PhysicsCBSE Class 12 · Board exam 2023, set 55/5/1 · 5 marks · Long answer

CBSE Class 12 Physics 2023: question 31(b)

Set in the CBSE Class 12 board exam of 2023, paper 55/5/1, section D, question 31(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(b) · 5 marks · Long answerCurrent Electricity

(i) A cell emf of (E) and internal resistance (r) is connected across a variable load resistance (R). Draw plots showing the variation of terminal voltage V with (i) R and (ii) the current (I) in the load. (2)
(ii) Three cells, each of emf E but internal resistances 2r, 3r and 6r are connected in parallel across a resistor R.
Obtain expressions for (i) current flowing in the circuit, and (ii) the terminal potential difference across the equivalent cell. (3)

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 55/5/1.

Hints

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

Hint 1

Write V in terms of R using I = E/(R + r), and in terms of I using V = E − Ir.

Hint 2

For cells in parallel: 1/r_eq = Σ1/r_i and E_eq/r_eq = ΣE_i/r_i.

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)(i) V vs R: $V = \frac{E R}{R + r}$ . The graph starts at $V = 0$ when $R = 0$ and rises steeply at first. It then bends over and approaches $E$ asymptotically as $R \to \infty$ , giving a saturating curve.
1 mark
(i)(ii) V vs I: $V = E - I r$ . The graph is a straight line with negative slope $- r$ . Its intercept on the V-axis is $E$ (at $I = 0$ ) and its intercept on the I-axis is $E / r$ (short circuit).
1 mark
(ii) Equivalent internal resistance: $\frac{1}{r _{e q}} = \frac{1}{2 r} + \frac{1}{3 r} + \frac{1}{6 r} = \frac{3 + 2 + 1}{6 r} = \frac{1}{r}$ , so $r_{e q} = r$ .
½ mark
Equivalent emf: $\frac{E _{e q}}{r _{e q}} = \frac{E}{2 r} + \frac{E}{3 r} + \frac{E}{6 r} = \frac{E}{r}$ , so $E_{e q} = E$ .
1 mark
(ii)(i) Current in the circuit: $I = \frac{E _{e q}}{R + r _{e q}} = \frac{E}{R + r}$ .
½ mark
(ii)(ii) Terminal potential difference: $V = E_{e q} - I r_{e q} = E - \frac{E r}{R + r}$ .
½ mark
Simplifying: $V = \frac{E R}{R + r}$ .
½ mark

Answer: (i) V–R: rising curve approaching E; V–I: straight line from E with slope −r; (ii) I = E/(R + r), V = ER/(R + r)

Where the marks are

Plot of V vs R (1 mark)
Plot of V vs I (1 mark)
r_eq = r (½ mark)
E_eq = E (1 mark)
I = E/(R + r) (½ mark)
V = E − Ir_eq (½ mark)
V = ER/(R + r) (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 paper2023 board paper, set 55/5/1: every question, by section.

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CBSE Class 12 Physics 2023: question 31(b)

Hints

Step-by-step solution

Revise the chapter

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