A set of n equal resistances, of value R each, are connected in...

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 $nR + R = ( n + 1 ) R$.

$∵ I = \dfrac{ε}{nR + R}$

→ $I = \dfrac{E}{( n + 1 ) R}$ .... (i)

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

Hence,

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

From (i) and (ii);

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

$n = 10$