Two identical cells each of emf 1.5 V are connected in parallel across a parallel combination of two resistors each of resistance 20 Ω. A voltmeter connected in the circuit measures 1.2 V. The internal resistance of each cell is: (a) 2.5 Ω (b) 4 Ω (c) 5 Ω (d) 10 Ω

Answer: (c). Two identical cells in parallel: Eeq = 1.5 V, req = r/2. External resistance = 20 ∥ 20 = 10 Ω, so I = V/R = 1.2/10 = 0.12 A. E = V + Ir/2 ⇒ 1.5 = 1.2 + 0.12 · r/2 r/2 = 2.5 ⇒ r = 5 Ω. So the answer is (c).

Two identical cells each of emf 1.5 V are connected in parallel… | JEE Main 2022 (24 Jun, Shift 1)

Q23JEE Main 2022 · 24 JUN · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations

Two identical cells each of emf 1.5 V are connected in parallel across a parallel combination of two resistors each of resistance 20 $\Omega$ . A voltmeter connected in the circuit measures 1.2 V. The internal resistance of each cell is:

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Answer: (c)

Two identical cells in parallel: $E_{eq} = 1.5\text{ V}$ , $r_{eq} = \dfrac{r}{2}$ .
External resistance $= 20 \parallel 20 = 10\ \Omega$ , so $I = \dfrac{V}{R} = \dfrac{1.2}{10} = 0.12\text{ A}$ .
$E = V + I\dfrac{r}{2} \Rightarrow 1.5 = 1.2 + 0.12 \cdot \dfrac{r}{2}$
$\dfrac{r}{2} = 2.5 \Rightarrow r = 5\ \Omega$ . So the answer is (c).

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