When a current of 2A flows in a battery from negative to positive terminal, the potential difference across it is 12V. If a current of 3A flowing in the opposite direction produces a potential difference of 15V, the EMF of the battery is
A. 12.6V
B. 13.2V
C. 13.5V
D. 14V

Hint: Note that, $V = E + Ir$, where V is voltage, E is emf and r is the internal resistance of the cell.
Putting the given values in this equation we get two new equations. Solve these equations to find the value of E.
Formula used: $V = E + Ir$, where V is voltage, E is emf and r is the internal resistance of the cell.

Complete step by step answer:
Let the electromotive force of the given cell be E
In the first case, a current of 2A flows in the battery from negative to positive terminal and the potential difference across it is 12V.
Therefore, ${V_1} = E - {I_1}r$ , where r is the internal resistance of the cell.
Now, $V_1$=12 V, $I_1$=2 A
∴ 12 = E − 2r …… (1)
In the second case, a current of 3A flowing in the opposite direction produces a potential difference of 15V.
As current flows in the opposite direction, the equation becomes
${V_2} = E + {I_2}r$
Now, $V_2$=15 V, $I_2$=3 A
∴ 15 = E + 3r …… (2)
 Subtracting equation (1) from (2) we get,
$ \Rightarrow 5r = 3$
$ \Rightarrow r = \dfrac{3}{5}$
Putting the value of r in equation (2) we get,
$E = 15 - 3 \times \dfrac{3}{5}$
$ \Rightarrow E = 15 - \dfrac{9}{5}$
$ \Rightarrow E = \dfrac{{75 - 9}}{5}$
And hence on solving,we have
$ \Rightarrow E = \dfrac{{66}}{5}$
$ \Rightarrow E = 13.2{\text{ V}}$
Therefore, the EMF of the battery is 13.2 V.
So, the correct option is (B).

Note:
Note that, $V = E + Ir$
In the first case, the current flows in the battery from negative to positive terminal.
Therefore ${V_1} = E - {I_1}r$
But in the second case, as current flows in the opposite direction, the equation becomes
${V_2} = E + {I_2}r$