Under what conditions current passing through a resistance $R$ can be increased by short-circuiting the battery of emf $E_{2}$ . The internal resistances of the two batteries are $r_{1}$ and $r_{2}$ respectively.

E_{2} r_{1} > E_{1} (R + r_{2})

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E_{1} r_{2} > E_{2} (R + r_{1})

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E_{2} r_{2} < E_{1} (R + r_{2})

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E_{1} r_{1} < E_{2} (R + r_{1})

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Solution

The correct option is B $E_{1} r_{2} > E_{2} (R + r_{1})$
From the given diagram, we observe that both cells are in series combination with same polarity. So,

Equivalent emf, $E_{e q} = E_{1} + E_{2}$

And equivalent resistance of the circuit will be

$R_{e q} = R + r_{1} + r_{2}$

The current through the circuit before the battery of emf $E_{2}$ is short-circuited is,

$I_{1} = \frac{E_{e q}}{R_{e q}} = \frac{E_{1} + E_{2}}{R + r_{1} + r_{2}}$

After short-circuiting the battery of emf $E_{2}$ , current through resistance $R$ would be,

$I_{2} = \frac{E_{1}}{R + r_{1}}$

Now, $I_{2} > I_{1}$

$∴ \frac{E_{1}}{R + r_{1}} > \frac{E_{1} + E_{2}}{R + r_{1} + r_{2}}$

$\Rightarrow E_{1} r_{2} > E_{2} (R + r_{1})$

Hence, the correct answer is option $(b)$ .

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(IIT-JEE 1997)