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 n equal resistors of resistance R are connected in series, then the current drawn is given as
`I=E/(nR+r)`
Where, nR=equivalent resistance of n resistors in series and r=internal resistance of battery.
given, r=R
`rArr I=E/(nR+R)=E/(R(n+1)`..........(i)
Similarly, when n equal resistors are connected in parallel, then the current drawn is given as
`I^(')=E/(nR+R) = E/(R(n+1))`
Where, R/n= equivalent resistance of n resistors in parallel.
Given, `I^(') = 10I rArr 10I = E/(R/n)+R=((nE)/((n+1)R))`,,,,,,,,,(i)
Substituting the value of I from Eq. (i) in Eq. (ii), we get
`10E/(R(n+1))=(nE)/(R(n+1)) rArr n=10`