Question

A potentiometer wire of length 100 cm has a resistance of 10$\Omega$ . It is connected in series with a resistance and a cell of emf 2 V and of negligible internal resistance. A source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire. What is the value of external resistance?

• A. $790\Omega$
• B. 890$\Omega$
• C. 990$\Omega$
• D. 1090$\Omega$

Answer 1

Answer: (A)

$790\Omega$

Answer 2

: The current in the potentiometer wire AC is

$I = \frac{2}{10 + R}$

The potential difference across the potentiometer wire is

V = current × resistance

$= \frac{2}{10 + R} \times 10$

The length of the wire is, l = 100 cm.

So, the potential gradient along the wire is

$k = \frac{V}{l}\left( \frac{2}{10 + R} \right) \times \frac{10}{100}$ …(i)

The source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire

i.e., $10 \times 10^{- 3} = k \times 40$

or $10 \times 10^{- 3} = \frac{2}{(10 + R)} \times \frac{40}{10}$ (Using (ii))

or $R = 790\Omega$.