A set of $n$ equal resistors, of value $R$ each, are connected in series to a battery of emf $E$ and internal resistance $R.$ The current drawn is $I.$ Now, the $n$ resistors are connected in parallel to the same battery. Then the current drawn from battery becomes $10\,I.$ The value of $n$ is
NEET 2018, Medium
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Current drawn from a battery when $n$ resistors are connected in series is
$I=\frac{E}{n R+R}$ .....$(i)$
Current drawn from same battery when $n$ resistors are connected in parallel is
$10 I=\frac{E}{R / n+R}$ ......$(ii)$
On dividing eqn. $(ii)$ by $(i),$ $10=\frac{(n+1) R}{(1 / n+1) R}$
After solving the equation, $n=10$.
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