Two cells of emf $E_{1}$ and $E_{2}$ are joined in opposition (such that $E_{1} > E_{2}$ ). If $r_{1}$ and $r_{2}$ be the internal resistance and $R$ be the external resistance, then the terminal potential difference is

\frac{E_{1} + E_{2}}{r_{1} + r_{2}} \times R

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\frac{E_{1} + E_{2}}{r_{1} + r_{2} + R} \times R

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\frac{E_{1} - E_{2}}{r_{1} + r_{2}} \times R

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\frac{E_{1} - E_{2}}{r_{1} + r_{2} + R} \times R

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Solution

The correct option is D $\frac{E_{1} - E_{2}}{r_{1} + r_{2} + R} \times R$
Equivalent emf of the circuit $E_{e q} = E_{1} - E_{2}$
Equivalent resistance (in series connection) of the circuit $R_{e q} = r_{1} + r_{2} + R$
Thus current flowing in the circuit $I = \frac{E_{e q}}{R_{e q}} = \frac{E_{1} - E_{2}}{r_{1} + r_{2} + R}$
$∴$ Terminal potential difference $V = I R$
$⟹$ $V = \frac{E_{1} - E_{2}}{r_{1} + r_{2} + R} \times R$

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