1. Given data

• Resistance: $R = 200\,\Omega$
• Supply voltage: $V = 220\,\text{V}$
• Frequency: $50\,\text{Hz}$

We need the power dissipated in the series $LCR$ circuit.

For a series $LCR$ circuit, $P = I^2 R = VI\cos\phi$ where $\phi$ is the phase angle of the complete circuit.

2. Case 1: Capacitor removed

Then the circuit becomes a series $LR$ circuit.

Given: current lags voltage by $30^\circ$. For an $LR$ circuit, $\tan \phi_L = \frac{X_L}{R}$ So, $\tan 30^\circ = \frac{X_L}{R}$ $\frac{1}{\sqrt{3}} = \frac{X_L}{200}$ Hence, $X_L = \frac{200}{\sqrt{3}}\,\Omega$

3. Case 2: Inductor removed

Then the circuit becomes a series $CR$ circuit.

Given: current leads voltage by $30^\circ$. For an $RC$ circuit, $\tan \phi_C = \frac{X_C}{R}$ (using magnitude of phase angle) Thus, $\tan 30^\circ = \frac{X_C}{R}$ $\frac{1}{\sqrt{3}} = \frac{X_C}{200}$ Hence, $X_C = \frac{200}{\sqrt{3}}\,\Omega$

4. Now consider the complete $LCR$ circuit

Net reactance in series $LCR$ circuit is $X = X_L - X_C$ But here, $X_L = X_C = \frac{200}{\sqrt{3}}$ So, $X = 0$

Therefore the circuit is in resonance, and the impedance is purely resistive: $Z = R = 200\,\Omega$

Hence current is $I = \frac{V}{R} = \frac{220}{200} = 1.1\,\text{A}$

5. Power dissipated

Since the circuit is purely resistive at resonance, $\cos\phi = 1$ Therefore, $P = I^2R = (1.1)^2 \times 200$ $P = 1.21 \times 200 = 242\,\text{W}$

6. Option check

• A: $305\,\text{W}$ — incorrect
• B: $210\,\text{W}$ — incorrect
• C: $0\,\text{W}$ — incorrect
• D: $242\,\text{W}$ — correct

Final Answer: $\boxed{242\,\text{W}}$