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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 two way key $K_{3}$ is open. The balance point is at a length $l_{1}$ cm from P when 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 :

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$\frac{l_{2}}{l_{2} - l_{1}}$

$\frac{l_{1}}{l_{1} + l_{2}}$

$\frac{l_{1}}{l_{2} - l_{1}}$

$\frac{l_{1}}{l_{1} - l_{2}}$

answer is D.

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Detailed Solution

When key is in between 1 and 2

$V_{1} = {IR}_{1} = {xl}_{1}$

When key is in between 1 and 3

$\begin{aligned} V_{2} = I (R_{1} + R_{2}) = {xl}_{2} \\ \frac{R_{1}}{R_{1} + R_{2}} = \frac{l_{1}}{l_{2}} \\ \frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2} - l_{1}} \end{aligned}$

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