An alternating voltage source is connected across a series combination of a 0.5 H inductor, a 2 F capacitor, and a 50 resistor. At what frequency of the AC source will the current amplitude in the circuit be maximum?

Step-by-step solution

The current amplitude in a series LCR circuit is maximum at the resonant frequency. The resonant frequency is given by f = 1 2 LC Given L = 0.5 H and C = 2 F = 2 10⁻⁶ F Substituting the values: f = 1 2 0.5 2 10⁻⁶ f = 1 2 10⁻⁶ f = 1 2 10⁻³ f = 1000 2 = 500 Hz Answer: 500 Hz