A potentiometer $$PQ$$ is set up to compare two resistances, as shown in the figure. The ammeter $$A$$ in the circuit reads 1.0 A when the two-way key $$K_3$$ is open. The balance point is at a length $$l_1$$ cm from $$P$$ when the two-way key $$K_3$$ is plugged in between 2 and 1, while the balance point is at a length $$l_2$$ cm from $$P$$ when the key $$K_3$$ is plugged in between 3 and 1. The ratio of two resistances $$\frac{R_1}{R_2}$$, is found to be:

When key $$K_3$$ is plugged between 2 and 1, the galvanometer compares the potential difference across the resistance $$R_1$$

Let the balance length be $$l_1$$. Then,

$$V_1\ ∝\ l_1$$ & $$V_1\ ∝\ R_1$$

When key $$K_3$$ is plugged between 3 and 1, now the galvanometer compares the potential difference across the combination of $$R_1$$ & $$R_2$$.

Let the balance length be $$l_2$$. Then,

$$V_2\ ∝\ l_2$$ & $$V_2\ ∝\ R_1+R_2$$

$$\frac{V_1}{V_2}=\frac{R_1}{R_1+R_2}=\frac{l_1}{l_2}$$

$$R_1l_2=\left(R_1+R_2\right)l_1$$

$$R_1\left(l_2-l_1\right)=R_2l_1$$

$$\therefore\ \frac{R_1}{R_2}=\frac{l_1}{l_2-l_1}$$