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 $\left( E _{1}\right)$ 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 $a =$ ...............
JEE MAIN 2021, Medium
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