ScienceCBSE Class 10 · Sample paper, session 2025-26 · 5 marks · Case-based

CBSE Class 10 Science sample paper 2025-26: question 39(A)

From the official CBSE sample question paper for 2025-26, section C, question 39(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q39(A) · 5 marks · Case-basedCompetency-basedElectricity

The arrangement of resistors shown in the above figure is connected to a battery. The power dissipation in the 100 Ω resistor is 81 W. Calculate
(i) the current in the circuit
(ii) the reading in the voltmeter V $_{2}$
(iii) the reading in the voltmeter V $_{1}$

Figure in the paper: A battery at the top with voltmeter V1 connected across its terminals. A 100 Ω resistor is in series on the left side. Two 25 Ω resistors are connected in parallel to each other in the right-hand branch. Voltmeter V2 is connected across the parallel 25 Ω combination (the parallel branch). See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

The power in the 100 Ω resistor tells you the current through it. Which formula links P, I and R?

Hint 2

The same current flows through the parallel pair. Find their combined resistance first.

Hint 3

V₁ is across the whole circuit, so add the voltage drops.

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) All the current from the battery flows through the 100 Ω resistor, because it is in series. Use $P = I^{2} R$ :
$81 = I^{2} \times 100$
1 mark
$I^{2} = \frac{81}{100} = 0.81$ , so $I = 0.9$ A. This is the current in the circuit.
1 mark
(ii) The two 25 Ω resistors are in parallel:
$\frac{1}{R _{p}} = \frac{1}{25} + \frac{1}{25} = \frac{2}{25}$ , so $R_{p} = 12.5 Ω$
1 mark
V₂ reads the voltage across this combination:
$V_{2} = I R_{p} = 0.9 \times 12.5 = 11.25$ V
1 mark
(iii) Voltage across the 100 Ω resistor: $0.9 \times 100 = 90$ V.
V₁ is across the battery, so it reads the total voltage:
$V_{1} = 90 + 11.25 = 101.25$ V
(Check: $I \times R_{t o t a l} = 0.9 \times 112.5 = 101.25$ V.)
1 mark

Answer: (i) 0.9 A; (ii) V₂ = 11.25 V; (iii) V₁ = 101.25 V

Where the marks are

Uses P = I²R for the 100 Ω resistor (1 mark)
Current I = 0.9 A (1 mark)
Equivalent resistance of the parallel 25 Ω pair = 12.5 Ω (1 mark)
V₂ = 11.25 V (1 mark)
V₁ = 90 V + 11.25 V = 101.25 V (1 mark)

Revise the chapter

Chapter: Electricity (notes, formulas, where marks are lost and every past question)
Topic: Electric Power and Energy

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 2025-26: every question, by section.

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CBSE Class 10 Science sample paper 2025-26: question 39(A)

Hints

Step-by-step solution

Revise the chapter

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