A series circuit resonates at 6 kHz. If at resonance the inductor reactance and capacitor reactance are each 4 kΩ in magnitude and the series resistance is R = 50 Ω, what is the quality factor Q of the circuit?

Introduction:
The quality factor Q characterizes the sharpness of resonance and the ratio of stored to dissipated energy per cycle. For a series RLC circuit, Q directly relates reactive impedance to resistance at the resonant frequency. This question checks the simple but crucial Q = X / R relationship at resonance.

Given Data / Assumptions:

• Series RLC at resonance f0 = 6 kHz (frequency value not needed for Q if X is given).
• Inductive and capacitive reactance magnitudes: XL = XC = 4 kΩ.
• Series resistance: R = 50 Ω.
• Idealized small-signal behavior.

Concept / Approach:
For series resonance, Q = XL / R = XC / R evaluated at f0. Because XL and XC are equal in magnitude at resonance, either may be used. This definition reflects how large the circulating reactive currents are compared to the resistive drop.

Step-by-Step Solution:
1) Use Q = X / R at resonance.2) Substitute X = 4000 Ω and R = 50 Ω.3) Compute Q = 4000 / 50 = 80.

Verification / Alternative check:
Bandwidth relation for series RLC: Q = f0 / BW. A Q of 80 implies BW ≈ 6000 / 80 = 75 Hz, consistent with a sharp resonance for a relatively small R compared to reactance.

Why Other Options Are Wrong:
0.001: Orders of magnitude too small; contradicts X ≫ R.50: Would imply X ≈ 2500 Ω with R = 50 Ω, not given here.4.0: Would require X ≈ 200 Ω at resonance, not 4000 Ω.

Common Pitfalls:
Mixing parallel and series definitions of Q, or mistakenly using total |Z| instead of the reactive magnitude at resonance.

Final Answer:
80