Question

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

• A. $790\Omega$
• B. $444\Omega$
• C. $512\Omega$
• D. $680\Omega$

Answer 1

The correct option is A $790\Omega$

Figure shows the situation described in the question.


As the source of emf, $E^{\prime }=10\text{mV}=$$10\times {10}^{-3}\text{V}$ is balanced by a length of $40\text{cm}$ of the potentiometer wire, so we have

$I{R}_{cb}=E^{\prime }=10\times {10}^{-3}\text{V}$

$\Rightarrow \left(\frac{2}{R+10}\right)R_{cb}=10\times {10}^{-3}\text{V}....\left(1\right)$

Given, resistance of $100\text{cm}$ is$R_{ab}=10\Omega$.

So, the resistance of $40\text{cm}$ of the potentiometer wire,

$\Rightarrow R_{cb}=\left(\frac{40}{100}\right)\times 10=4\Omega$

Substituting $R_{cb}$ in $\left(1\right)$,

$\left(\frac{2}{R+10}\right)\times 4=10\times {10}^{-3}\text{V}$

$\Rightarrow R=790\Omega$

Hence, $\left(\text{A}\right)$ is the correct answer.