7

With $v = v_{m} sin ω t$ and $i = i_{m} sin (ω t + ϕ)$ , the instantaneous power is $v i = v_{m} i_{m} sin ω t sin (ω t + ϕ) = \frac{v _{m} i _{m}}{2} [cos ϕ - cos (2 ω t + ϕ)]$ . The second term averages to zero over a cycle, so

$P = \frac{v _{m} i _{m}}{2} cos ϕ = V I cos ϕ$

with V and I rms values. $cos ϕ$ is the power factor; for a series LCR circuit $cos ϕ = \frac{R}{Z}$ .

Pure R: φ = 0, cos φ = 1, P = VI (maximum).
Pure L or C: φ = ±90°, cos φ = 0, P = 0.
Resonance: Z = R, cos φ = 1.

Resolve the current into $I cos ϕ$ (in phase with V, does all the work) and $I sin ϕ$ (at 90° to V, does no work on average). The second part is the wattless current. Power companies want a power factor close to 1, because a low power factor means a larger current (and larger line losses) for the same useful power.

Worked example

A device draws 5 A (rms) from a 220 V (rms) supply, with the current lagging the voltage by $6 0^{\circ}$ . Find (a) the power factor, (b) the average power, (c) the wattless component of current. [2 marks]

Try it first, then show the solutionSolution

(a) $cos 6 0^{\circ} = 0.5$
(b) $P = V I cos ϕ = 220 \times 5 \times 0.5 = 550 W$
(c) Wattless current $= I sin ϕ = 5 \times 0.866 \approx 4.33 A$

Answer: (a) 0.5 (b) 550 W (c) about 4.33 A

AC generator (in the syllabus with this unit)

NCERT section 6.8

An AC generator works by electromagnetic induction. A coil of N turns and area A rotates at angular speed ω in a uniform field B. The flux linkage $N B A cos ω t$ changes continuously, inducing

$ε = N B A ω sin ω t = ε_{0} sin ω t$

The emf is greatest when the plane of the coil is parallel to B and zero when it is perpendicular. Slip rings and brushes take the alternating emf to the external circuit. The full derivation and diagram are on the Electromagnetic Induction page, where NCERT treats it (section 6.8).

Worked example

An AC generator's emf is $ε = 300 sin (100 π t)$ volts. Find the peak emf, the rms emf and the frequency. If the coil is made to rotate twice as fast, what is the new equation? [2 marks]

Try it first, then show the solutionSolution

Peak $ε_{0} = 300 V$ ; rms $= \frac{300}{2} \approx 212 V$
$ω = 100 π$ , so $ν = \frac{ω}{2 π} = 50 H z$
$ε_{0} = N B A ω \propto ω$ : doubling ω doubles both: $ε = 600 sin (200 π t)$

Answer: 300 V peak, about 212 V rms, 50 Hz; new emf $600 sin (200 π t)$ V

Transformer

NCERT section 7.8

Principle: mutual induction. Two coils, the primary ( $N_{p}$ turns) and the secondary ( $N_{s}$ turns), are wound on a common soft iron core. An alternating current in the primary produces an alternating flux in the core, which links the secondary and induces an emf in it.

Ratios (ideal transformer): with negligible resistance and no flux leakage, the same flux Φ links each turn: $ε_{s} = - N_{s} \frac{d Φ}{d t}$ and $v_{p} = - N_{p} \frac{d Φ}{d t}$ (back emf equals the applied voltage). So

$\frac{v _{s}}{v _{p}} = \frac{N _{s}}{N _{p}}$

If there is no energy loss, $i_{p} v_{p} = i_{s} v_{s}$ , so $\frac{i _{s}}{i _{p}} = \frac{N _{p}}{N _{s}}$ . A step-up transformer ( $N_{s} > N_{p}$ ) raises the voltage and lowers the current; a step-down transformer does the opposite. A transformer does not work on steady DC (no changing flux).

Energy losses in real transformers (as listed in NCERT): flux leakage (not all primary flux links the secondary), resistance of the windings (Joule heating), eddy currents in the core (reduced by a laminated core) and hysteresis in the core. Large-scale transmission uses step-up transformers at the power station and step-down transformers near users, so that the line current, and hence the $I^{2} R$ loss, is small.

Worked example

A step-down transformer reduces 4400 V to 220 V. Its primary has 2000 turns. (a) How many turns does the secondary have? (b) If it delivers 11 kW, find the secondary and primary currents, assuming it is ideal. [2 marks]

Try it first, then show the solutionSolution

(a) $N_{s} = N_{p} \frac{V _{s}}{V _{p}} = 2000 \times \frac{220}{4400} = 100$ turns
(b) $I_{s} = \frac{11 000}{220} = 50 A$
$I_{p} = \frac{11 000}{4400} = 2.5 A$ (check: $\frac{I _{s}}{I _{p}} = 20 = \frac{N _{p}}{N _{s}}$ )

Answer: (a) 100 turns (b) 50 A in the secondary, 2.5 A in the primary

Physics in real life: Alternating Current

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

rms and peak values

Suppose the mains supply at home is labelled 220 V, 50 Hz, and you wonder how high the voltage actually rises in each cycle.

Find the peak voltage of a 220 V (rms) supply and how many times per second the voltage passes through zero.

Try it first, then show the solutionSolution

$v_{m} = 2 \times 220 \approx 311 V$
A 50 Hz voltage completes 50 cycles each second and passes through zero twice in each cycle.
So it is zero 100 times a second.

Answer: About 311 V peak; it passes through zero 100 times every second.

Memory hook: The label shows rms: the real peak is √2 times higher.

