JEE Physics 2024 Question with Solution

Given: $L = \frac{100}{\pi} \, \text{mH} = \frac{100}{\pi} \times 10^{-3} \, \text{H}$, $C = 10^{-3} \, \text{F}$, $R = 10 \, \Omega$, $f = 50 \, \text{Hz}$.

Find: The power factor of the series LCR circuit.

First, calculate the inductive reactance:

$$X_L = 2\pi f L = 2\pi \times 50 \times \frac{100}{\pi} \times 10^{-3} = 10 \, \Omega$$

Now calculate the capacitive reactance:

$$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 10^{-3}} = 10 \, \Omega$$

Since $X_L = X_C$, the circuit is in resonance, so the impedance becomes:

$$Z = R = 10 \, \Omega$$

The power factor is:

$$\cos\phi = \frac{R}{Z} = \frac{10}{10} = 1$$

Therefore, the power factor of the circuit is $1$.

Given: $L = \frac{100}{\pi} \, \text{mH}$, $C = 10^{-3} \, \text{F}$, $R = 10 \, \Omega$, $f = 50 \, \text{Hz}$.

Find: The power factor.

Using the reactance formulas:

$$X_L = 2\pi f L$$

and

$$X_C = \frac{1}{2\pi f C}$$

Substituting the values:

$$X_L = 10 \, \Omega, \quad X_C = 10 \, \Omega$$

Thus the net reactance is zero, so only resistance contributes to impedance.

$$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{10^2 + 0^2} = 10 \, \Omega$$

Hence,

$$\text{Power Factor} = \frac{R}{Z} = 1$$

The correct answer is $1$.