Question

Two cells of e.m.f $E_1$ and $E_2(E_1>E_2)$ are connected as shown in figure. When a potentiometer is connected between $\text{A}$ and $\text{B}$, the balancing length of the potentiometer wire is $300\text{cm}$. On connecting the same potentiometer between $\text{A}$ and $\text{C}$, the balancing length is $100\text{cm}$. The ratio $E_1/E_2$ is

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

Answer 1

The correct option is D $3:2$

Since we know that in a potentiometer,

$\text{emf}\propto \text{Balancing length}$

In the given problem,

$E_1\propto 300$ and $E_1-E_2\propto 100$

$\therefore \frac{E_1}{E_1-E_2}=\frac{300}{100}$

$E_1=3E_1-3E_2\Rightarrow 3E_2=2E_1$

$\therefore \frac{E_1}{E_2}=\frac{3}{2}$

Hence, option $\left(d\right)$ is correct.

Key concept: If a constant current flows through a wire of uniform area of cross-section, then potential drop along the wire is directly proportional to the length of the wire.