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

Answer: (B)

120 cm

Answer 2

: 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 ×resistance $= 0.2 \times 20 = 4V$

The length of the wire is $l = 20cm.$ So, the potential gradient along the wire is

$K = \frac{V}{l} = \frac{4}{200} = 0.02Vcm^{- 1}$

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

i.e., $2.4 = kL$

or $L = \frac{2.4}{k} = \frac{2.4}{0.02} = 120cm$