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A resistance of $4 Ω$ and a wire of length 5 metres and resistance $5 Ω$ are joined in series and connected to a cell of e.m.f. $10 V$ and internal resistance $1 Ω$ . A parallel combination of two identical cells is balanced across $300 c m$ of the wire. The e.m.f. $E$ of each cell is

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$3.0 V$

$1.33 V$

$1.5 V$

$0.67 V$

answer is B.

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Detailed Solution

$E = (\frac{e}{R + R_{h} + r}) \times \frac{R}{L} \times l = (\frac{10}{5 + 4 + 1}) \times \frac{5}{5} \times 3 = 3 V$

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