Question

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

• A. $480\mathrm{ohm}$
• B. $1120\mathrm{ohm}$
• C. $790\mathrm{ohm}$
• D. $640\mathrm{ohm}$

Answer 1

The correct option is C

$790\mathrm{ohm}$

Step 1: Given data

Length of the wire, $l=100\mathrm{cm}$

Resistance, $R=10\Omega$

Emf, $E=2\mathrm{V}$

Souce emf, $E\text{'}=10\mathrm{mV}$

Balancing length,$l\text{'}=40cm$

We have to find the value of external resistance of the wire.

Step 2: Formula to be used

It is an instrument for measuring an electromotive force by balancing it against the potential difference produced by passing a known current through a known variable resistance or potentiometer.

It is defined as the rate of change of potential with respect to displacement in the direction of electric field.

We can find the external resistance by using the formula,

$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$

Here, $\mathrm{V}$ is potential difference.

Step 3: Find the external resistance of the wire.

If $\mathrm{I}$ is current via the potentiometer wire, then we get,

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}+10} \\ \Rightarrow \mathrm{I}=\frac{2}{\mathrm{R}+10}------\left(1\right) \end{gathered}$$

So,

The resistance of $40\mathrm{cm}$ of the potentiometer wire,

$$\begin{gathered} R\text{'}=\frac{R}{l}\times l\text{'} \\ R\text{'}=\frac{10}{100}\times 40 \\ R\text{'}=4\Omega \end{gathered}$$

The emf of the cell balanced by a length of the wire,

$$\begin{gathered} \mathrm{E}\text{'}=10\mathrm{mV} \\ \mathrm{E}\text{'}=10\times {10}^{-3}\mathrm{V} \\ \mathrm{E}\text{'}={10}^{-2}\mathrm{V} \end{gathered}$$

So, by using $\mathrm{V}=\mathrm{IR}$, we get,

$$\begin{gathered} E\text{'}=I\times R\text{'} \\ \Rightarrow {10}^{-2}=\mathrm{I}\times 4 \\ \Rightarrow I=2.5\times {10}^{-3}A \end{gathered}$$

Step 4: Solve the equation

Put the values in equation $\left(1\right)$. we get,

$0.0025=\frac{2}{10+\mathrm{R}\text{'}\text{'}}$

$0.025+0.0025\mathrm{R}\text{'}\text{'}=2$

$0.0025\mathrm{R}\text{'}\text{'}=2-0.025$

$\mathrm{R}\text{'}\text{'}=\frac{1.975}{0.0025}$

$\mathrm{R}\text{'}\text{'}=790\Omega$

Therefore, the external resistance is $790\Omega$.

Thus, option C is correct answer.