PhysicsCBSE Class 12 · Board exam 2026, set 55/1/1 · 1 mark · MCQ

CBSE Class 12 Physics 2026: question 6

Set in the CBSE Class 12 board exam of 2026, paper 55/1/1, section A, question 6, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q6 · 1 mark · MCQAlternating Current

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

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

Equal $V_{L}$ and $V_{C}$ tells you something about the circuit.

Hint 2

Find the new impedance with only R and L in the circuit.

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.

$V_{L} = V_{C}$ , 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)

Revise the chapter

Chapter: Alternating Current (notes, formulas, where marks are lost and every past question)
Topic: Series LCR Circuit and Resonance

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 6

Hints

Step-by-step solution

Revise the chapter

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