PhysicsCBSE Class 12 · Board exam 2026, set 55/1/1 · 3 marks · Short answer

CBSE Class 12 Physics 2026: question 22(b)

Set in the CBSE Class 12 board exam of 2026, paper 55/1/1, section C, question 22(b), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q22(b) · 3 marks · Short answerCurrent Electricity

(a) Obtain the condition of balance of a Wheatstone bridge.
(b) Find net resistance of the network of resistors connected between A and B, as shown in figure.

Figure in the paper: Network between A and B: series path A-(2R)-M-(R)-O... precisely: A to M through 2R; M to P through R (middle resistor, with point O between); P to N through R; N to B through 3R. An extra resistor R is connected in parallel across M to P via a top loop (from M above to P), and another resistor R is connected from the point between M and P region below, forming a loop from O side to N (bottom resistor R joins the junction near M-P section to N). Labels: A, M, O, P, N, B; resistors 2R, R, R, R, 3R, R (top), R (bottom). See it in the official paper.

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

Hints

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

Hint 1

At balance no current flows through the galvanometer. Apply Kirchhoff's loop rule to two loops.

Hint 2

Check whether the network between M and N is a balanced Wheatstone bridge before you combine resistors.

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) Wheatstone bridge set-up:
- Arms: $P$ (A to B), $Q$ (B to C), $R$ (A to D), $S$ (D to C).
- Battery between A and C.
- Galvanometer $G$ between B and D.
At balance $I_{g} = 0$ , so the current $I_{1}$ in P also flows through Q, and the current $I_{2}$ in R also flows through S.
½ mark
Kirchhoff's loop rule:
- Loop ABDA: $I_{1} P + I_{g} G - I_{2} R = 0 \Rightarrow I_{1} P = I_{2} R$
- Loop BCDB: $I_{1} Q - I_{2} S - I_{g} G = 0 \Rightarrow I_{1} Q = I_{2} S$
½ mark
Dividing the two equations gives the balance condition $\frac{P}{Q} = \frac{R}{S}$ .
½ mark
(b) Reading of the figure: between M and N lie the resistors
- M–O = R, M–P (top) = R, O–N (bottom) = R, P–N = R
- a bridge O–P = R
This is a Wheatstone bridge with $\frac{R}{R} = \frac{R}{R}$ , so it is balanced and no current flows in the O–P resistor.
½ mark
Ignoring O–P, the two paths M–O–N and M–P–N are each $2 R$ and are in parallel: $R_{M N} = \frac{2 R \times 2 R}{4 R} = R$ .
½ mark
Net resistance: $R_{A B} = 2 R + R + 3 R = 6 R$
½ mark

Answer: (a) $P / Q = R / S$ ; (b) $R_{A B} = 6 R$

Where the marks are

Circuit with the condition $I_{g} = 0$ (½ mark)
Loop equations for the two meshes (½ mark)
Balance condition P/Q = R/S (½ mark)
Identifies the balanced bridge between M and N (½ mark)
R_MN = R (½ mark)
R_AB = 6R (½ 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 55/1/1: 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 22(b)

Hints

Step-by-step solution

Revise the chapter

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