A current of `2 A` flows through a `2 Omega` resistor when connected across a battery. The same battery supplies a current of `0.5 A` when connected across a `9 Omega` resistor. The internal resistance of the battery is

Options

A. `0.5Omega`
B. `1//3 Omega`
C. `1//4Omega`
D. `1Omega`

Answer

2

Explanation

Let internal resistance of b attery be r then according to question `(E)/(2+r)=2` and `(E)/(9+r)=0.5`
`rArr(2)/(0.5)=(9+r)/(2+r)rArrr=(1)/(3)Omega`