Question

A set of $‘n’$ equal resistors, of value ${}^{\prime }{R}^{\prime }$ each, are connected in series to a battery of emf $E’$ and internal resistance ${}^{\prime }R$ ' The current drawn is $I$ . Now, the $‘n^{\prime }$ resistors are connected in parallel to the same battery. Then the current drawn from battery becomes $10I$ . The value of $‘n’$ is

• A. 9
• B. 10
• C. 20
• D. 11

Answer 1

Answer: (B)

$I=\frac{E}{nR+R}$ .....(i)
$10I=\frac{E}{\frac{R}{n}+R}$ ....(ii)
Dividing (ii) by (i),
$10=\frac{\left(n+1\right)R}{\left(\frac{1}{n}+1\right)R}$
After solving the equation, $n=10$