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A student is asked to connected four cells of emf of 1 V and internal resistance 0.5 ohm in series with an external resistance of 1 ohm. But one cell is wrongly connected by
him with its terminal reversed, the current in the circuit is

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$\frac{1}{3} A$

$\frac{2}{3} A$

$\frac{3}{4} A$

$\frac{4}{3} A$

answer is B.

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Detailed Solution

$i = \frac{(N - 2 n) E}{R + Nr} = \frac{(4 - 2) \times 1}{1 + 4 \times 0 .5} = \frac{2 \times 1}{3}$

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