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(A) When $n$ resistors of resistance $R$ are connected in series with a battery of $emf$ $E$ and internal resistance $R$,the total resistance is $R_{e

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$10$

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(A) When $n$ resistors of resistance $R$ are connected in series with a battery of $emf$ $E$ and internal resistance $R$,the total resistance is $R_{eq} = R + nR = R(1+n)$.The current $I$ is given by $I = \frac{E}{R(1+n)}$ ... $(1)$When the $n$ resistors are connected in parallel,the equivalent resistance is $R_p = \frac{R}{n}$.The new current $I' = 10I$ is given by $I' = \frac{E}{R + R/n} = \frac{nE}{R(n+1)}$ ... $(2)$Dividing equation $(2)$ by equation $(1)$:$\frac{10I}{I} = \frac{nE}{R(n+1)} 	imes \frac{R(n+1)}{E}$$10 = n$Thus,the value of $n$ is $10$.

First,a set of $n$ equal resistors of $R$ each are connected in series to a battery of $emf$ $E$ and internal resistance $R$. $A$ current $I$ is observed to flow. Then,the $n$ resistors are connected in parallel to the same battery. It is observed that the current increases $10$ times. What is $n$?

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