In a series LCR circuit, R = 200 Ω and the voltage and frequency of the main supply are 220 V and 50 Hz respectively. On taking out the capacitance from the circuit, the current lags behind the voltage by 30°. On taking out the inductor from the circuit, the current leads the voltage by 30°. The power dissipated in the LCR circuit is?

The given circuit is under resonance as XL = XC. In a series LCR circuit, the impedance is given by:

Z = √(R² + (XL - XC)²)

At resonance, XL = XC, so the impedance becomes:

Z = R

The current in the circuit is given by:

I = V/Z = V/R

The power dissipated in the LCR circuit is given by:

P = I²R = (V/R)²R = V²/R

Given that V = 220 V and R = 200 Ω, the power dissipated is:

P = (220 V)² / 200 Ω = 48400 / 200 = 242 W

Therefore, the power dissipated in the LCR circuit is 242 W.