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Simple Circuit

Two resistanc...

Question

Two resistances $R_{1}$ and $R_{2}$ are joined as shown in the figure to two batteries of e.m.f. $E_{1}$ and $E_{2}$ . If $E_{2}$ is short-circuited, the current through $R_{1}$ is

A

\frac{E_{1}}{R_{1}}

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B

\frac{E_{2}}{R_{1}}

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C

\frac{E_{2}}{R_{2}}

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D

\frac{E_{1}}{(R_{2} + R_{1})}

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Solution

The correct option is A $\frac{E_{1}}{R_{1}}$
After short circuiting, $R_{2}$ becomes meaningless.

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