Question

Two batteries of emf's ${\epsilon }_1$ and ${\epsilon }_2$ are connected in a circuit as shown in figure. The galvanometer is kept undeflected always. Here $AB=l$. When the switch $S$ is connected to point $S_1$, $BP=x_1$ and when the switch $S$ is connected to point $S_2$, $BP=x_2$. Here, the ratio of emf's of the two batteries, i.e., ${\epsilon }_1/{\epsilon }_2$ equal/s :

• A. $x_1/x_2$
• B. $x_2/x_1$
• C. $\frac{(l-x_1)}{(l-x_2)}$
• D. $\frac{(l-x_2)}{(l-x_1)}$

Answer 1

The correct option is C $\frac{(l-x_1)}{(l-x_2)}$

$AP=l-x$
We know that EMF$\propto$length of wire across battery(i.e., AP)
$E\propto AP$
For first battery
$E_1=(l-x_1)$
For second battery
$E_2=(l-x_2)$
$\Rightarrow \frac{E_1}{E_2}=\frac{l-x_1}{l-x_2}$
Answer C