Question

In the given circuit of potentiometer, the potential difference $E$ across $AB$ (10m length) is larger than $E_1$ and $E_2$ as well. For key $K_1$ (closed), the jockey is adjusted to touch the wire at point $J_1$ so that there is no deflection in the galvanometer. Now the first battery $\left(E_1\right)$ is replaced by second battery $\left(E_2\right)$ for working by making $K_1$ open and $K_2$ closed. The galvanometer gives then null deflection at $J_2$. The value of $\frac{E_1}{E_2}$ is $\frac{a}{b},$ where a $=$___

Answer 1

Answer: 1

Length of $AB=10m$
For battery $E_1$, balancing length is $l_1$
$l_1=380cm[$ from end $A]$
For battery $E_2$, balancing length is $l_2$
$l_2=760cm[$ from end $A]$
Now, we know that $\frac{E_1}{E_2}=\frac{l_1}{l_2}$
$\Rightarrow \frac{E_1}{E_2}=\frac{380}{760}=\frac{1}{2}=\frac{a}{b}$
$\therefore a=1\&b=2$
$a=1$