10 resistors, each of resistance R are connected in series to a battery of emf E and negligible internal…

When the resistors are connected in series with the battery, the equivalent resistance $\left(R_s\right)$ can be written as

$\begin{array}{rcl}{R}_s& =& R+R+....10 \mathrm{times}\\ & =& 10R ...\left(1\right)\end{array}$

Hence, the current $\left(i_s\right)$ is given by

$\begin{array}{rcl}{i}_s& =& \frac{E}{R_s}\\ & =& \frac{E}{10R} ...\left(2\right)\end{array}$

When the resistors are connected in parallel, the equivalent resistance $\left(R_p\right)$ is given by

$\begin{array}{rcl}{R}_p& =& \frac{1}{{\displaystyle \frac{1}{R}}+{\displaystyle \frac{1}{R}}+.... 10 \mathrm{times}}\\ & =& \frac{R}{10} ...\left(3\right)\end{array}$

Hence, the current $\left(i_p\right)$ is given by

$\begin{array}{rcl}{i}_p& =& \frac{E}{R_p}\\ & =& \frac{E}{{\displaystyle \frac{R}{10}}}\\ & =& \frac{10E}{R} ...\left(4\right)\end{array}$

Divide equation (4) by equation (2) to obtain the required value of the multiplier $n$.

$\begin{array}{rcl}\frac{i_p}{i_s}& =& \frac{{\displaystyle \frac{10E}{R}}}{{\displaystyle \frac{E}{10R}}}\\ & =& 100\end{array}$