Final answer:
The quality factor (Q) for a series resonant circuit with a resonant frequency of 6000 Hz and bandwidth of 500 Hz is approximately 1.9. The inductive reactance at resonance is 60000Ω. From this, we can calculate the inductance (L) and capacitance (C) values required for the resonant frequency.
Step-by-step explanation:
The question asks about the quality factor (Q), inductive reactance at resonance, and the inductance and capacitance values for a series resonant circuit with given parameters.
To find the quality factor (Q) of the circuit, use the formula Q = ω₀/(2πΔω), where ω₀ is the resonant frequency and Δω is the bandwidth. From the given resonant frequency of 6000 Hz and a bandwidth of 500 Hz, we calculate Q as:
Q = 6000/(2π×500) = 6000/3140 ≈ 1.9
The inductive reactance at resonance (Xₜ) can be found using the formula Xₜ = ΩR, where R is the resistance and Ω is the resonant frequency. With R given as 10Ω and Ω as 6000 Hz, the inductive reactance is:
Xₜ = 10Ω × 6000 = 60000Ω
To calculate the value of the inductor (L), we use the formula Xₜ = 2πΩL. With the inductive reactance found above, we solve for L to find the inductance of the circuit. Finally, to find the value of the capacitor (C), the resonance condition ΩL = 1/(Ω2πΩC) is used, and C can be solved for algebraically. These calculations result in the inductance and capacitance values needed for the circuit to resonate at the given frequency.