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 the battery becomes $10 \,I$ . The value of ‘n’ is

• A. 9
• B. 10
• C. 20
• D. 11

Answer 1

Answer: (B)

10

Answer 2

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