Question

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?

• A. $54$V
• B. $13.2$V
• C. $65$V
• D. $50$V

Answer 1

Answer: (B)

$13.2$V

Let the emf of cell $=E$

In 1st case: $V_1=E-I_{1}r$ where $r=$ internal resistance of cell.

$⟹12=E-2r......\left(1\right)$

In 2nd case as current in the opposite direction so the equation becomes

$V_2=E+I_{2}r$

$⟹15=E+3r....\left(2\right)$

Using $\left(1\right)$

$15=E+3(\frac{E}{2}-6)⟹2.5E=33⟹E=13.2V$