An ac source of 200 V (rms) is connected across a series combination of a resistor of 30 , an inductor of reactance 80 , and a capacitor of reactance 40 . The power dissipated in the circuit, and the additional capacitive reactance that must be connected in series to achieve maximum power dissipation are, respectively:

Step-by-step solution

The net impedance of the series LCR circuit is given by: Z = R^2 + (X_L - X_C)^2 Z = 30^2 + (80 - 40)^2 = 30^2 + 40^2 = 50 The rms current in the circuit is: I_ rms = V_ rms Z = 200 50 = 4 A The power dissipated in the circuit is: P = I_ rms ^2 R = (4)^2 30 = 16 30 = 480 W For maximum power dissipation, the circuit must be in resonance, which requires the total capacitive reactance to equal the inductive reactance: X_ C( total ) = X_L = 80 Since the existing capacitive reactance is 40 , the additional capacitive re