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. $\frac{E_{0}l}{L}$
• B. $\frac{L{E}_{0}r}{(r+r_1)l}$
• C. $\frac{L{E}_{0}r}{l{r}_1}$
• D. $\frac{E_{0}r}{(r+r_1)}.\frac{l}{L}$

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 $V=lr=\frac{E_{0}r}{\left(r+r_1\right)}$
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,
$\therefore E=kl=\frac{E_{0}r}{\left(r+r_1\right)}\frac{l}{L}$