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

(c ) By using `i= (E)/(R + r)`
`implies 0.5 = (E)/(11 + r) implies E = 5.5 + 0.5 r` …..(i)
and `0.9 = (E)/(5 + r) implies E = 4.5 + 0.9 r` ……..(ii)
On solving these equaitons, we have `r = 2.5 Omega`