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

Explanation

When `n` resistors are in series, then current in the circuit, `I =epsilon/(R+nR) = epsilon/((n+1)R)` …(i)
When `n` resistors are in parallel, then effective resistance is `R_("eff")=R//n`.
current in circuit, `I' = epsilon/(R+R//n) = (n epsilon)/(R(n+1))`
As per question, `I' = 10 I :. (n epsilon)/(R(n+1)) = (10 epsilon)/((n+1)R) or n=10`