Question

A potentiometer wire of length 200 cm has a resistance of $20\Omega$ . It is connected in series with a resistance of $10\Omega$ and an accumulator of emf 6 V having negligible internal resistance. A source of 2.4 V is balanced against a length L of the potentiometer wire. The value of L is

• A. 100 cm
• B. 120 cm
• C. 110 cm
• D. 140 cm

Answer 1

The correct option is B

120 cm

The current in the potentiometer wire AB is

$I=\frac{6}{20+10}=0.2A$

The potential difference across the potentiometer wire is

V = current $\times$ resistance = $0.2\times 20=4V$

The potential gradient along the wire is $k=\frac{V}{l}=\frac{4}{200}=0.02Vc{m}^{-1}$

The emf 2.4 V is balanced against a length L of the potentiometer wire.

i.e. 2.4 = kL
or $L=\frac{2.4}{k}=\frac{2.4}{0.02}=120cm$