Transformer

Suppose a small adaptor uses a step-down transformer to turn the 220 V mains into 11 V for a device.

Find the turns ratio. If the device draws 2 A, what current does the primary take (assume an ideal transformer)?

Try it first, then show the solutionSolution

$\frac{N _{p}}{N _{s}} = \frac{V _{p}}{V _{s}} = \frac{220}{11} = 20$
Power out $= 11 \times 2 = 22 W$ = power in (ideal)
$I_{p} = \frac{22}{220} = 0.1 A$

Answer: Turns ratio 20 : 1; primary current 0.1 A.

Memory hook: What the transformer gives in voltage, it takes back in current.

Power factor

Imagine a workshop motor that runs from a 220 V supply and draws 10 A, but the electricity meter shows it uses less power than 220 × 10 W.

Suppose the motor's power factor is 0.8 (current lagging). Find the actual power and the wattless current.

Try it first, then show the solutionSolution

$P = V I cos ϕ = 220 \times 10 \times 0.8 = 1760 W$
$sin ϕ = 1 - 0.64 = 0.6$
Wattless current $= 10 \times 0.6 = 6 A$ , which flows in the wires but does no work.

Answer: 1760 W; wattless current 6 A.

Memory hook: Power factor is the share of the current that actually works.

Topics in this chapter

Alternating Current formula sheet

rms values: $I = \frac{i _{m}}{2}, V = \frac{v _{m}}{2}$
Average of sinusoidal current over half a cycle: $i_{a v} = \frac{2 i _{m}}{π}$; Zero over a full cycle.
Inductive reactance: $X_{L} = ω L = 2 π ν L$; Current lags V by π/2.
Capacitive reactance: $X_{C} = \frac{1}{ω C} = \frac{1}{2 π ν C}$; Current leads V by π/2.
Impedance of series LCR: $Z = R^{2} + (X_{C} - X_{L})^{2}$
Phase angle: $tan ϕ = \frac{X _{C} - X _{L}}{R}$
Ohm's law for AC: $I = \frac{V}{Z}, i_{m} = \frac{v _{m}}{Z}$
Resonant frequency: $ω_{0} = \frac{1}{L C}, ν_{0} = \frac{1}{2 π L C}$; At resonance Z = R and cos φ = 1.
Average power: $P = V I cos ϕ = I^{2} R$; V, I rms.
Power factor: $cos ϕ = \frac{R}{Z}$
Wattless current: $I_{w a ttl ess} = I sin ϕ$
AC generator: $ε = N B A ω sin ω t$
Transformer (ideal): $\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}} = \frac{I _{p}}{I _{s}}$
Efficiency of a transformer: $η = \frac{V _{s} I _{s}}{V _{p} I _{p}} \times 100%$

Alternating Current: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Phasor diagram of a series LCR circuit

Impedance derivation (3-5 marks).

Draw

Draw the current phasor I as a horizontal arrow (reference).
Draw $V_{R}$ along I, $V_{L}$ straight up (90° ahead) and $V_{C}$ straight down (90° behind).
Draw the resultant $V_{L} - V_{C}$ (vertical) and complete the right-angled triangle with $V_{R}$ to get V.
Mark φ between V and I.

Label

I
$V_{R}$
$V_{L}$
$V_{C}$
$V_{L} - V_{C}$
V (resultant)
φ

Watch out: Also draw the impedance triangle (R, $X_{L} - X_{C}$ , Z): it makes $cos ϕ = R / Z$ obvious.

Step-up transformer

Transformer principle and working (3-5 marks).

Draw

Draw a rectangular laminated soft iron core.
Wind the primary coil (fewer turns) on one limb and the secondary coil (more turns) on the other, or both on the same limb one over the other.
Connect the primary to an AC source and the secondary to a load.

Label

Primary coil ( $N_{p}$ )
Secondary coil ( $N_{s} > N_{p}$ )
Laminated soft iron core
AC input $V_{p}$
Output $V_{s}$

Watch out: Make the secondary visibly have more turns for a step-up transformer; mark 'laminated' on the core.

How the board asks Alternating Current

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
rms and peak values	Find rms, peak or average values, frequency and times from an equation such as $i = 5 sin (100 π t)$ .	Read $i_{m}$ and ω from the equation; rms = peak/√2; ν = ω/2π; T = 1/ν.
Pure L or C circuits	Reactance, current, phase relation and power; effect of changing frequency or C; lamp in series with L or C.	Use $X_{L} = ω L$ and $X_{C} = 1/ ω C$ ; state the 90° phase relation; average power is zero.
Series LCR with phasors	Draw the phasor diagram and derive Z and tan φ; numericals on current and voltages; RC and RL circuits.	Start with the current phasor, draw $V_{R}$ , $V_{L}$ , $V_{C}$ , combine $V_{L}$ and $V_{C}$ , then use Pythagoras.
Resonance	Condition and frequency of resonance; graph of current against frequency; tuning a radio; voltages across L and C at resonance.	$X_{L} = X_{C}$ , $ω_{0} = 1/ L C$ , Z = R; sketch a peaked I-ω graph.
Power and power factor	Derive $P = V I cos ϕ$ ; find power, power factor or wattless current; why power factor matters.	Write the product of v and i, average over a cycle; $cos ϕ = R / Z$ .
Transformer	Principle, labelled diagram, derivation of the voltage and current ratios, energy losses, numericals.	Mutual induction; same flux per turn in both coils; $V_{s} / V_{p} = N_{s} / N_{p}$ ; list four losses and how each is reduced.

Tricks to remember Alternating Current

Short memory aids. Say each one aloud once, then use it on a question the same day.

