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

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

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

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

An electric heater consists of three similar heating elements A, B and C, connected as shown in the figure above. Each heating element is rated as 1.2 kW, 240 V and has constant resistance. S $_{1}$ , S $_{2}$ and S $_{3}$ are respective switches. The circuit is connected to a 240 V supply.

(i)
Calculate the resistance of one heating element.
1 mark
(ii)
Calculate the current in each resistor when only S $_{1}$ and S $_{3}$ are closed.
2 marks
(iii)
Calculate the power dissipated across A when S $_{1}$ , S $_{2}$ and S $_{3}$ are closed.
2 marks

Figure in the paper: A 240 V supply line at the top left with switch S1 on the top right branch. Heating element A is in the middle branch on the left, connected in series with element B to its right. Switch S2 connects the junction between A and B to the right-hand return wire. Element C is in the bottom branch, in series with switch S3. The branches are joined at both ends to form a parallel arrangement. 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

Use the rating (1.2 kW at 240 V) to find the resistance of one element.

Hint 2

Work out which elements lie in the same branch when each switch is open or closed. Every branch gets 240 V.

Hint 3

When S₂ closes, check whether any element is bypassed by a plain wire.

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) Use $P = \frac{V ^{2}}{R}$ , so $R = \frac{V ^{2}}{P} = \frac{240 \times 240}{1200} = 48 Ω$
1 mark
(ii) With S₁ and S₃ closed and S₂ open, A and B are in series in one branch. C is in a parallel branch. Each branch has 240 V across it.
Current in C: $I_{C} = \frac{V}{R} = \frac{240}{48} = 5$ A
1 mark
A and B in series: $R = 48 + 48 = 96 Ω$ , so $I_{A} = I_{B} = \frac{240}{96} = 2.5$ A
1 mark
(iii) With S₂ also closed, the junction between A and B is joined straight to the return wire. B is short-circuited, so A alone has 240 V across it.
$I_{A} = \frac{240}{48} = 5$ A
1 mark
$P_{A} = I^{2} R = 5^{2} \times 48 = 1200$ W $= 1.2$ kW
1 mark
For visually impaired students (alternative, 5 marks).
The resistors are in parallel, so each has 24 V across it.
$I_{1} = \frac{24}{6} = 4$ A and $I_{2} = \frac{24}{12} = 2$ A [1 mark]
Time $t = 5 \times 60 = 300$ s. Using Joule's law $H = I^{2} R t$ :
$H_{1} = 4^{2} \times 6 \times 300 = 28800$ J
$H_{2} = 2^{2} \times 12 \times 300 = 14400$ J [1 mark]
Total heat $= 28800 + 14400 = 43200$ J [1 mark]
Power in each resistor:
$P_{1} = V I_{1} = 24 \times 4 = 96$ W
$P_{2} = 24 \times 2 = 48$ W
Both are below the 100 W rating, so both resistors are safe to use [2 marks].

Answer: (i) 48 Ω; (ii) C: 5 A, A and B: 2.5 A each; (iii) 1200 W (1.2 kW). VI: 43200 J; both resistors safe (96 W and 48 W < 100 W)

Where the marks are

R = V²/P = 48 Ω (1 mark)
Current in C = 5 A (1 mark)
Current in A and B = 240/96 = 2.5 A (1 mark)
Recognises that B is shorted, so the current in A is 5 A (1 mark)
P_A = 1200 W (1 mark)
VI: currents 4 A and 2 A (1 mark); heats 28800 J and 14400 J (1 mark); total 43200 J (1 mark); powers 96 W and 48 W, both safe (2 marks)

Revise the chapter

Chapter: Electricity (notes, formulas, where marks are lost and every past question)
Topic: Heating Effect of Electric Current

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.

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 sample paper 2025-26: question 39(B)

Hints

Step-by-step solution

Revise the chapter

More questions from Electricity