JEE Physics 2025 Question with Solution

Given: $X_L = 100 \, \Omega$, $X_C = 50 \, \Omega$, $R = 50 \, \Omega$, $V_{\text{rms}} = 10 \, V$, and frequency $f = 50 \, Hz$.

Find: Average power dissipated by the circuit.

In a series LCR circuit, the total impedance is

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

Substituting the given values,

$$Z = \sqrt{50^2 + (100 - 50)^2}$$ $$Z = \sqrt{50^2 + 50^2} = \sqrt{2500 + 2500} = \sqrt{5000}$$ $$Z = 50\sqrt{2} \, \Omega$$

The RMS current is

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}$$ $$I_{\text{rms}} = \frac{10}{50\sqrt{2}} = \frac{1}{5\sqrt{2}} \, \text{A}$$

Only the resistor dissipates average power, so

$$P = I_{\text{rms}}^2 R$$

Substituting,

$$P = \left(\frac{1}{5\sqrt{2}}\right)^2 \times 50$$ $$P = \frac{1}{50} \times 50 = 1 \, \text{W}$$

Therefore, the average power dissipated by the circuit is $1 \, \text{W}$.

Given: Inductive reactance $X_L = 100 \, \Omega$, capacitive reactance $X_C = 50 \, \Omega$, resistance $R = 50 \, \Omega$, RMS voltage $V_{\text{rms}} = 10 \, V$, and frequency $f = 50 \, Hz$.

Find: Average power dissipated by the circuit.

1. First calculate the net reactance:

$$X_L - X_C = 100 - 50 = 50 \, \Omega$$

2. Now find the impedance of the series circuit:

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$ $$Z = \sqrt{(50)^2 + (50)^2} = \sqrt{2500 + 2500} = \sqrt{5000}$$ $$Z = 50\sqrt{2} \, \Omega$$

3. Calculate the RMS current:

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{10}{50\sqrt{2}} = \frac{1}{5\sqrt{2}} \, \text{A}$$

4. Average power is dissipated only in the resistor:

$$P_{\text{avg}} = I_{\text{rms}}^2 R$$ $$P_{\text{avg}} = \left(\frac{1}{5\sqrt{2}}\right)^2 \times 50$$ $$P_{\text{avg}} = \frac{1}{25 \times 2} \times 50 = \frac{1}{50} \times 50 = 1 \, \text{W}$$

Therefore, the required average power dissipated is $1 \, \text{W}$.