PhysicsCBSE Class 12 · Board exam 2026, set 55/1/1 · 5 marks · Long answer

CBSE Class 12 Physics 2026: question 31(b)

Set in the CBSE Class 12 board exam of 2026, paper 55/1/1, section E, 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

(a) Two cells of emf $E_{1}$ and $E_{2}$ with internal resistances $r_{1}$ and $r_{2}$ respectively, are connected in parallel by connecting their positive terminals together and negative terminals together. Deduce an expression for equivalent emf and equivalent internal resistance of the combination.
(b) A parallel combination, as stated in (a) above, of two cells of emfs E and 3E and internal resistances R each is connected across a resistance 2R. Find the current that flows through resistance 2R.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 55/1/1.

Hints

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

Hint 1

Both cells share the same terminal voltage V, and the external current is $I_{1} + I_{2}$ .

Hint 2

Rearrange your result into the form $V = E_{e q} - I r_{e q}$ .

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.

(a) Let currents $I_{1}$ and $I_{2}$ leave the positive terminals of cells 1 and 2, with total current $I = I_{1} + I_{2}$ . Let $V$ be the common terminal p.d. Then
$V = E_{1} - I_{1} r_{1} \Rightarrow I_{1} = \frac{E _{1} - V}{r _{1}}$ , and $V = E_{2} - I_{2} r_{2} \Rightarrow I_{2} = \frac{E _{2} - V}{r _{2}}$
1 mark
$I = \frac{E _{1}}{r _{1}} + \frac{E _{2}}{r _{2}} - V (\frac{1}{r _{1}} + \frac{1}{r _{2}})$ , which rearranges to
$V = \frac{E _{1} r _{2} + E _{2} r _{1}}{r _{1} + r _{2}} - I \frac{r _{1} r _{2}}{r _{1} + r _{2}}$
1 mark
Comparing with $V = E_{e q} - I r_{e q}$ :
$E_{e q} = \frac{E _{1} r _{2} + E _{2} r _{1}}{r _{1} + r _{2}}$ , $r_{e q} = \frac{r _{1} r _{2}}{r _{1} + r _{2}}$
1 mark
(b) $E_{e q} = \frac{E \cdot R + 3 E \cdot R}{R + R} = 2 E$ and $r_{e q} = \frac{R \cdot R}{2 R} = \frac{R}{2}$
1 mark
$I = \frac{E _{e q}}{2 R + r _{e q}} = \frac{2 E}{2 R + R /2} = \frac{4 E}{5 R}$
1 mark

Answer: $E_{e q} = \frac{E _{1} r _{2} + E _{2} r _{1}}{r _{1} + r _{2}}$ , $r_{e q} = \frac{r _{1} r _{2}}{r _{1} + r _{2}}$ ; current through 2R $= \frac{4 E}{5 R}$

Where the marks are

Writes the terminal voltage equations for each cell (1 mark)
Combines them using I = I₁ + I₂ (1 mark)
Expressions for E_eq and r_eq (1 mark)
E_eq = 2E and r_eq = R/2 (1 mark)
Current 4E/5R (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 paper2026 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 2026: question 31(b)

Hints

Step-by-step solution

Revise the chapter

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