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The cells of emf’s $E_{1} and E_{2}$ be connected in parallel in a circuit. Let $r_{1} and r_{2}$ be the internal resistance of the cells. Then the current through the circuit is

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$\frac{E_{1} + E_{2}}{(r_{1} + r_{2})}$

$\frac{2 (E_{1} + E_{2})}{(r_{1} + r_{2})}$

$\frac{E_{1} - E_{2}}{(r_{1} + r_{2})}$

$\frac{E_{1} + E_{2}}{(r_{1} - r_{2})}$

answer is B.

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Detailed Solution

Net Emf = E₁ -E₂

Net resistance = r₁+r₂

Current in the circuit = $\frac{E_{1} - E_{2}}{r_{1} + r_{2}}$

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