ScienceCBSE Class 10 · Sample paper, session 2026-27 · 3 marks · Short answer

CBSE Class 10 Science sample paper 2026-27: question 37

From the official CBSE sample question paper for 2026-27, 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

The figure shows a circuit with a 15 V battery connected to several resistors. Using the given circuit diagram: A. Find the total current supplied by the battery. B. Calculate the potential difference between points A and B.

Figure in the paper: Circuit: a triangular arrangement of two 1 Ω resistors meeting at the top, with the two lower corners at A (left) and B (right). A 2 Ω resistor connects A to B horizontally. A 5 Ω resistor is in the left branch from A down to the battery. To the right of B a 3 Ω resistor leads to a junction, from which a second triangle of a 3 Ω resistor is connected in parallel. Battery V = 15 V at the bottom. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

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

Hint 1

Start with the small groups. Which resistors in each triangle are in series and which are in parallel?

Hint 2

After simplifying, the 5 Ω, the A–B group and the 3 Ω group all carry the same current.

Hint 3

For V_AB, multiply the total current by the equivalent resistance between A and B.

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. In the left triangle, the two 1 Ω resistors are in series: $1 + 1 = 2 Ω$ . This 2 Ω path is in parallel with the 2 Ω resistor joining A and B: $\frac{1}{R _{A B}} = \frac{1}{2} + \frac{1}{2} = 1$ , so $R_{A B} = 1 Ω$ .
½ mark
In the right-hand portion, the two 3 Ω resistors are in parallel: $\frac{1}{R ^{'}} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ , so $R^{'} = 1.5 Ω$ .
½ mark
The 5 Ω resistor, $R_{A B}$ and $R^{'}$ are in series: $R_{s} = 5 + 1 + 1.5 = 7.5 Ω$ .
½ mark
By Ohm's law, $I = \frac{V}{R _{s}} = \frac{15 V}{7.5 Ω} = 2$ A. The total current supplied by the battery is 2 A.
½ mark
B. The whole 2 A passes through the combination between A and B, whose resistance is $R_{A B} = 1 Ω$ . So $V_{A B} = I \times R_{A B} = 2 \times 1 = 2$ V.
1 mark

Answer: A. I = 2 A (R_total = 7.5 Ω); B. V_AB = 2 V

Where the marks are

Reduces the A–B triangle to 1 Ω (series 1+1 in parallel with 2 Ω) (0.5 mark)
Reduces the two 3 Ω in parallel to 1.5 Ω (0.5 mark)
Total resistance 7.5 Ω (0.5 mark)
Current I = 2 A using Ohm's law (0.5 mark)
V_AB = 2 A × 1 Ω = 2 V (1 mark)

Revise the chapter

Chapter: Electricity (notes, formulas, where marks are lost and every past question)
Topic: Ohm's Law

Practise this chapterBoard-style questions from Electricity, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2026-27: every question, by section.

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CBSE Class 10 Science sample paper 2026-27: question 37

Hints

Step-by-step solution

Revise the chapter

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