A series LCR circuit containing a resistor of 5 , an inductor of 0.2 H and a capacitor of 20 F is connected to a variable-frequency 200 V (rms) AC source. The frequency of the source is tuned until the current in the circuit is in phase with the applied voltage. The rms voltage across the inductor is x 10^2 V . The value of x is _________.

Step-by-step solution

The current is in phase with the applied voltage, which means the circuit is at resonance. At resonance, the angular frequency ₀ is given by: ₀ = 1 LC = 1 0.2 20 10⁻⁶ = 1 4 10⁻⁶ = 1000 2 = 500 rad/s The inductive reactance is: X_L = ₀ L = 500 0.2 = 100 At resonance, the impedance of the circuit is equal to its resistance ( Z = R = 5 ). The rms current in the circuit is: I_ rms = V_ rms R = 200 5 = 40 A The rms voltage across the inductor is: V_L = I_ rms X_L = 40 100 = 4000 V This can be written as 40 10^2 V . Thus