A cell E1 of emf 6V and internal resistance 2 Ω is connected with another cell E2 of emf 4V and internal resistance 8 Ω (as shown in the figure). The potential difference across points X and Y is: (a) 10.0 V (b) 2.0 V (c) 5.6 V (d) 3.6 V

Answer: (c). From the figure both negative terminals face X, so the cells oppose: Enet = 6 - 4 = 2 V. I = (2)/(2 + 8) = 0.2 A, driven by E₁. Across E₁ (discharging): VP - VX = 6 - 0.2 × 2 = 5.6 V, and VP = VY (same wire). Check with E₂ (being charged): 4 + 0.2 × 8 = 5.6 V. So the answer is (c) 5.6 V.

A cell E1 of emf 6V and internal resistance 2 Ω is connected with… | JEE Main 2021 (24 Feb, Shift 1)

Q25JEE Main 2021 · 24 FEB · SHIFT 1Medium

Cells: EMF, Internal Resistance & Combinations

A cell E1 of emf 6V and internal resistance 2 $\Omega$ is connected with another cell E2 of emf 4V and internal resistance 8 $\Omega$ (as shown in the figure). The potential difference across points X and Y is:

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Answer: (c)

From the figure both negative terminals face $X$ , so the cells oppose: $E_{net} = 6 - 4 = 2\text{ V}$ .
$I = \frac{2}{2 + 8} = 0.2\text{ A}$ , driven by $E_1$ .
Across $E_1$ (discharging): $V_P - V_X = 6 - 0.2 \times 2 = 5.6\text{ V}$ , and $V_P = V_Y$ (same wire).
Check with $E_2$ (being charged): $4 + 0.2 \times 8 = 5.6\text{ V}$ . So the answer is (c) 5.6 V.

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