PhysicsCBSE Class 12 · Sample paper, session 2025-26 · 5 marks · Long answer

CBSE Class 12 Physics sample paper 2025-26: question 33(II)

From the official CBSE sample question paper for 2025-26, section E, question 33(II), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(II) · 5 marks · Long answerAlternating Current

(A) An input potential $V_{in} = 200 sin 100 π t$ V is provided to an ideal transformer having 1000 turns in primary coil and 100 turns in secondary coil as shown in figure. The load circuit has a resistance of 4Ω, a capacitive reactance of 2Ω and an inductive reactance of 6Ω.
Find:
(i) the output voltage across the load circuit
(ii) the current flowing through the load circuit
(iii) the power supplied to the load circuit by the transformer
(B) State the working principle of a transformer and explain how it is a key component in the transfer of electrical power over long distances.

Figure in the paper: AC source connected to primary coil of a transformer (many turns); secondary coil (fewer turns) connected to a series load of R, L and C. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

Open one hint at a time and try again before the next.

Hint 1

For an ideal transformer, the voltage ratio equals the turns ratio.

Hint 2

Find the impedance from R and $(X_{L} - X_{C})$ ; whether $X_{L}$ or $X_{C}$ is larger decides if the current leads or lags.

Hint 3

Average power uses rms values: $P = V_{r m s} I_{r m s} cos ϕ$ , or simply $I_{r m s}^{2} R$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(A)(i) For an ideal transformer:
$\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}} = \frac{100}{1000} = 0.1$
So $V_{o u t} = 0.1 \times 200 sin 100 π t = 20 sin 100 π t$ V, i.e. peak value 20 V.
1 mark
(ii) Impedance:
$Z = R^{2} + (X_{L} - X_{C})^{2} = 4^{2} + (6 - 2)^{2} = 42 Ω$
Peak current:
$I_{m} = \frac{V _{m}}{Z} = \frac{20}{4 2} = \frac{5}{2} \approx 3.54 A$
Phase: $tan ϕ = \frac{X _{L} - X _{C}}{R} = 1$ , so $ϕ = π /4$ . Since $X_{L} > X_{C}$ , the circuit is inductive and the current lags the voltage.
$i = \frac{5}{2} sin (100 π t - \frac{π}{4}) A$
1 mark
(iii) Average power:
$P = V_{r m s} I_{r m s} cos ϕ = \frac{V _{m} I _{m}}{2} cos ϕ = \frac{20 \times \frac{5}{2}}{2} \times \frac{1}{2} = 25 W$
Check: $P = I_{r m s}^{2} R = (\frac{5}{2})^{2} \times 4 = 25$ W.
1 mark
(B) 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, with $\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$ .
1 mark
Role in power transmission. At the power station, a step-up transformer raises the voltage to a very high value. For the same power $P = V I$ , the current in the lines becomes very small. The line loss $I^{2} R$ therefore drops greatly. Near the cities, step-down transformers reduce the voltage to safe values for use.
1 mark

Answer: (i) $V_{o u t} = 20 sin 100 π t$ V; (ii) $i = \frac{5}{2} sin (100 π t - π /4)$ A ( $Z = 42 Ω$ , current lags); (iii) P = 25 W; (B) mutual induction; step-up gives high V, low I, so low $I^{2} R$ loss

Where the marks are

Uses the turns ratio to get $V_{o u t} = 20 sin 100 π t$ V (1 mark)
Finds $Z = 42 Ω$ , $I_{m} = 5/ 2$ A and phase $π /4$ to write i(t) (1 mark)
Computes average power = 25 W (1 mark)
States the principle of mutual induction (1 mark)
Explains that step-up gives high voltage and low current, reducing $I^{2} R$ losses, with step-down at the user end (1 mark)

Revise the chapter

Chapter: Alternating Current (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Alternating Current, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2025-26: every question, by section.

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CBSE Class 12 Physics sample paper 2025-26: question 33(II)

Hints

Step-by-step solution

Revise the chapter

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