PhysicsCBSE Class 12 · Board exam 2026, set 55/1/1 · 5 marks · Long answer

CBSE Class 12 Physics 2026: question 33(b)

Set in the CBSE Class 12 board exam of 2026, paper 55/1/1, section E, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerAlternating Current

(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.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 55/1/1.

Hints

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

Hint 1

The same changing flux links both coils. Write the induced emf in each using Faraday's law.

Hint 2

In an ideal transformer, power in equals power out, which links the currents to the voltages.

Hint 3

Primary current is power divided by primary voltage. For the output voltage, multiply the primary voltage by the turns ratio $N_{s} / N_{p}$ .

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) 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)

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 paper2026 board paper, set 55/1/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Physics 2026: question 33(b)

Hints

Step-by-step solution

Revise the chapter

More questions from Alternating Current