Question

A potentiometer wire of length $10m$ and resistance $10\Omega$ per metre is connected in series with a resistance box and a $2volt$ battery. If a potential difference of $100mV$ is balanced across the whole length of potentiometer wire, then the resistance introduced in the resistance box will be:

• A. $1900\Omega$
• B. $900\Omega$
• C. $190\Omega$
• D. $90\Omega$

Answer 1

The correct option is A $1900\Omega$

length of potentiometer $l=10m$

resistance of wire $R=10\Omega$

Emf of the cell $E=2V$

balancing length is $l=10m$ (whole length or wire )

let $I$ be the current through the potentiometer then,

$I=\frac{E}{R+R^{\prime }}$.........................(1)

EMf of the cell balanced by whole length of the wire

$V=100mV$

$100\times {10}^{-3}={10}^{-1}$

by using $V=IR$

$\begin{array}{c}{10}^{-1}=I\times 10\\ I={10}^{-2}\\ I=0.01A\end{array}$

putting the value in equation (1),we get

$\begin{array}{c}0.01=\frac{2}{10+R}\\ 0.1+0.01R=2\\ 0.012R=1.9\\ R^{\prime }=1900\Omega \end{array}$

Hence, the option $\left(A\right)$ is correct.