Question

Two cells of emfs $E_1$ and $E_2(E_1>E_2)$ are connected as shown in fig.


When a potentiometer is connected between $A$ and $B$, the balancing length of the potentiometer wire is $300\text{cm}$. On connecting the same potentiometer between $A$ and $C$, the balancing length is $100\text{cm}$. The ratio $\frac{E_1}{E_2}$ is

• A. $3:1$
• B. $1:3$
• C. $2:3$
• D. $3:2$

Answer 1

The correct option is D $3:2$

For balanced potentiometer,
$\frac{E_{AB}}{E_{AC}}=\frac{l_{AB}}{l_{AC}}$
$\Rightarrow \frac{E_1}{E_1-E_2}=\frac{300}{100}=3$

or $E_1=3E_1-3E_2$
or $3E_2=2E_1$
or $\frac{E_1}{E_2}=\frac{3}{2}$