Question

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

• A. 3 : 2
• B. 4 : 3
• C. 5 : 4
• D. 2 : 1

Answer 1

The correct option is A 3 : 2

For balance potentiometer : $\frac{E_{ab}}{E_{ac}}=\frac{l_{ab}}{l_{ac}}$

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

or $3E_1-3E_2=E_1$

or $2E_1=3E_2$ or $E_1/E_2=3/2$

or $E_1:E_2=3:2$