In a series L-C-R circuit, the resonant angular frequency is 500 rad s ⁻¹ . The difference between the two angular frequencies at which the current amplitude is 1 2 of its maximum value is 20 rad s ⁻¹ . If the capacitance in the circuit is 40 F , what are the values of inductance and resistance respectively?

Step-by-step solution

The resonant angular frequency is given by ₀ = 1 LC . Squaring both sides: ₀^2 = 1 LC Rearranging for L : L = 1 ₀^2 C = 1 (500)^2 40 10⁻⁶ L = 1 250000 40 10⁻⁶ = 1 10 = 0.1 H The difference between the half-power frequencies is the bandwidth, which is given by R L . Bandwidth = R L = 20 rad s ⁻¹ Substituting the value of L : R 0.1 = 20 R = 20 0.1 = 2 . Answer: 0.1 H and 2