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

Answer

B

Explanation

Let the internal resistance of the battery be r. Then the current flowing through the circuit in first case is given by
as `i= (E )/( R + r) rArr 2 = (E )/( 2 + r)" "…(1)`
In second case,
`0.5 ( E)/( 9+ r)" "…(2)`
From `(1) & (2)`.
`4+2 r= 4.5 +0.5 rArr 1.5 r=0.5 rArr r= (1)/(3) Omega`.