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

When resistances are connected in series then current can be written as
`I = (E )/(nR + R)` …(i)
When resistances are connected in parallel then current can be written as
`10I = (E )/((R )/(n) + R)` …(ii)
Dividing (ii) by (i), `10 = ((n+1))/(((1)/(n)+1))`
On solving we get
n= 10
Hence option (c ) is correct