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. 11
C. 10
D. 9

Answer

C

Explanation

`I=(E )/(nR+R)=(E )/((n+1)R)" " …(1)`
`101=(E )/((R )/(n) +R ) =(nE)/((n+1)R)" " ……(2)`
For Eqs. (1) and (2)
n = 10