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

Answer 1

Answer: (C)

$\frac{E_{0}r}{(r+r_1)}.\frac{l}{L}$

The current in the circuit , $I=\frac{E_0}{r+r_1}$
The potential drop across the wire is $V=Ir=\frac{E_{0}r}{r+r_1}$
Potential gradient, $k=\frac{V}{L}=\frac{E_{0}r}{(r+r_1)L}$
Thus, emf $E=kl=\frac{E_{0}rl}{(r+r_1)L}$