An inductance of 2.0 ext H , a capacitance of 18 ext F and a resistance of 10 ext k are connected to an AC source of 20 ext V with an adjustable frequency. Determine the value of the maximum current in the circuit.

Step-by-step solution

The current in a series LCR circuit is given by: I = V Z where the impedance Z is: Z = R^2 + (X_L - X_C )^2 The current in the circuit is maximum when the impedance is minimum. Since the frequency is adjustable, the minimum impedance occurs at the resonant frequency where the inductive reactance equals the capacitive reactance ( X_L = X_C ). At resonance, the impedance is: Z = R Therefore, the maximum current is: I_ max = V R Given: V = 20 V R = 10 k = 10^4 Substituting the values: I_ max = 20 10^4 A I_ max = 2 10⁻