Question

The circuit shown here is used to compare the e.m.f.'s of the two cells $E_1$ and $E_2(E_1>E_2)$. The null point is at J when the galvanometer is connected to $E_1$. When the galvanometer is connected to $E_2$, the null point will be

• A. to the left of C
• B. to the right of C
• C. at C itself
• D. nowhere on AB

Answer 1

The correct option is A to the left of C

We know that in this scenerio:

$E=IR⟹E\propto R$ and $R\propto l$

$\therefore E\propto l$

so if $E_2<E_1$ l would be less and shift to the left of current position.