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 . On connecting the same potentiometer between A and C, the balancing length is 100 cm. The ratio $\frac{E_1}{E_2}$ is

• A. 3:01
• B. 1:03
• C. 2:03
• D. 3:02

Answer 1

Answer: (D)

3:02

Answer 2

When potentiometer is connected between A and B, then it measures only $E_1$ and when connected between A and C,then it measures $E_1 - E_2$ . $\therefore \, \, \, \frac{E_1}{E_1 - E_2} = \frac{l_1}{l_2}$ $\, \, \, \, \, \, \, \, \, \, \, \, \, \frac{E_1 - E_2}{E_1} = \frac{l_2}{l_1}$ $\Rightarrow 1 - \frac{E_2}{E_1} = \frac{100}{300}$ $\Rightarrow \frac{E_2}{E_1} = 1 - \frac{1}{3}$ $\Rightarrow \frac{E_2}{E_1} = \frac{2}{3}$ $\Rightarrow \frac{E_1}{E_2} = \frac{3}{2}$