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Potentiometer

A potentiomet...

Question

A potentiometer PQ is set up to compare two resistance 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 between 2 and 1, while the balance point is at a length $l_{2}$ cm from P when key $K_{3}$ is plugged between 3 and 1. The ratio of two resistances $\frac{R_{1}}{R_{2}}$ , is found to be

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

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

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

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

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Solution

The correct option is D $\frac{l_{1}}{l_{2} - l_{1}}$
When key is a t point
$V_{1} = I R_{1} = x l_{1}$
when key is at (3)
$V_{2} = I (R_{1} + R_{2}) = x l_{2}$
$\frac{R_{1}}{R_{1} + R_{2}} = \frac{l_{1}}{l_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2} - l_{1}}$

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The correct option is D l1l2−l1When key is a t point V1=IR1=xl1when key is at (3)V2=I(R1+R2)=xl2R1R1+R2=l1l2R1R2=l1l2−l1

The correct option is D $\frac{l_{1}}{l_{2} - l_{1}}$
When key is a t point
$V_{1} = I R_{1} = x l_{1}$
when key is at (3)
$V_{2} = I (R_{1} + R_{2}) = x l_{2}$
$\frac{R_{1}}{R_{1} + R_{2}} = \frac{l_{1}}{l_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2} - l_{1}}$