Two sources of current of equal emf are connected in series and have different internal resistances $r_{1}$ and $r_{2} (r_{2} > r_{1})$ Find the external resistance $R$ at which the potential difference across the terminals of one of the sources becomes equal to zero.

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$R = r_{2} + r_{1}$

$R = r_{1} - r_{2}$

$R = r_{1} + r_{2}$

$R = r_{2} - r_{1}$

answer is B.

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Detailed Solution

$V = E - i r$

$E$ and $i$ for both the sources are equal. Therefore, potential difference ( $V$ will be zero for a source having greater internal resistance, i.e. $r_{z}$

$∴$ $0 = E - i r_{2}$

or $E = i r_{2} = (\frac{2 E}{R + r_{1} + r_{2}}) \cdot r_{2}$

$2 r_{2} = R + r_{1} + r_{2}$

$R = r_{2} - r_{1}$

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