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A battery of emf E and… (Class 12 Physics 2025, Q17)

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PhysicsCBSE Class 12 · Board exam 2025, set 55/1/1 · 2 marks · Very short answer

CBSE Class 12 Physics 2025: question 17

Set in the CBSE Class 12 board exam of 2025, paper 55/1/1, section B, question 17, for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q17 · 2 marks · Very short answerCurrent Electricity

A battery of emf E and internal resistance r is connected to a rheostat. When a current of 2A is dawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of E and r.

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

Hints

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

Hint 1

The rheostat voltage is the terminal voltage of the battery, which is less than E because of the drop across r.

Hint 2

Write once for each reading. You get two equations in the two unknowns E and r.

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.

  1. Terminal voltage of a cell delivering current : .

    ½ mark

  2. Case 1 ( A, V): ...(1). Case 2 ( A, V): ...(2).

    ½ mark

  3. Equating (1) and (2): .

    ½ mark

  4. From (1): V.

    ½ mark

Answer: E = 6 V, r = 0.5 Ω

Where the marks are

  • Uses and writes the two equations and (1 mark)
  • Solves to get (½ mark)
  • Gets V (½ mark)

Revise the chapter

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 55/1/1: 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.