JEE Physics 2023 Question with Solution

Given: A $3 \, \text{V}$ source, a $2 \, \Omega$ resistor, and a branch containing a $5 \, \Omega$ resistor in parallel with a voltmeter of resistance $R$. The voltmeter reads $2 \, \text{V}$.

Find: The resistance $R$ of the voltmeter.

Since the $5 \, \Omega$ resistor and the voltmeter are in parallel, the potential difference across each is $2 \, \text{V}$.

Current through the $5 \, \Omega$ resistor is

$$I_1 = \frac{2}{5}$$

The source is $3 \, \text{V}$, so the remaining potential drop across the $2 \, \Omega$ resistor is

$$3 - 2 = 1 \, \text{V}$$

Hence the total current through the series $2 \, \Omega$ resistor is

$$I = \frac{1}{2}$$

Using junction law at the parallel branch,

$$I = I_1 + I_2$$

So,

$$\frac{1}{2} = \frac{2}{5} + I_2$$ $$I_2 = \frac{1}{2} - \frac{2}{5} = \frac{1}{10}$$

This $I_2$ is the current through the voltmeter. Therefore,

$$R = \frac{V}{I_2} = \frac{2}{1/10} = 20 \, \Omega$$

Therefore, the resistance of the voltmeter is $20 \, \Omega$.

Given: The voltmeter of resistance $R$ is in parallel with the $5 \, \Omega$ resistor, and this parallel combination is in series with $2 \, \Omega$.

Find: The value of $R$.

The equivalent resistance of the parallel part is

$$\frac{5R}{5+R}$$

So total resistance is

$$R_{eq} = 2 + \frac{5R}{5+R}$$

The total current from the $3 \, \text{V}$ source is

$$I = \frac{3}{R_{eq}}$$

Because the parallel branch has a voltage of $2 \, \text{V}$, the series $2 \, \Omega$ resistor has a drop of $1 \, \text{V}$, so total current is also

$$I = \frac{1}{2}$$

Equating and solving gives $R = 20 \, \Omega$.

Therefore, the resistance of the voltmeter is $20 \, \Omega$.