Question

A potentiometer wire of length $L$ and of resistance $r$ is 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. $\frac{E_{0}r}{(r+r_1)}\times \frac{l}{L}$
• B. $\frac{E_{0}l}{L}$
• C. $\frac{L{E}_{0}r}{(r+r_1)l}$
• D. $\frac{L{E}_{0}r}{l{r}_1}$

Answer 1

The correct option is A $\frac{E_{0}r}{(r+r_1)}\times \frac{l}{L}$

$I=\frac{E_o}{R_{eq}}=\frac{E_o}{r+r_1}$
potential difference across potentiometer $=Ir=\frac{E_{o}r}{(r+r_1)}$
Potential gradient($\varphi$) is given by $\varphi =\frac{E_o}{L}=\frac{E_{o}r}{(r+r_1)}\times \frac{1}{L}$
we know that $\text{e.m.f.}$ = $\varphi l$
$l=$ balancing length
$\text{e.m.f.}=\left(\frac{E_{o}r}{r+r_1}\right)\frac{l}{L}=\left(\frac{E_{o}r}{r+r_1}\right)\frac{l}{L}$