Physics - Current Electricity Question with Solution | TestHub

Let $k$ be the potential gradient of the potentiometer wire.

When the cell of e.m.f. $E_1$ is connected, the balancing length is ${\ell }_1$.

$E_1=k{\ell }_1$

When the two cells are connected such that they oppose each other, the effective e.m.f. is $E_1-E_2$ (since $E_1>E_2$). The balancing length is ${\ell }_2$.

$E_1-E_2=k{\ell }_2$

Substituting the value of $E_1$ from the first equation into the second equation:

$k{\ell }_1-E_2=k{\ell }_2$

$E_2=k{\ell }_1-k{\ell }_2=k({\ell }_1-{\ell }_2)$

The ratio of $E_1$ to $E_2$ is:

$\frac{E_1}{E_2}=\frac{k{\ell }_1}{k({\ell }_1-{\ell }_2)}=\frac{{\ell }_1}{{\ell }_1-{\ell }_2}$

Answer: $\frac{{\ell }_1}{{\ell }_1-{\ell }_2}$