ELI the ICE man
In an inductor (L), E (voltage) comes before I (current). In a capacitor (C), I comes before E.
Use it when: Lead and lag questions.
L loves high frequency, C loves low… for blocking
$X_{L}$ grows with frequency (inductor blocks high frequencies); $X_{C}$ falls with frequency (capacitor blocks low frequencies and DC).
Use it when: Graphs of reactance against frequency and filter-type reasoning.
Resonance = balance
At resonance the L and C reactances balance, leaving only R: Z = R, φ = 0, cos φ = 1, current largest.
Use it when: Resonance MCQs and 5-mark answers.
cos φ = R/Z
Power factor is the 'resistive share' of the impedance triangle (R adjacent, Z hypotenuse).
Use it when: Power factor numericals.
Turns follow voltage, current goes the other way
$V \propto N$ ; $I \propto 1/ N$ . More turns, more volts, fewer amps.
Use it when: Transformer numericals.
√2 bridge
Peak to rms: divide by √2 (×0.707). rms to peak: multiply by √2 (×1.414).
Use it when: Every AC numerical.

Alternating Current: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: rms of the mains · 1 mark · MCQBoard level

An AC voltage is $v = 311 sin (100 π t)$ volts. Its rms value and frequency are:
(a) 311 V, 100 Hz
(b) 220 V, 50 Hz
(c) 220 V, 100 Hz
(d) 156 V, 50 Hz

Try it first, then show the solutionSolution

$V = \frac{311}{2} \approx 220 V$ ; $ν = \frac{100 π}{2 π} = 50 H z$

Answer: (b)

Example 2: Power in an inductor · 1 mark · Assertion-reasonBoard level

Assertion (A): The average power consumed by a pure inductor over a complete cycle of AC is zero.
Reason (R): In a pure inductor, the current and voltage differ in phase by $\frac{π}{2}$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

$P = V I cos ϕ$ and $cos 9 0^{\circ} = 0$ . Both true; R explains A.

Answer: (a)

Example 3: Reading an AC equation · 2 marks · Very short answerHarder board level

A current is $i = 5 sin (100 π t)$ A. Find its rms value, frequency and the least time it takes to fall from its maximum to zero. [2 marks]

Try it first, then show the solutionSolution

$I = \frac{5}{2} \approx 3.54 A$ ; $ν = 50 H z$ , T = 0.02 s
From maximum to zero is a quarter period: $\frac{0.02}{4} = 5 \times 1 0^{- 3} s$

Answer: 3.54 A; 50 Hz; 5 ms

Example 4: Derivation: impedance of a series LCR circuit · 3 marks · Short answerHarder board level

A resistor R, an inductor L and a capacitor C are in series with $v = v_{m} sin ω t$ . Using a phasor diagram, derive the impedance and the phase angle. When does the current lead the voltage? [3 marks]

Try it first, then show the solutionSolution

Same current through all; take I as the reference phasor. $V_{R} = i_{m} R$ along I; $V_{L} = i_{m} X_{L}$ 90° ahead; $V_{C} = i_{m} X_{C}$ 90° behind.
$V_{L}$ and $V_{C}$ are opposite, giving $i_{m} (X_{C} - X_{L})$ perpendicular to $V_{R}$ .
$v_{m}^{2} = i_{m}^{2} [R^{2} + (X_{C} - X_{L})^{2}]$ , so $Z = \frac{v _{m}}{i _{m}} = R^{2} + (X_{C} - X_{L})^{2}$ and $tan ϕ = \frac{X _{C} - X _{L}}{R}$ .
The current leads when $X_{C} > X_{L}$ (capacitive circuit).

Answer: $Z = R^{2} + (X_{C} - X_{L})^{2}$ , $tan ϕ = \frac{X _{C} - X _{L}}{R}$ ; current leads when $X_{C} > X_{L}$ .

Example 5: Resonance and power factor · 5 marks · Long answerHarder board level

(a) What is resonance in a series LCR circuit? Derive the resonant frequency. (b) Sketch the variation of current with frequency. (c) A circuit has L = 0.1 H, $C = 10 μ F$ , R = 20 Ω and a 220 V variable-frequency source. Find the current and power at resonance and at ω = 500 rad s⁻¹. (d) State one use of resonance. [5 marks]

Try it first, then show the solutionSolution

(a) When $X_{L} = X_{C}$ , Z = R (minimum) and the current is maximum: resonance. $ω_{0} L = \frac{1}{ω _{0} C} \Rightarrow ω_{0} = \frac{1}{L C}$ .
(b) Current rises to a peak at $ω_{0}$ and falls on either side (a peaked curve).
(c) $ω_{0} = \frac{1}{1 0 ^{- 6}} = 1000 r a d s^{- 1}$ ; $I = \frac{220}{20} = 11 A$ ; $P = I^{2} R = 121 \times 20 = 2420 W$ .
At 500 rad s⁻¹: $X_{L} = 50 Ω$ , $X_{C} = 200 Ω$ , $Z = 400 + 22500 \approx 151 Ω$ ; $I \approx 1.45 A$ ; $P = I^{2} R \approx 42 W$ .
(d) Tuning a radio or TV receiver to the frequency of the wanted station.

Answer: $ω_{0} = 1/ L C$ ; at resonance 11 A and 2420 W; at 500 rad s⁻¹ about 1.45 A and 42 W; used in tuning circuits.

