Quality factor to reactance: If a coil has a quality factor Q = 60 and a winding resistance of 5 Ω at a given frequency, what is its inductive reactance at that frequency?

Introduction / Context:
The quality factor Q of an inductor quantifies how reactive energy storage compares to resistive loss at a particular frequency. It is frequently used in RF design, filters, and resonant circuits to predict selectivity and loss. Converting Q and winding resistance to reactance is a common quick calculation.

Given Data / Assumptions:

• Coil winding resistance R = 5 Ω (series model).
• Quality factor Q = XL / R = 60 at the operating frequency.
• We seek the inductive reactance XL at that frequency.

Concept / Approach:
For a series R–L model at a specific frequency, Q = XL / R. Rearranging gives XL = Q * R. This relation holds at the specified frequency because Q is frequency-dependent via XL = 2 * pi * f * L, but R is typically considered the effective series resistance at that same frequency.

Step-by-Step Solution:
Use Q definition: Q = XL / R.Solve for XL: XL = Q * R.Substitute values: XL = 60 * 5 Ω = 300 Ω.Therefore, the inductive reactance is 300 Ω.

Verification / Alternative check:
If L were known, XL = 2 * pi * f * L could be equated to 300 Ω to back-calculate the operating frequency or inductance—consistent with the same Q value.

Why Other Options Are Wrong:
0.083 Ω: reciprocal-style error; not from Q * R.12 Ω or 30 Ω: result from incorrect multiplication or mixing R and XL.

Common Pitfalls:
Confusing series Q with parallel Q or using DC resistance instead of effective AC resistance. Always use values specified for the operating frequency.

Final Answer:
300 Ω