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 : 1
• B. 1 : 3
• C. 2 : 3
• D. 3 : 2

Answer 1

Answer: (D)

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