Question

Two cells of emf $E_1$ and $E_2(E_1>E_2)$ are connected shown in figure. When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is $300cm.$ On connecting the same potentiometer between A and C, the balancing length is $100cm.$ 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 $3:2$

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

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

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