A set of n equal resistors, of value 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 10I. The value of n is

Options

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

Answer

C

Explanation

Equivalent resistances in series,
`R_1=nR+R=(n+1)R`
`thereforeI=E/R_1=E/((n+1)R)`......(1)
Equivalent resistance in parallel,
`R_2=(R/n+R)=(1+1/n)R`
`therefore10I=E/R_2=E/((1+1/n)R)`......(2)
From equation (1) and (2) `(10I)/I=((n+1)R)/((1+1/n)R)or,10((n+1)/n)=(n+1)`
or,`n=10`