What should be the value of $E_{1} / E_{2}$ so that current flowing through $7 Ω$ resistor could be increased by short-circuiting the battery of emf $E_{2} ?$

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$\frac{E_{1}}{E_{2}} < 4$

$\frac{E_{1}}{E_{2}} > 4$

$\frac{E_{1}}{E_{2}} > 3$

$E_{1} = E_{2}$

answer is C.

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Detailed Solution

Initially $I_{1} = \frac{E_{1} + E_{2}}{(1 + 2 + 7)} = \frac{E_{1} + E_{2}}{10}$

When $E_{2}$ is shorted $I_{2} = \frac{E_{1}}{(1 + 7)} = \frac{E_{1}}{8}$

If $I_{2} > I_{2}$

$\frac{E_{1}}{8} > \frac{E_{1} + E_{2}}{10} o r \frac{5}{4} > \frac{E_{1} + E_{2}}{E_{1}}$

or $\frac{5}{4} > 1 + \frac{E_{2}}{E_{1}} o r \frac{E_{2}}{E_{1}} < \frac{1}{4}$

or $\frac{E_{1}}{E_{2}} > 4$

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