Two cells are connected between points A and B as shown. Cell 1 has emf of 12 V and internal resistance of 3 Ω. Cell 2 has emf of 6V and internal resistance of 6 Ω. An external resistor R of 4 Ω is connected across A and B. The current flowing through R will be ____________ A. Numerical answer type. Work it out, then open the answer.

Answer: 1. From the figure the two cells are in parallel with opposite polarity. Eeq = (E₁/r₁ - E₂/r₂)/(1/r₁ + 1/r₂) = (12/3 - 6/6)/(1/3 + 1/6) = 3/1/2 = 6 V req = (3 × 6)/(3 + 6) = 2 Ω I = (6)/(2 + 4) = 1 A

Two cells are connected between points A and B as shown. Cell 1 has… | JEE Main 2023 (25 Jan, Shift 2)

Q18JEE Main 2023 · 25 JAN · SHIFT 2Medium

Cells: EMF, Internal Resistance & Combinations

Two cells are connected between points A and B as shown. Cell 1 has emf of 12 V and internal resistance of 3 $\Omega$ . Cell 2 has emf of 6V and internal resistance of 6 $\Omega$ . An external resistor R of 4 $\Omega$ is connected across A and B. The current flowing through R will be ____________ A.

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

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Answer: 1

From the figure the two cells are in parallel with opposite polarity.
$E_{eq} = \dfrac{E_1/r_1 - E_2/r_2}{1/r_1 + 1/r_2} = \dfrac{12/3 - 6/6}{1/3 + 1/6} = \dfrac{3}{1/2} = 6\text{ V}$
$r_{eq} = \dfrac{3 \times 6}{3 + 6} = 2\ \Omega$
$I = \dfrac{6}{2 + 4} = 1\text{ A}$

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