10 resistors, each of resistance R are connected in series to a...

When the resistors are connected in series with the battery, the equivalent resistance $\left({R}_{s}\right)$ can be written as

$\begin{cases}{R}_{s} & = & R + R + . . . . 10times \\ & = & 10R . . . \left(1\right)\end{cases}$

Hence, the current \left({i}_{s}\right)$is given by</p><p>$\begin{cases}{i}_{s} & = & \dfrac{E}{{R}_{s}} \\ & = & \dfrac{E}{10R} . . . \left(2\right)\end{cases}$</p><p>When the resistors are connected in parallel, the equivalent resistance \left({R}_{p}\right)$ is given by

$\begin{cases}{R}_{p} & = & \dfrac{1}{\dfrac{1}{R} + \dfrac{1}{R} + . . . . 10times} \\ & = & \dfrac{R}{10} . . . \left(3\right)\end{cases}$

Hence, the current \left({i}_{p}\right)$is given by</p><p>$\begin{cases}{i}_{p} & = & \dfrac{E}{{R}_{p}} \\ & = & \dfrac{E}{\dfrac{R}{10}} \\ & = & \dfrac{10E}{R} . . . \left(4\right)\end{cases}$</p><p>Divide equation (4) by equation (2) to obtain the required value of the multiplier$n$.</p><p>$\begin{cases}\dfrac{{i}_{p}}{{i}_{s}} & = & \dfrac{\dfrac{10E}{R}}{\dfrac{E}{10R}} \\ & = & 100\end{cases}