When a current of 2 A flows in a battery from negative to positive terminal, the potential difference across it is 12 V. If a current of 3 A flowing in the opposite direction produces a potential difference of 15 V, the emf of the battery is

Options

A. 12.6 V
B. 13.2 V
C. 13.5 V
D. 14.0 V

Answer

B

Explanation

Let `epsilon` be emf and r be internal resistance of the battery.
In first case, `12=epsilon-2r" "…(i)`
In second case, `15=epsilon+3r" "…(ii)`
Subtract (i) from (ii), we get `r=(3)/(5)Omega`
Putting this value of r in eqn. (i), we get
`epsilon=12+(2xx3)/(5)=(60+6)/(5)=(66)/(5)=13.2V`