A set of `'n'` equal resistor, of value of `'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 10.1. The value of `'n'` is

Options

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

Answer

A

Explanation

Current drawn from a battery when n resistors are connected in series is
`I=(E)/(nR+R)" "…(i)`
Current drawn from same battery when n resistors are connected in parallel is
`10I=(E)/((R)/(n)+R)" "...(ii)`
On dividing eqn. (ii) by (i), `10=((n+1)R)/(((1)/(n)+1)R)`
After solving the equation, `n=10`