Question:

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 \(l_1\) cm from P when two way key K3 is plugged in between 2 and 1, while the balance point is at a length \(l_2\) cm from P when key K3 is plugged in between 3 and 1. The ratio of two resistances \(\dfrac{R_1}{R_2}\), is found to be :

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The same current 1.0 A flows through \(R_1\) and \(R_2\) in series; the balance length is proportional to the potential difference being measured.
Updated On: Nov 11, 2026
  • \(\dfrac{l_1}{l_1 + l_2}\)
  • \(\dfrac{l_2}{l_2 - l_1}\)
  • \(\dfrac{l_1}{l_1 - l_2}\)
  • \(\dfrac{l_1}{l_2 - l_1}\)
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The Correct Option is D

Solution and Explanation

The current in the upper circuit is the same (1.0 A) through \(R_1\) and \(R_2\) in series.
Plug between 2 and 1: the potential difference across \(R_1\) is balanced, so \(IR_1 = k l_1\).
Plug between 3 and 1: the potential difference across \(R_1 + R_2\) is balanced, so \(I(R_1 + R_2) = k l_2\).
Subtracting: \(IR_2 = k(l_2 - l_1)\).
Dividing: \(\dfrac{R_1}{R_2} = \dfrac{l_1}{l_2 - l_1}\).
Answer: (4) \(\dfrac{l_1}{l_2 - l_1}\)
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