Example 6: Active and wattless current · 3 marks · Short answerHarder board level

A coil draws 5 A (rms) from a 220 V (rms), 50 Hz supply with the current lagging by $6 0^{\circ}$ . Find the average power, the wattless current and the resistance and inductive reactance of the coil. [3 marks]

Try it first, then show the solutionSolution

$P = 220 \times 5 \times cos 6 0^{\circ} = 550 W$ ; wattless current $= 5 sin 6 0^{\circ} \approx 4.33 A$
$Z = \frac{220}{5} = 44 Ω$
$R = Z cos ϕ = 22 Ω$ ; $X_{L} = Z sin ϕ = 44 \times 0.866 \approx 38.1 Ω$

Answer: 550 W; 4.33 A; R = 22 Ω; $X_{L} \approx 38.1 Ω$

Example 7: Short-circuiting the capacitor · 3 marks · Short answerHOTS / competency

In a series LCR circuit the rms voltages across R, L and C are 30 V, 80 V and 40 V. (a) Find the source voltage. (b) The capacitor is now short-circuited, the source being unchanged. Find the new voltage across R. [3 marks]

Try it first, then show the solutionSolution

(a) $V = 3 0^{2} + (80 - 40)^{2} = 50 V$
(b) The voltages are in the ratio of R : $X_{L}$ : $X_{C}$ , so take R = 30k, $X_{L}$ = 80k (k a constant). Without C: $Z^{'} = k 3 0^{2} + 8 0^{2} = 85.4 k$
Current falls in the ratio $\frac{50 k}{85.4 k}$ ; $V_{R}^{'} = 30 \times \frac{50}{85.4} \approx 17.6 V$

Answer: (a) 50 V (b) about 17.6 V

Example 8: A real transformer · 2 marks · Very short answerHOTS / competency

A step-up transformer takes 5 A from a 220 V supply and gives an output of 2200 V. Its efficiency is 90%. Find the output current. Explain, in terms of energy, why the output current must be smaller than the input current. [2 marks]

Try it first, then show the solutionSolution

Input power $= 220 \times 5 = 1100 W$ ; output power $= 0.9 \times 1100 = 990 W$
$I_{s} = \frac{990}{2200} = 0.45 A$
Energy cannot be created: if the voltage is raised ten times, the current must fall at least ten times so that output power does not exceed input power.

Answer: 0.45 A; power out cannot exceed power in, so higher voltage means lower current.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Tuning a radio

Suppose the tuning circuit of a radio is a series combination of a coil of inductance $10 μ H$ , a variable capacitor and a small resistance. Turning the knob changes the capacitance, and the station whose frequency equals the circuit's resonant frequency is heard loudly.

(i) · 1 mark
What is the condition for resonance in the circuit?
Answer
Answer: $X_{L} = X_{C}$
Then the impedance is minimum (equal to R) and the current is maximum.
(ii) · 1 mark
What capacitance tunes the circuit to 1 MHz?
Answer
Answer: About 2.5 nF
$C = \frac{1}{4 π ^{2} ν ^{2} L} = \frac{1}{39.5 \times 1 0 ^{12} \times 1 0 ^{- 5}} \approx 2.5 \times 1 0^{- 9} F$
(iii) · 1 mark
If the capacitance is increased, does the resonant frequency go up or down?
Answer
Answer: Down
$ν_{0} = \frac{1}{2 π L C}$ falls as C rises.
(iv) · 1 mark
What is the power factor of the circuit at resonance?
Answer
Answer: 1
Z = R, so $cos ϕ = R / Z = 1$ .

Power from the station to the town

Suppose a power station generates electricity at 11 kV. A step-up transformer raises it to 220 kV for transmission over long lines, and a step-down transformer near the town lowers it again. Assume the transformers are ideal.

(i) · 1 mark
What is the turns ratio $N_{s} : N_{p}$ of the step-up transformer?
Answer
Answer: 20 : 1
$\frac{220}{11} = 20$
(ii) · 1 mark
If the station sends 22 MW, what is the current in the transmission line?
Answer
Answer: 100 A
$I = \frac{22 \times 1 0 ^{6}}{220 \times 1 0 ^{3}} = 100 A$
(iii) · 1 mark
How would the line loss change if the power were sent at 11 kV instead?
Answer
Answer: It would be 400 times larger.
The current would be 20 times larger, and loss $\propto I^{2}$ .
(iv) · 1 mark
On what principle does a transformer work?
Answer
Answer: Mutual induction
The changing flux of the primary links the secondary and induces an emf in it.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

An alternating current is given by $i = 5 sin (100 π t)$ A. Find (a) its rms value, (b) its frequency, (c) the least time it takes to fall from its peak to zero. [2 marks] (2 marks)

Step	Marks
(a) $I = \frac{5}{2} \approx 3.54 A$ ; (b) $ν = \frac{100 π}{2 π} = 50 H z$	1
(c) Peak to zero is a quarter period: $\frac{T}{4} = \frac{1}{4 \times 50} = 5 \times 1 0^{- 3} s$	1

Using a phasor diagram, derive the impedance of a series LCR circuit and the phase angle between voltage and current. [3 marks] (3 marks)

Step	Marks
Phasor diagram: I as reference; $V_{R}$ along I, $V_{L}$ 90° ahead, $V_{C}$ 90° behind	1
$v_{m}^{2} = v_{R}^{2} + (v_{C} - v_{L})^{2} = i_{m}^{2} [R^{2} + (X_{C} - X_{L})^{2}]$	1
$Z = R^{2} + (X_{C} - X_{L})^{2}$ ; $tan ϕ = \frac{X _{C} - X _{L}}{R}$	1

(a) State the principle of a transformer and draw a labelled diagram of a step-up transformer. (b) Derive the ratio of secondary to primary voltage. (c) List two energy losses and how each is reduced. (d) An ideal transformer has 100 turns in the primary and 2000 in the secondary; the primary is connected to 220 V AC. Find the secondary voltage. [5 marks] (5 marks)

