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 ($E_1$) is replaced by the second battery ($E_2$) for working by making $K_1$ open and $E_2$ closed. The galvanometer gives then null deflection at $J_2$. The value of $\frac{E_1}{E_2}$ is the smallest fraction of $\frac{a}{b}$, Then the value of $a$ is ____.

Answer 1

Step 1: Find the balancing length from end A, when $K_1$ is closed and

Length of

$\begin{array}{rcl}AB& =& 10m\\ & =& 10\times 100cm\\ & =& 1000cm\end{array}$

For battery $E_1$, the balancing length is $l_1$

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, that is balancing length is $l_1=380cm$

Step 2: Find the balancing length from end A when $K_2$ is closed.

For battery $E_2$, the balancing length is $l_2$

For key $K_2$ (closed), the jockey is adjusted to touch the wire at point $J_2$ so that there is no deflection in the galvanometer, which is balancing at $l_2=760cm$.

Step 3: Find the value of $\frac{E_1}{E_2}$ is the smallest fraction of $\frac{a}{b}$

Know that

$\begin{array}{c}\frac{E_1}{E_2}=\frac{l_1}{l_2}\\ =\frac{380}{760}\\ =\frac{1}{2}\end{array}$

Equating with the smallest fraction, the value of a would-be

$\begin{array}{l}\frac{E_1}{E_2}=\frac{a}{b}=\frac{1}{2}\\ \Rightarrow a=1,b=2\end{array}$

Therefore, the correct answer is 1.