When a resistor of `11 Omega` is connected in series with an electric cell, the current flowing in it is `0.5 A`. Instead, when a resistor of `5 Omega` is connected to the same electric cell in series, the current increases by `0.4 A` The internal resistance of the cell is

Options

A. `1.5 Omega`
B. `2 Omega`
C. `2.5 Omega`
D. `3.5 Omega`

Answer

C

Explanation

`0.5(r + 11) = 0.9(r + 5)`
`rArr 5r + 55 = 9r + 45`
`rArr 4r =10 rArr r = 10/4 = 2.5 Omega`