Question

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

Answer 1

$I=\frac{E}{R+n R} ; I'$
$=\frac{E}{R+\frac{R}{n}}=10 I$
$\frac{1+n}{1+\frac{1}{n}}=10$
$ \Rightarrow \frac{1+n}{n+1} \times n=10$
$\therefore n=10$