In the given circuit of potentiometer, the potential difference $E$ across $AB$ (10 m 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 second battery $(E_2)$ for working by making $K_1$ open and $K_2$ closed. The galvanometer gives then null deflection at $J_2$. The value of $\dfrac{E_1}{E_2}$ is $\dfrac{a}{2}$, where $a$ = ______.
Numerical value type. Enter your answer.
Answer: 1
Measure lengths along the wire from $A$. Each straight length is 1 m and the wire turns back at each end.
$J_1$ is on the 4th length (which runs right to left), 20 cm from its left end, i.e. 80 cm from its start: $l_1 = 3\ \text{m} + 0.8\ \text{m} = 3.8$ m.
$J_2$ is on the 8th length (also running right to left), 60 cm from its right end, i.e. 60 cm from its start: $l_2 = 7\ \text{m} + 0.6\ \text{m} = 7.6$ m.
$$\frac{E_1}{E_2} = \frac{l_1}{l_2} = \frac{3.8}{7.6} = \frac12 \Rightarrow a = 1$$
Solution by Sreeraj P, M.Sc Physics