PhysicsCBSE Class 12 · Board exam 2023, set 55/1/1 · 1 mark · MCQ

CBSE Class 12 Physics 2023: question 3

Set in the CBSE Class 12 board exam of 2023, paper 55/1/1, section A, question 3, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q3 · 1 mark · MCQCurrent Electricity

The potential difference across a cell in an open circuit is 8 V. It falls to 4 V when a current of 4 A is drawn from it. The internal resistance of the cell is :

(A) $4 Ω$
(B) $3 Ω$
(C) $2 Ω$
(D) $1 Ω$

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 55/1/1.

Hints

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

Hint 1

The open-circuit voltage is the emf.

Hint 2

Use $V = ε - I r$ .

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.

In open circuit the terminal voltage equals the emf, so $ε = 8$ V. With current drawn, $V = ε - I r$ .
½ mark
$4 = 8 - 4 r \Rightarrow r = 1 Ω$ . This is option (D).
½ mark

Answer: $1 Ω$ , option (D)

Where the marks are

Applies $V = ε - I r$ and gets $1 Ω$ (1 mark)

Revise the chapter

Chapter: Current Electricity (notes, formulas, where marks are lost and every past question)
Topic: EMF, Internal Resistance and Cells

Practise this chapterBoard-style questions from Current Electricity, one at a time, marked against the answer key and marking points.Back to the full paper2023 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.

CBSE Class 12 Physics 2023: question 3

Hints

Step-by-step solution

Revise the chapter

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