Step	Marks
(a) Mutual induction; diagram with primary (fewer turns), secondary (more turns), laminated soft iron core	1
(b) Same flux per turn: $ε_{s} = - N_{s} \frac{d Φ}{d t}$ , $v_{p} = - N_{p} \frac{d Φ}{d t}$	1
$\frac{v _{s}}{v _{p}} = \frac{N _{s}}{N _{p}}$	1
(c) Any two: flux leakage (wind coils over each other), copper loss (thick wire), eddy currents (laminated core), hysteresis (soft iron core)	1
(d) $V_{s} = 220 \times \frac{2000}{100} = 4400 V$	1

Where students lose marks in Alternating Current

The slips examiners see most often, and what to write instead.

Mistake: Adding V_R, V_L and V_C as numbers
Treating AC like DC.
Wrong working
$V_{R} = 30 V$ , $V_{L} = 80 V$ , $V_{C} = 40 V$ : source = 150 V
Right working
$V = 3 0^{2} + (80 - 40)^{2} = 50 V$
Do this: Combine with the phasor diagram: $V_{L}$ and $V_{C}$ subtract, the result adds to $V_{R}$ at right angles.
Usually costs: Full marks of the numerical.
Mistake: Using frequency ν in place of ω in reactances
Forgetting the 2π.
Wrong working
$X_{L} = ν L = 50 \times 0.1 = 5 Ω$
Right working
$X_{L} = 2 π ν L = 31.4 Ω$
Do this: Write $ω = 2 π ν$ explicitly before substituting.
Usually costs: 1-2 marks.
Mistake: Mixing peak and rms values in power
Using $v_{m} i_{m}$ for average power.
Do this: $P = V I cos ϕ$ with rms values, or $\frac{1}{2} v_{m} i_{m} cos ϕ$ with peak values.
Usually costs: 1 mark.
Mistake: Not starting the phasor diagram with the current
Drawing V first in a series circuit.
Do this: In series the current is common: draw I first, then $V_{R}$ along it, $V_{L}$ ahead by 90°, $V_{C}$ behind by 90°.
Usually costs: 1 mark for the diagram.
Mistake: Saying a transformer increases power
Thinking step-up means more of everything.
Do this: Power out ≤ power in. Step-up raises voltage and lowers current in the same ratio.
Usually costs: 1 mark in reasoning items.
Mistake: Writing about the Q factor or LC oscillations
Older syllabi included them.
Do this: Sharpness of resonance and LC oscillations are not in the 2026-27 syllabus; keep the resonance answer to the condition, frequency, graph and use.
Usually costs: Time.

NCERT exercise guide: Alternating Current

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 7.1 and 7.2 (resistor; rms and peak values)

What it tests: rms and peak current through a resistor on AC; peak voltage of a supply; rms current from peak current.

How to approach it: Convert with √2; I = V/R using rms values.

Same kind as Exercise 7.1

A 55 Ω resistor is connected to a 220 V, 50 Hz supply. Find the rms and peak currents and the average power.

Show the solutionSolution

$I = \frac{220}{55} = 4 A$ (rms)
$i_{m} = 42 \approx 5.66 A$
$P = I^{2} R = 16 \times 55 = 880 W$

Answer: 4 A rms, 5.66 A peak, 880 W

Exercises 7.3 to 7.5 (pure inductor and capacitor)

What it tests: rms current through an inductor or a capacitor on a given supply, and the power each absorbs.

How to approach it: $X_{L} = 2 π ν L$ , $X_{C} = 1/2 π ν C$ , I = V/X; average power zero.

Same kind as Exercise 7.4

A $100 μ F$ capacitor is connected to a 110 V, 50 Hz supply. Find the rms current and the average power.

Show the solutionSolution

$X_{C} = \frac{1}{2 π \times 50 \times 1 0 ^{- 4}} = \frac{1}{0.0314} \approx 31.8 Ω$
$I = \frac{110}{31.8} \approx 3.46 A$
Average power = 0 (current leads by 90°)

Answer: About 3.46 A; zero power

Exercises 7.7 and 7.8 (series LCR circuit; resonance)

What it tests: Impedance, current and power factor of a series LCR circuit, the resonant frequency of a given circuit, and the current at resonance.

How to approach it: Compute $X_{L}$ and $X_{C}$ , then Z, I = V/Z, cos φ = R/Z; resonance from $ω_{0} = 1/ L C$ . (Exercise 7.6, on LC oscillations, is outside the 2026-27 syllabus.)

Same kind as Exercise 7.8

A series circuit has L = 0.1 H, $C = 10 μ F$ and R = 20 Ω on a 220 V variable-frequency source. Find the current at resonance and at ω = 500 rad s⁻¹, and the power factor at 500 rad s⁻¹.

