Question

A potentiometer wire of length $1m$ has a resistance of $100\Omega$. It is connected in series with a resistance and a battery of emf $2V$ of negligible resistance. A source of emf $10mV$ is balanced against a length of $40cm$ of the potentiometer wire. What is the value of the external resistance.

• A. $890\Omega$
• B. $7900\Omega$
• C. $680\Omega$
• D. $740\Omega$

Answer 1

The correct option is B $7900\Omega$

Given: A potentiometer wire of length $1m$ has a resistance of $100\Omega$. It is connected in series with a resistance and a battery of emf $2V$ of negligible resistance. A source of emf $10mV$ is balanced against a length of $40cm$ of the potentiometer wire.

To find the value of the external resistance

Solution:

If $J$ is current through the potentiometer wire then

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

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

Now resistance of $40cm$ of the potentiometer wire

$=\frac{10}{100}\times 40=4\Omega$

So,

$10\times {10}^{-3}=\frac{2}{R+10}\times 4⟹R=790\Omega$

is the value of the external resistance.