Question

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

The current in the potentiometer wire $AC$ is $I=\frac{2}{10+R}$

The potential difference across the potentiometer wire is $V=$ current $\times$ resistance
$=\frac{2}{10+R}\times 10$
The length of the wire is $l=100cm$.
So, the potential gradient along the wire is
$k=\frac{V}{l}=\left(\frac{2}{10+R}\right)\times \frac{10}{100}\dots \left(i\right)$
The source of emf $10mV$ is balanced against a length of $40cm$ of the potentiometer wire
i.e. $10\times 10^{-3}=k\times 40$
or $10\times 10^{-3}=\frac{2}{\left(10+R\right)}\times \frac{40}{10}$ (Using $(i)$)
or $R=790\Omega$.