Show the solutionSolution

Resonance ( $ω_{0} = 1000 r a d s^{- 1}$ ): $I = \frac{220}{20} = 11 A$
At 500 rad s⁻¹: $X_{L} = 50 Ω$ , $X_{C} = \frac{1}{500 \times 1 0 ^{- 5}} = 200 Ω$
$Z = 2 0^{2} + 15 0^{2} = 22900 \approx 151 Ω$ ; $I = \frac{220}{151} \approx 1.45 A$
$cos ϕ = \frac{20}{151} \approx 0.13$

Answer: 11 A at resonance; about 1.45 A and power factor 0.13 at 500 rad s⁻¹

Alternating Current: previous year questions

Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 4.4 marks per paper; 18 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Alternating Current

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 55/1/1, Q6 · 1 mark · MCQ

In a series LCR circuit, the voltage across the resistor, capacitor and inductor is 10 V each. If the capacitor is short circuited, the voltage across the inductor will be

(A)10 V
(B) $52$ V
(C) $\frac{5}{2}$ V
(D) $102$ V

Try it first, then show our solutionSolution

$V_{L} = V_{C}$ , so the circuit is at resonance and the source voltage is $V = V_{R} = 10$ V. Also $R = X_{L}$ .
With C shorted, $Z = R^{2} + X_{L}^{2} = 2 R$ , so $I^{'} = \frac{10}{2 R}$ .
$V_{L} = I^{'} X_{L} = \frac{10}{2} = 52$ V.
1 mark

Answer: $52$ V, option (B)

Where the marks are

Uses resonance to get V = 10 V and R = X_L, then finds $V_{L} = 52$ V (1 mark)

See it in the full 2026 board paper, set 55/1/1

2026, set 55/1/1, Q33(b) · 5 marks · Long answer

(a) Draw a labelled diagram of a step-up transformer. State the principle on which it works and obtain the ratio of secondary voltage to primary voltage in terms of number of turns and currents in the two coils.
(b) The ratio of the number of turns in the primary to the secondary of an ideal transformer is 1 : 5. If 5 kW power at 200 V is supplied to the primary, find
(i) current in the primary, and
(ii) output voltage.

Try it first, then show our solutionSolution

(a) Diagram (described in words). Draw a closed rectangular laminated soft-iron core.
- On one arm, wind the primary coil $P$ with few turns ( $N_{p}$ ) and connect it to the ac input voltage $V_{p}$ .
- On the other arm, wind the secondary coil $S$ with many more turns ( $N_{s} > N_{p}$ ) and connect it to the load across the output $V_{s}$ .
- Label the core, the primary, the secondary, the input and the output.
1 mark
Principle. A transformer works on mutual induction. An alternating current in the primary produces a changing magnetic flux in the core. This changing flux links the secondary and induces an emf in it.
1 mark
Ratio in terms of turns. Assume an ideal transformer with no flux leakage. The same flux $Φ$ links each turn of both coils.
$ε_{s} = - N_{s} \frac{d Φ}{d t}, ε_{p} = - N_{p} \frac{d Φ}{d t}$
With negligible coil resistance, $V_{p} = ε_{p}$ and $V_{s} = ε_{s}$ . Therefore:
$\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$
1 mark
Ratio in terms of currents. For an ideal transformer with no energy losses, input power equals output power:
$V_{p} I_{p} = V_{s} I_{s}$
Therefore:
$\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}} = \frac{I _{p}}{I _{s}}$
(b)(i) Current in the primary.
$I_{p} = \frac{P}{V _{p}} = \frac{5000 W}{200 V} = 25 A$
1 mark
(b)(ii) Output voltage.
$V_{s} = V_{p} \frac{N _{s}}{N _{p}} = 200 \times \frac{5}{1} = 1000 V$
1 mark

Answer: (a) $\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}} = \frac{I _{p}}{I _{s}}$ ; (b)(i) $I_{p} = 25$ A; (ii) $V_{s} = 1000$ V

Where the marks are

Labelled step-up transformer diagram with $N_{s} > N_{p}$ on a soft-iron core (1 mark)
States the principle of mutual induction (1 mark)
Derives $V_{s} / V_{p} = N_{s} / N_{p}$ and, using power conservation, $= I_{p} / I_{s}$ (1 mark)
Calculates the primary current as 25 A (1 mark)
Calculates the output voltage as 1000 V (1 mark)

See it in the full 2026 board paper, set 55/1/1

2025, set 55/1/1, Q25 · 3 marks · Short answer

Answer the following.

(a)
A parallel plate capacitor is charged by an ac source. Show that the sum of conduction current ( $I_{c}$ ) and the displacement current ( $I_{d}$ ) has the same value at all points of the circuit.
0 marks
(b)
In case (a) above, is Kirchhoff's first rule (junction rule) valid at each plate of the capacitor ? Explain.
0 marks

Try it first, then show our solutionSolution

(a) The total current is $I = I_{c} + I_{d}$ . In the connecting wires (outside the capacitor) there is no changing electric field, so $I_{d} = 0$ and $I = I_{c}$ .
½ mark
Between the plates no charge crosses the gap, so $I_{c} = 0$ and $I = I_{d} = ε_{0} \frac{d Φ _{E}}{d t} = ε_{0} \frac{d ( E A )}{d t}$ .
½ mark
The field between the plates is $E = \frac{Q}{ε _{0} A}$ , so $I_{d} = ε_{0} A \frac{d}{d t} (\frac{Q}{ε _{0} A}) = \frac{d Q}{d t}$ .
½ mark
$\frac{d Q}{d t}$ is exactly the conduction current $I_{c}$ in the wires. So $I_{d}$ (between the plates) equals $I_{c}$ (in the wires), and $I_{c} + I_{d}$ has the same value at every point of the circuit.
½ mark
(b) Yes. At each plate the conduction current $I_{c}$ entering the plate equals the displacement current $I_{d}$ leaving it through the gap. With $I_{d}$ included, current into the junction equals current out of it, so the junction rule holds at each plate.
1 mark

Answer: (a) I_d = dQ/dt = I_c, so I_c + I_d is the same everywhere; (b) Yes, because I_c into a plate equals I_d leaving it.

