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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 K3 is open.  The balance point i

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 K3 is open.  The balance point is at a length l1 cm from P when two way key K3 is plugged in between 2 and 1, while the balance point is at a length l2 cm from P when key K3 is plugged in between 3 and 1.The ratio of two resistances \frac{R_{1}}{R_{2}}    is  found to be  1:2, then find the ratio of   \frac{l_1}{l_2}   

Option: 1

\frac{1}{2}


Option: 2

\frac{1}{3}


Option: 3

\frac{1}{4}


Option: 4

1


Answers (1)

 

 

When key is at point (1)

V_{1}=iR_{1}=Xl_{1}

When key is at 3 

V_{2}=i(R_{1}+R_{2})=Xl_{2}

\frac{R_{1}}{R_{2}}=\frac{l_{i}}{l_{2}}=\frac{R_{1}}{R_{2}}=\frac{l_{1}}{l_{2}-l_{1}}

 

So we get a ratio of resistances as

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

\\ \frac{R_2}{R_1} = 2 = \frac{l_2-l_1}{l_1} \\ \\ \\ \frac{l_1}{l_2} = \frac{1}{3}

Posted by

Kshitij

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