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(A) When $n$ resistors of resistance $R$ are connected in series,the equivalent resistance is $n R$. The total resistance of the circuit including the

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$10$

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(A) When $n$ resistors of resistance $R$ are connected in series,the equivalent resistance is $n R$. The total resistance of the circuit including the internal resistance $R$ is $(n R + R)$.Thus,the current $I$ is given by $I = \frac{E}{n R + R} = \frac{E}{R(n + 1)}$ .....$(i)$When $n$ resistors are connected in parallel,the equivalent resistance is $R/n$. The total resistance of the circuit is $(R/n + R)$.Thus,the new current $10I$ is given by $10I = \frac{E}{R/n + R} = \frac{E}{R(\frac{1+n}{n})} = \frac{nE}{R(n+1)}$ .....$(ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{10I}{I} = \frac{nE / R(n+1)}{E / R(n+1)}$$10 = n$Therefore,the value of $n$ is $10$.

$A$ set of $n$ equal resistors,each of value $R$,are connected in series to a battery of emf $E$ and internal resistance $R$. The current drawn is $I$. Now,the $n$ resistors are connected in parallel to the same battery. The current drawn from the battery becomes $10I$. The value of $n$ is:

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