Question

A potentiometer wire of length 100 cm has a resistance of 10 ohm. it is connected in series with a resistance and an accumulator of e.m.f 2 V and of negligible internal resistance . A source of e.m.f. 10 m V is balanced against a length of 40 cm of the potentiometer wire. What is the value of external resistance?

• A. 480 ohm
• B. 1120 ohm
• C. 790 ohm
• D. 640 ohm

Answer 1

Answer: (C)

790 ohm

Given,

As the source of e.m.f. $E^{\prime }=10mV=10\times {10}^{-3}V$ is balanced by a length of 40$cm$ of the potentiometer wire, it follows that $10\times {10}^{-3}=\text{J}$ resistance of 40$cm$ of the potentiometer wire.

If I is current through the potentiometer wire then

$J=\frac{E}{R+10}=\frac{2}{R+10}$

Now resistance of 40$cm$ of the potentiometer wire $=\frac{10}{100}\times 40=4\Omega$

$10\times {10}^{-3}=\frac{2}{R+10}\times 4$

$\Rightarrow R=790\Omega$

Hence, resistance is $790\Omega$