In a series LCR circuit, the resistance is 8 , the inductive reactance is 10 , and the net impedance of the circuit is 10 . If the alternating current in the circuit leads the applied voltage, what is the value of the capacitive reactance?

Step-by-step solution

The net impedance Z of a series LCR circuit is given by: Z = R^2 + (X_C - X_L)^2 Given Z = 10 , R = 8 , and X_L = 10 , we can substitute these values into the equation: 10 = 8^2 + (X_C - 10)^2 Squaring both sides: 100 = 64 + (X_C - 10)^2 (X_C - 10)^2 = 36 X_C - 10 = 6 This gives two possible values for X_C : X_C = 10 + 6 = 16 or X_C = 10 - 6 = 4 The problem states that the alternating current leads the applied voltage, which means the circuit is capacitive in nature. For a capacitive circuit, the capacitive reactan