Q.In potentiometer experiment, if $l_1$ is the balancing length for e.m.f. of a cell of internal resistance r and $l_2$ is the balancing length for its terminal potential difference when shunted with resistance R then: (A) $l_1 = l_2\left(\dfrac{R+r}{R}\right)$
(B) $l_1 = l_2\left(\dfrac{R}{R+r}\right)$
(C) $l_1 = l_2\left(\dfrac{R}{R-r}\right)$
(D) $l_1 = l_2\left(\dfrac{R-r}{R}\right)$

Comparing the balancing lengths for a cell on open circuit and on a known resistance box gives the internal resistance of the cell.

In a potentiometer experiment to find the internal resistance $r$ of a cell of emf $E$: with the key in the resistance-box circuit open, the balancing length $l_1$ gives $E \propto l_1$. With a known resistance $R$ connected across the cell (so the terminal potential difference is $V$), the balancing length $l_2$ gives $V \propto l_2$. Since $l_1/l_2 = E/V$, and the current in the circ …

In a potentiometer, balancing length is proportional to the potential difference being measured; e.m.f. gives length $l_1$, terminal p.d. (when shunted by $R$) gives length $l_2$.

In a potentiometer experiment, the balancing length is directly proportional to the potential difference (or e.m.f.) being measured (since the potential gradient along the wire is constant):

$\text{p.d.}\propto l$

When the cell (e.m.f. $\varepsilon$, internal resistance $r$) is balanced directly (open circuit), the balancing length $l_1$ corresponds to its full e.m.f.:

$\varepsilon\propto l_1$

When the cell is shunted by an external resistance $R$, it drives a current $i=\dfrac{\varepsilon}{R+r}$ through the circuit, and its terminal potential difference becomes …