Three cell, $(A, B$ & $C)$ each of emf $E$ and internal resistance $(r)$ are connected in series as shown in figure. The potential difference across the cell $^{'} C^{'}$ is

\frac{E r}{R}

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\frac{E (R + r)}{R + 3 r}

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\frac{E (R + 4 r)}{R + 3 r}

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\frac{E (R + 3 r)}{R + 4 r}

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Solution

The correct option is C $\frac{E (R + 4 r)}{R + 3 r}$
$So, circuit current, I$

$= \frac{E}{R + 3 r}$

$P.D. across C$

$= I r + E$

$= \frac{E r}{R + 3 r} + E$

$= E (\frac{R + 4 r}{R + 3 r})$

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