10 resistors, each of resistance $R$ are connected in series to a battery of emf $E$ and negligible internal r

Question

10 resistors, each of resistance $R$ are connected in series to a battery of emf $E$ and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased $n$ times. The value of $n$ is

Accepted Answer

In series, total resistance $R_s = 10R$. Current $I_s = E / (10R)$. In parallel, total resistance $R_p = R/10$. Current $I_p = E / (R/10) = 10E/R$. The ratio $I_p / I_s = (10E/R) / (E/10R) = 100$. Thus $n = 100$.

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