Where the marks are

Shows $I_{d} = 0$ outside and $I_{c} = 0$ inside the plates (1 mark)
Uses $E = Q / ε_{0} A$ to show $I_{d} = d Q / d t = I_{c}$ (1 mark)
States that the junction rule is valid, with the $I_{c} = I_{d}$ reason (1 mark)

See it in the full 2025 board paper, set 55/1/1

Every Alternating Current question, by year

For 2027

Sample paper 2026-27: Q33(A) (5 marks, solved), Q33(B) (5 marks, solved)

2026

Set 55/1/1: Q6 (1 mark, solved), Q33(b) (5 marks, solved)

Set 55/5/1: Q5 (1 mark, solved), Q12 (1 mark, solved)

Sample paper 2025-26: Q3 (1 mark, solved), Q10 (1 mark, solved), Q33(II) (5 marks, solved)

2025

Set 55/1/1: Q6 (1 mark, solved), Q25 (3 marks, solved)

Set 55/7/1: Q4 (1 mark, solved), Q5 (1 mark, solved)

Sample paper 2024-25: Q6 (1 mark, solved), Q32(I) (5 marks, solved), Q32(II) (5 marks, solved)

2024

Set 55/1/1: Q7 (1 mark, solved), Q25 (3 marks, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 55/5/1: Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Sample paper 2023-24: Q9 (1 mark, solved), Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

2023

Set 55/1/1: Q27(a) (3 marks, solved), Q27(b) (3 marks, solved)

Set 55/5/1: Q13 (1 mark, solved), Q27 (3 marks, solved)

Practice questions on Alternating Current

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The capacitance in an AC circuit is doubled and the frequency is halved. The capacitive reactance:
(a) doubles
(b) halves
(c) stays the same
(d) becomes four times
Hint
$X_{C} = 1/2 π ν C$ .
Answer
(c) stays the same
Q2 · 1 mark · MCQBoard level
The power factor of a series LCR circuit at resonance is:
(a) 0
(b) 0.5
(c) 1
(d) 0.707
Hint
Z = R at resonance.
Answer
(c) 1
Q3 · 1 mark · Very short answerBoard level
Can a transformer be used to step up a DC voltage? Give a reason.
Hint
What does induction need?
Answer
No. A steady DC current gives a constant flux in the core, so no emf is induced in the secondary.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): At resonance, the current in a series LCR circuit is maximum.
Reason (R): At resonance the impedance of the circuit equals its resistance.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Z is minimum when $X_{L} = X_{C}$ .
Answer
(a)
Q5 · 2 marks · Very short answerHarder board level
A coil of inductance 0.2 H and resistance 50 Ω is connected to a 220 V, 50 Hz supply. Find (a) the impedance, (b) the rms current, (c) the power factor and the phase angle, (d) the average power consumed.
Hint
A real coil is R and L in series: $Z = R^{2} + X_{L}^{2}$ ; only R uses power.
Answer
(a) $X_{L} = 2 π \times 50 \times 0.2 \approx 62.8 Ω$ ; $Z = 5 0^{2} + 62. 8^{2} \approx 80.3 Ω$ . (b) $I = \frac{220}{80.3} \approx 2.74 A$ . (c) $cos ϕ = \frac{R}{Z} = \frac{50}{80.3} \approx 0.62$ , so $ϕ \approx 51. 5^{\circ}$ , current lagging. (d) $P = I^{2} R \approx 2.7 4^{2} \times 50 \approx 375 W$ (check: $V I cos ϕ = 220 \times 2.74 \times 0.62 \approx 375 W$ ).
Q6 · 3 marks · Short answerHarder board level
A 40 Ω resistor and a $50 μ F$ capacitor are in series across a 220 V, 50 Hz supply. Find the impedance, current and phase angle. Does the current lead or lag?
Hint
No inductor: $Z = R^{2} + X_{C}^{2}$ .
Answer
$X_{C} \approx 63.7 Ω$ ; $Z = 1600 + 4053 \approx 75.2 Ω$ ; $I = 220/75.2 \approx 2.93 A$ ; $tan ϕ = 63.7/40$ , φ ≈ 58°; current leads.
Q7 · 3 marks · Short answerHarder board level
Derive an expression for the average power consumed in a series LCR circuit, and show that it is zero for a purely inductive circuit.
Hint
Average $cos (2 ω t + ϕ)$ over a cycle.
Answer
$p = v i = v_{m} i_{m} sin ω t sin (ω t + ϕ) = \frac{v _{m} i _{m}}{2} [cos ϕ - cos (2 ω t + ϕ)]$ . The second term averages to zero, so $P = \frac{v _{m} i _{m}}{2} cos ϕ = V I cos ϕ$ . For a pure inductor φ = −90°, cos φ = 0, so P = 0.
Q8 · 5 marks · Long answerHarder board level
(a) Explain the principle and working of a transformer with a labelled diagram. (b) Show that for an ideal transformer $\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}} = \frac{I _{p}}{I _{s}}$ . (c) Give the main energy losses. (d) A transformer with 1100 turns in its primary runs from a 220 V supply and lights a 12 V, 24 W lamp at full brightness. Find the number of secondary turns, the lamp current and, if the transformer is 80% efficient, the primary current.
Hint
Same flux per turn; power in = power out.
Answer
(a) Mutual induction; primary and secondary on a laminated iron core. (b) $ε_{s} = - N_{s} d Φ/ d t$ , $v_{p} = - N_{p} d Φ/ d t$ gives the voltage ratio; $V_{p} I_{p} = V_{s} I_{s}$ gives the current ratio. (c) Flux leakage, winding resistance, eddy currents, hysteresis. (d) $N_{s} = 1100 \times \frac{12}{220} = 60$ turns; lamp current $= \frac{24}{12} = 2 A$ ; input power $= \frac{24}{0.8} = 30 W$ , so $I_{p} = \frac{30}{220} \approx 0.14 A$ .
Q9 · 2 marks · Very short answerHOTS / competency
A bulb and an inductor (a coil) are connected in series to an AC supply. A soft iron rod is slowly pushed into the coil. What happens to the brightness of the bulb? Explain. What would happen with a DC supply?
Hint
Iron raises L.
Answer
The iron raises L, so $X_{L} = ω L$ and Z increase and the current falls: the bulb dims. On DC, $X_{L} = 0$ in steady state, so inserting the rod makes no lasting difference to the brightness.
Q10 · 2 marks · Very short answerHOTS / competency
What capacitance must be connected in series with a 1 H inductor (negligible resistance) for resonance at 50 Hz? ( $π^{2} \approx 9.87$ )
Hint
$C = 1/ ω^{2} L$ .
Answer
$ω = 100 π$ ; $C = \frac{1}{( 100 π ) ^{2} \times 1} = \frac{1}{98 700} \approx 1.01 \times 1 0^{- 5} F \approx 10 μ F$ .
Q11 · 2 marks · Very short answerHarder board level
An alternating current has a peak value of 10 A. Find its rms value and its average value over a half cycle. What is its average over a full cycle?
Hint
Half-cycle average is $2 i_{m} / π$ .
Answer
rms $= 10/ 2 \approx 7.07 A$ ; half-cycle average $= 20/ π \approx 6.37 A$ ; full-cycle average zero.
Q12 · 2 marks · Very short answerHarder board level
Sketch graphs of $X_{L}$ and $X_{C}$ against frequency. At what point on your graph does resonance occur?
Hint
One is a straight line, the other a hyperbola.
Answer
$X_{L}$ is a straight line through the origin (∝ ν); $X_{C}$ is a rectangular hyperbola (∝ 1/ν). Resonance is at the frequency where the two graphs cross ( $X_{L} = X_{C}$ ).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Relation between rms and peak current?Show answer
$I = i_{m} / 2$ .
2.Phase relation of current and voltage in a pure inductor and a pure capacitor?Show answer
Inductor: current lags by π/2. Capacitor: current leads by π/2.
3.Write $X_{L}$ and $X_{C}$ and how each varies with frequency.Show answer
$X_{L} = ω L$ (∝ ν); $X_{C} = 1/ ω C$ (∝ 1/ν).
4.Write the impedance of a series LCR circuit.Show answer
$Z = R^{2} + (X_{C} - X_{L})^{2}$ .
5.Condition and frequency of resonance?Show answer
$X_{L} = X_{C}$ ; $ν_{0} = \frac{1}{2 π L C}$ .
6.Define power factor. What is it at resonance?Show answer
cos φ, the ratio of average power to VI (= R/Z); 1 at resonance.
7.What is wattless current?Show answer
The component $I sin ϕ$ of current at 90° to the voltage, which consumes no average power.
8.Write the transformer voltage and current ratios.Show answer
$V_{s} / V_{p} = N_{s} / N_{p}$ ; $I_{s} / I_{p} = N_{p} / N_{s}$ (ideal).
9.List four energy losses in a transformer.Show answer
Flux leakage, resistance of windings, eddy currents in the core, hysteresis.

