Question

In potentiometer experiment, if $l_1$ is the balancing length for e.m.f of cell of internal resistance rand $l_2$ is the balancing length for its terminal potential difference when shunted with resistance R then :

• A. $l_1=l_2\left(\frac{R+r}{R}\right)$
• B. $l_1=l_2\left(\frac{R}{R+r}\right)$
• C. $l_1=l_2\left(\frac{R}{R-r}\right)$
• D. $l_1=l_2\left(\frac{R-r}{R}\right)$

Answer 1

Answer: (A)

$l_1=l_2\left(\frac{R+r}{R}\right)$

The internal resistance of a cell is given as $r=\left(\frac{l_1-l_2}{l_2}\right)R$
$\frac{r}{R}=\frac{l_1-l_2}{l_2}$
$\therefore r{l}_2=R{l}_1-R{l}_2$
$\therefore l_2(R+r)=R{l}_1$
$l_1=l_2\left(\frac{R+r}{R}\right)$