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

CBSE Class 12 Physics 2026: question 31(a)

Set in the CBSE Class 12 board exam of 2026, paper 55/5/1, section E, question 31(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(a) · 5 marks · Long answerCurrent Electricity

(i) Derive the condition for which a Wheatstone Bridge is balanced.
(ii) Determine the current in 3 $Ω$ branch of a Wheatstone Bridge in the circuit shown in the figure.

Figure in the paper: Wheatstone bridge: top junction to left junction has 20 Ω, top junction to right junction has 2 Ω, left junction to bottom junction has 12 Ω, right junction to bottom junction has 1 Ω, and a 3 Ω resistor connects the top and bottom junctions vertically. A 6 V battery is connected between the left junction and the right junction through an external wire. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 55/5/1.

Hints

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

Hint 1

For the derivation, apply Kirchhoff's loop rule to the two meshes containing the galvanometer and put I_g = 0.

Hint 2

Check 20/2 against 12/1 — if unequal, current flows in 3 Ω; assign currents using the junction rule and write three loop equations.

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.

(i) Bridge: resistances P (A–B), Q (B–C), R (A–D), S (D–C); battery across A and C; galvanometer G (resistance $R_{g}$ ) between B and D. Let $I_{1}$ flow through P, $I_{2}$ through R, $I_{g}$ through G (B to D).
½ mark
Kirchhoff's loop rule for loop ABDA: $I_{1} P + I_{g} R_{g} - I_{2} R = 0$ ; for loop BCDB: $(I_{1} - I_{g}) Q - (I_{2} + I_{g}) S - I_{g} R_{g} = 0$ .
1 mark
For balance $I_{g} = 0$ : $I_{1} P = I_{2} R$ and $I_{1} Q = I_{2} S$ . Dividing: $\frac{P}{Q} = \frac{R}{S}$ — the balance condition.
½ mark
(ii) Here $\frac{20}{2} = 10 \neq = \frac{12}{1} = 12$ , so the bridge is not balanced and current flows through 3 Ω. Let $I_{1}$ flow in 20 Ω, $I_{2}$ in 12 Ω, $I_{3}$ in 3 Ω (top to bottom). Then 2 Ω carries $(I_{1} - I_{3})$ and 1 Ω carries $(I_{2} + I_{3})$ .
½ mark
Loop (left–top–bottom–left): $20 I_{1} + 3 I_{3} - 12 I_{2} = 0$ ...(1)
Loop (top–right–bottom–top): $2 (I_{1} - I_{3}) - 1 (I_{2} + I_{3}) - 3 I_{3} = 0 \Rightarrow 2 I_{1} - I_{2} - 6 I_{3} = 0$ ...(2)
Loop with battery (left–top–right): $20 I_{1} + 2 (I_{1} - I_{3}) = 6 \Rightarrow 22 I_{1} - 2 I_{3} = 6$ ...(3)
1½ marks
From (2): $I_{2} = 2 I_{1} - 6 I_{3}$ . Put in (1): $20 I_{1} + 3 I_{3} - 24 I_{1} + 72 I_{3} = 0 \Rightarrow I_{1} = \frac{75}{4} I_{3}$ .
Put in (3): $22 \cdot \frac{75}{4} I_{3} - 2 I_{3} = 6 \Rightarrow 821 I_{3} = 12 \Rightarrow I_{3} = \frac{12}{821} \approx 0.0146$ A $\approx 14.6$ mA (from the top junction to the bottom junction).
1 mark

Answer: (i) P/Q = R/S; (ii) I(3 Ω) = 12/821 A ≈ 0.0146 A (≈ 14.6 mA)

Where the marks are

Labelled bridge and current distribution (½ mark)
Kirchhoff's loop equations for the two meshes (1 mark)
Setting I_g = 0 to obtain P/Q = R/S (½ mark)
Shows bridge unbalanced and assigns currents (½ mark)
Three correct Kirchhoff loop equations (1½ marks)
Solves to get current in 3 Ω ≈ 0.0146 A (1 mark)

Revise the chapter

Chapter: Current Electricity (notes, formulas, where marks are lost and every past question)
Topic: Kirchhoff's Rules and Wheatstone Bridge

Practise this chapterBoard-style questions from Current Electricity, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 55/5/1: 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 31(a)

Hints

Step-by-step solution

Revise the chapter

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