Question

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 $10I$. The value of $n$ is

• A. $20$
• B. $10$
• C. $9$
• D. $11$

Answer 1

The correct option is B $10$

Electrical resistance in series = $nR$
$I=\frac{E}{(n+1)R}$ ...............(1)
And in parallel $10I=\frac{E}{\frac{R}{n}+R}=\frac{nE}{R(n+1)}$ ......(2)
Equating (1) and (2)
we get $n=10$