Two identical cells each of emf 1.5 V are connected in series across a 10 Ω resistance. An ideal voltmeter connected across 10 Ω resistance reads 1.5 V. The internal resistance of each cell is __________ Ω. Numerical answer type. Work it out, then open the answer.

Answer: 5. I = V/R = 1.5/10 = 0.15 A. Series cells: 2E = V + I(2r) ⇒ 3 = 1.5 + 0.15 × 2r 2r = 10 ⇒ r = 5 Ω

Two identical cells each of emf 1.5 V are connected in series across… | JEE Main 2023 (11 Apr, Shift 2)

Q16JEE Main 2023 · 11 APR · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations

Two identical cells each of emf $1.5\text{ V}$ are connected in series across a $10\ \Omega$ resistance. An ideal voltmeter connected across $10\ \Omega$ resistance reads $1.5\text{ V}$ . The internal resistance of each cell is __________ $\Omega$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 5

$I = \dfrac{V}{R} = \dfrac{1.5}{10} = 0.15\text{ A}$ .
Series cells: $2E = V + I(2r) \Rightarrow 3 = 1.5 + 0.15 \times 2r$
$2r = 10 \Rightarrow r = 5\ \Omega$

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