JEE Physics 2024 Question with Solution

Given: The source voltage is $4\,\text{V}$, the resistors are $100\,\Omega$ and $200\,\Omega$, and the voltmeter reads $1\,\text{V}$ across the $100\,\Omega$ resistor. The voltmeter is connected in parallel with the $100\,\Omega$ resistor.

Find: The resistance of the voltmeter.

Let the voltmeter resistance be $R_V$ and let the equivalent resistance of the parallel combination of $100\,\Omega$ and $R_V$ be $R_p$.

The voltage across the parallel branch is $1\,\text{V}$, so the voltage across the $200\,\Omega$ resistor is

$$V_{200} = 4 - 1 = 3\,\text{V}$$

Hence the circuit current is

$$I = \frac{V_{200}}{200} = \frac{3}{200}\,\text{A}$$

Now this same current flows into the series branch containing the parallel combination, so

$$R_p = \frac{V}{I} = \frac{1}{3/200} = \frac{200}{3}\,\Omega$$

For the parallel combination,

$$R_p = \frac{100R_V}{100 + R_V}$$

So,

$$\frac{100R_V}{100 + R_V} = \frac{200}{3}$$

Cross-multiplying,

$$300R_V = 20000 + 200R_V$$ $$100R_V = 20000$$ $$R_V = 200\,\Omega$$

Therefore, the resistance of the voltmeter is $200\,\Omega$. The correct option is B.

The solution is inconsistent with the question data because it uses a source voltage of $5\,\text{V}$ and concludes option A. Using the given question values, the correct answer is B.

Given: A voltmeter is placed across the $100\,\Omega$ resistor, so it does not merely read voltage; it also changes the effective resistance of that branch.

Find: The voltmeter resistance.

Because the voltmeter is in parallel with the $100\,\Omega$ resistor, the branch resistance becomes less than $100\,\Omega$. The branch has a measured voltage of $1\,\text{V}$, while the total battery voltage is $4\,\text{V}$. Therefore the remaining drop across the series $200\,\Omega$ resistor is $3\,\text{V}$.

Using Ohm's law on the $200\,\Omega$ resistor,

$$I = \frac{3}{200}\,\text{A}$$

This is the total current supplied by the battery.

So the equivalent resistance of the parallel branch is

$$R_p = \frac{1}{3/200} = \frac{200}{3}\,\Omega$$

Now apply the parallel-resistance relation:

$$\frac{1}{R_p} = \frac{1}{100} + \frac{1}{R_V}$$

Substitute $R_p = \frac{200}{3}\,\Omega$:

$$\frac{3}{200} = \frac{1}{100} + \frac{1}{R_V}$$ $$\frac{3}{200} - \frac{2}{200} = \frac{1}{R_V}$$ $$\frac{1}{200} = \frac{1}{R_V}$$ $$R_V = 200\,\Omega$$

Therefore, the voltmeter resistance is $200\,\Omega$.