$A$ 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 key $K_3$ is plugged in between $3$ and $1$. The ratio of two resistances $\frac{R_1}{R_2}$ is found to be
- A$\frac{l_1}{l_1 + l_2}$
- B$\frac{l_2}{l_2 - l_1}$
- C$\frac{l_1}{l_1 - l_2}$
- D$\frac{l_1}{l_2 - l_1}$
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