ScienceCBSE Class 10 · Board exam 2026, set 31/3/1 · 3 marks · Short answer

CBSE Class 10 Science 2026: question 37

Set in the CBSE Class 10 board exam of 2026, paper 31/3/1, section C, question 37, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 3 marks · Short answerElectricity

Consider the given electric circuit :

Calculate the following :
(a) Total resistance of the circuit
(b) The electric current drawn from the battery
(c) Potential difference between points P and Q

Figure in the paper: Rectangular circuit with a 10 V battery between a key and point P. From P to Q: a 2 Ω resistor in series with a 3 Ω resistor (along the bottom and right side) in parallel with a 5 Ω resistor connected directly between P and Q. From Q the current goes through an ammeter A up to point S. Between S and R: a 5 Ω resistor in parallel with a series combination of 1 Ω and 4 Ω resistors (via the top-left). R connects down to the battery. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 31/3/1.

Hints

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

Hint 1

Combine the series pairs first (4 Ω + 1 Ω, and 2 Ω + 3 Ω).

Hint 2

Each series pair is in parallel with a 5 Ω resistor. Find each group's resistance, then add the two groups.

Hint 3

For (c), use V = IR with the resistance of the P–Q group only.

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) Between R and S: the 4 Ω and 1 Ω resistors are in series, giving $4 + 1 = 5 Ω$ . This 5 Ω is in parallel with the other 5 Ω:
$\frac{1}{R _{1}} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$ , so $R_{1} = \frac{5}{2} Ω$ .
Between P and Q: the 2 Ω and 3 Ω resistors are in series, giving $2 + 3 = 5 Ω$ . This 5 Ω is in parallel with the other 5 Ω:
$R_{2} = \frac{5}{2} Ω$ .
The two groups are in series:
$R = R_{1} + R_{2} = \frac{5}{2} + \frac{5}{2} = 5 Ω$ .
1 mark
(b) $I = \frac{V}{R} = \frac{10 V}{5 Ω} = 2 A$
1 mark
(c) The whole current of 2 A flows through the P–Q combination, so
$V_{P Q} = I R_{2} = 2 A \times \frac{5}{2} Ω = 5 V$ .
1 mark

Answer: (a) 5 Ω; (b) 2 A; (c) 5 V

Where the marks are

Finds R₁ = 5/2 Ω and R₂ = 5/2 Ω, total resistance 5 Ω (1 mark)
Current I = V/R = 2 A (1 mark)
Potential difference across PQ = 2 × 5/2 = 5 V (1 mark)

Revise the chapter

Chapter: Electricity (notes, formulas, where marks are lost and every past question)
Topic: Resistors in Series and Parallel

Practise this chapterBoard-style questions from Electricity, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 31/3/1: 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 10 Science 2026: question 37

Hints

Step-by-step solution

Revise the chapter

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