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

In the first case, it is given that 'n' resistances are arranged in series, each having a resistance R.

In that case, the nett reistance of the external circuit is 'nR'. The internal resistance itself is R. Hence, the resultant resistance of the circuit is $nR+R=(n+1)R$.

$\because I=\frac{\epsilon }{nR+R}$

⇒ $I=\frac{E}{(n+1)R}$ .... (i)

In the case of parallel combination, these external n resistances are arranged in parallel,thereby their nett external resistance is $\frac{R}{n}$. This is in series with the internal resistance R.

Hence,

$10I$$=\frac{E}{\frac{R}{n}+R}=\frac{nE}{R+nR}$ .... (ii)

From (i) and (ii);

$n\frac{E}{R+nR}=10\left(\frac{E}{nR+R}\right)$

$n=10$