A series L-C-R circuit is connected to a variable frequency ac source of voltage amplitude 20 V . The maximum current amplitude in the circuit is 2 A . The angular frequencies at which the current amplitude drops to 1 2 of its maximum value are 95 rad s ⁻¹ and 105 rad s ⁻¹ . The Quality factor ( Q ) of the circuit and the voltage amplitude across the inductor at resonance are respectively:

Step-by-step solution

The resonant angular frequency ₀ is the average of the half-power frequencies: ₀ = 95 + 105 2 = 100 rad s ⁻¹ The bandwidth is the difference between the half-power frequencies: = 105 - 95 = 10 rad s ⁻¹ The Quality factor Q is given by: Q = ₀ = 100 10 = 10 At resonance, the voltage amplitude across the inductor V_L is Q times the source voltage amplitude: V_L = Q V_ source = 10 20 V = 200 V Answer: 10 and 200 V