A potentiometer P Q is set up to compare two resistances, as shown in the figure. The ammeter A in the…

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 K3 is open. The balance point is at a length l1 cm from P when the two-way key K3 is plugged in between 2 and 1, while the balance point is at a length l2 cm from P when the key K3 is plugged in between 3 and 1 . The ratio of two resistances R1R2, is found to be 
  1. l1l1-l2
  2. l2l2-l1
  3. l1l1+l2
  4. l1l2-l1

Solution

When the key is between 2 and 1

Then, emf V1= IR1=xl1

When the key is between 3 and 1

Emf, V2=IR1+R2=xl2

The ratio of both emf is R1R1+R2=l1l2

On simplifying the above relation, we get the ratio of resitances, R1R2=l1l2-l1

Asked in: JEE Main 2017 (08 Apr Online)

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