PhysicsCBSE Class 12 · Board exam 2025, set 55/7/1 · 3 marks · Short answer

CBSE Class 12 Physics 2025: question 22

Set in the CBSE Class 12 board exam of 2025, paper 55/7/1, section C, question 22, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q22 · 3 marks · Short answerCurrent Electricity

Three cells A, B and C of emfs 2 V, 3 V and 5 V respectively are connected in parallel to each other. Their internal resistances are 5 $Ω$ , 5 $Ω$ and 1 $Ω$ respectively. Calculate the currents flowing through the cells A, B and C.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 55/7/1.

Hints

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

Hint 1

With no external resistor, the strongest cell drives current backwards through the other two.

Hint 2

Assign two unknown currents, use the junction rule for the third, and write two loop equations.

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.

All positive terminals are joined at one junction and all negative terminals at the other. Let $I_{1}$ flow through cell C (5 V, 1 Ω) in the direction of its emf. Let $I_{2}$ flow through cell B (3 V, 5 Ω) against its emf. By the junction rule, $I_{1} - I_{2}$ flows through cell A (2 V, 5 Ω) against its emf.
½ mark
Loop C–B (Kirchhoff's loop rule): $5 - 3 = 1 \cdot I_{1} + 5 I_{2} \Rightarrow I_{1} + 5 I_{2} = 2$ ...(1)
½ mark
Loop C–A: $5 - 2 = 1 \cdot I_{1} + 5 (I_{1} - I_{2}) \Rightarrow 6 I_{1} - 5 I_{2} = 3$ ...(2)
½ mark
Adding (1) and (2): $7 I_{1} = 5 \Rightarrow I_{1} = \frac{5}{7}$ A through cell C.
½ mark
From (1): $5 I_{2} = 2 - \frac{5}{7} = \frac{9}{7} \Rightarrow I_{2} = \frac{9}{35}$ A through cell B.
½ mark
Current through A: $I_{1} - I_{2} = \frac{25}{35} - \frac{9}{35} = \frac{16}{35}$ A. Cells A and B are being charged by C. (Check: common terminal voltage $= 5 - \frac{5}{7} \times 1 = \frac{30}{7}$ V.)
½ mark

Answer: I_C = 5/7 A ≈ 0.71 A; I_B = 9/35 A ≈ 0.26 A; I_A = 16/35 A ≈ 0.46 A (A and B are being charged).

Where the marks are

Correct current distribution using the junction rule (½ mark)
Correct loop equation for loop C–B (½ mark)
Correct loop equation for loop C–A (½ mark)
Current through C = 5/7 A (½ mark)
Current through B = 9/35 A (½ mark)
Current through A = 16/35 A (½ mark)

Revise the chapter

Chapter: Current Electricity (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Current Electricity, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 55/7/1: every question, by section.

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CBSE Class 12 Physics 2025: question 22

Hints

Step-by-step solution

Revise the chapter

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