An LCR series circuit is connected to an AC source. The circuit has a resistance R = 30 , and an inductive reactance X_L = 60 , . If the power factor of the circuit is 0.6 and the current leads the voltage, the value of the capacitive reactance X_C is:

Step-by-step solution

The power factor of an LCR series circuit is given by = R Z . Substituting the given values: 0.6 = 30 Z Z = 30 0.6 = 50 , The impedance Z is also given by Z = R^2 + (X_C - X_L)^2 . Squaring both sides: Z^2 = R^2 + (X_C - X_L)^2 50^2 = 30^2 + (X_C - 60)^2 2500 = 900 + (X_C - 60)^2 (X_C - 60)^2 = 1600 X_C - 60 = 40 , Since the current leads the voltage, the circuit is capacitive, which means X_C > X_L . Therefore, we must take the positive root: X_C - 60 = 40 X_C = 100 , Answer: 100 ,