In the given circuit of potentiometer, the potential difference $E$ across $A B(10 m$ length $)$ is larger than $E_1$ and $E_2$ as well. For key $\mathrm{K}_1$ (closed), the jockey is adjusted to touch the wire at point $\mathrm{J}_1$ so that there is no deflection in the galvanometer. Now the first battery $\left(\mathrm{E}_1\right)$ is replaced by second battery $\left(\mathrm{E}_2\right)$ for working by making $\mathrm{K}_1$ open and $\mathrm{K}_2$ closed. The galvanometer gives then null deflection at $\mathrm{J}_2$.
The value of $\frac{E_1}{E_2}$ is $\frac{a}{b}$, where $a=$ $\_\_\_\_$ .