Best order to study it

Start with AC through a resistor, rms values and the idea of phasors.
Learn AC through an inductor and a capacitor: reactance, phase and zero average power.
Draw the series LCR phasor diagram and derive Z and tan φ; practise numericals.
Study resonance: condition, frequency, current-frequency graph and radio tuning.
Derive the average power, learn the power factor and wattless current.
Finish with the AC generator (recap) and the transformer: principle, ratios, losses and transmission.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you convert between peak and rms values and read ω and ν from an AC equation?
Can you show that current lags in an inductor and leads in a capacitor by π/2?
Can you sketch $X_{L}$ and $X_{C}$ against frequency?
Can you draw the series LCR phasor diagram and derive Z and tan φ?
Can you find the current, voltages and phase in an LCR, RL or RC circuit?
Can you derive the resonant frequency and sketch current against frequency?
Can you derive $P = V I cos ϕ$ and explain wattless current?
Can you say why power factor should be close to 1?
Can you derive the transformer ratios and list its losses?
Can you explain why power is transmitted at high voltage?

Alternating Current: frequently asked questions

Is the analytical solution of the LCR circuit in the 2026-27 syllabus?

No. The syllabus says 'LCR series circuit (phasors only)'. Derive the impedance and phase from the phasor diagram.

Are LC oscillations and the Q factor asked?

No. LC oscillations and the sharpness of resonance (Q factor) are not in the 2026-27 syllabus.

Is the AC generator in this chapter?

The CBSE syllabus lists 'AC generator, Transformer' in this unit. NCERT explains the generator in Chapter 6 (section 6.8). Learn its principle, diagram and \varepsilon = NBA\omega\sin\omega t.

Why does a pure inductor consume no power?

Current and voltage are 90° out of phase, so cos φ = 0. Energy is stored in the magnetic field during one quarter cycle and returned to the source in the next.

Which 5-mark questions are most common from this chapter?

In the board papers in our question bank: the series LCR circuit (phasor diagram, impedance, resonance) and the transformer (principle, ratio, losses). Practise both as full derivations with a numerical.

Can a transformer work with DC?

No. It needs a changing flux in the core to induce an emf in the secondary; a steady DC current gives a constant flux.

Based on: CBSE Physics (042) syllabus 2026-27; NCERT Physics Part I, Textbook for Class XII, Chapter 7 (Reprint 2026-27)