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

Answer

D

Explanation

For seriese combination, `I=E/((n+1)R)....(1)`
For parallel combination `10 I=E/(R(1+1/n)).....(2)`
Solvings eqns. `(1)` and `(2)`, we get
`n+1=10(1+1/n) implies n=10`