Two batteries, one of e.m.f. 12 V and internal resistance 2 Ω and other of e.m.f. 6 V and internal resistance 1 Ω, are connected as shown in the figure. What will be the reading of the voltmeter 'V'? (a) 12 V (b) 8 V (c) 6 V (d) 4 V

Answer: (b). From the figure both batteries have their positive terminals on the same side, so they are in parallel. V = (E₁/r₁ + E₂/r₂)/(1/r₁ + 1/r₂) = (12/2 + 6/1)/(1/2 + 1) = 12/1.5 V = 8 V. So the answer is (b) 8 V.

Two batteries, one of e.m.f. 12 V and internal resistance 2 Ω and… | MHT CET 2023 (11 May, Shift 2)

Q15MHT CET 2023 · 11 MAY · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations

Two batteries, one of e.m.f. $12\text{ V}$ and internal resistance $2\ \Omega$ and other of e.m.f. $6\text{ V}$ and internal resistance $1\ \Omega$ , are connected as shown in the figure. What will be the reading of the voltmeter 'V'?

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Answer: (b)

From the figure both batteries have their positive terminals on the same side, so they are in parallel.
$V = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{12/2 + 6/1}{1/2 + 1} = \frac{12}{1.5}$
$V = 8\text{ V}$ . So the answer is (b) 8 V.

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