Question

$AB$ is a potentiometer wire of length $100\text{cm}$ and its resistance is $10\Omega$. It is connected in series with a resistance $R=40\Omega$ and a battery of emf $2\text{V}$ and negligible internal resistance. If a source of unknown emf $E$ is balanced by $40\text{cm}$ length of the potentiometer wire, then the value of $E$ is

• A. $0.8\text{V}$
• B. $1.6\text{V}$
• C. $0.08\text{V}$
• D. $0.16\text{V}$

Answer 1

The correct option is D $0.16\text{V}$

Suppose source of unknown emf is not connected,
Then, $I=\frac{2}{10+40}\text{A}=\frac{2}{50}\text{A}=\frac{1}{25}\text{A}$
Resistance per unit length of wire$=\frac{10}{100}\Omega \text{/m}$
Now, voltage drop across $40\text{cm}$ length of wire
$=\frac{1}{25}\times \frac{10}{100}\times 40=\frac{4}{25}\text{V}=0.16\text{V}$
We know that in balanced condition
$E=$voltage drop across $40\text{cm}$ length of wire$=0.16\text{V}$