Question

A potentiometer wire of length $L$ and a resistance $r$ are connected in series with a battery of e.m.f. $E_0$ and a resistance $r_1$ An unknown e.m.f. $E$ is balanced at a length $l$ of the potentiometer wire. The e.m.f. $E$ will be given by

• A. $\frac{E_0l}{L}$
• B. $\frac{LE_0r}{(r+r_1)l}$
• C. $\frac{LE_0r}{lr_1}$
• D. $\frac{E_0r}{(r+r_1)}.\frac{l}{L}$

Answer 1

Answer: (D)

The current through the potentiometer wire is $l=\frac{E_{0}}{\left(r+r_{1}\right)}$
and the potential difference across the wire is $V=l r=\frac{E_{0} r}{\left(r+r_{1}\right)}$
The potential gradient along the potentiometer wire is $k=\frac{V}{L}=\frac{E_{0} r}{\left(r+r_{1}\right) L}$
As the unknown e.m.f. E is balanced against length $l$ of the potentiometer wire,
$\therefore E=k l=\frac{E_{0} r}{\left(r+r_{1}\right)} \frac{l}{L}$