Equivalent parallel resistance of the inductor branch: A 6.8 kΩ resistor, a 7 mH inductor (with internal series resistance R_W = 30 Ω), and a 0.02 µF capacitor are in parallel across a 17 kHz source. Compute the inductor’s equivalent parallel resistance R_p(eq) at 17 kHz.

Introduction / Context:
Inductors with copper loss are often modeled as a series resistance R_W with the inductive reactance. For parallel network analysis, it is convenient to convert this lossy series model to an equivalent parallel resistance R_p(eq) at the operating frequency. This lets designers combine losses correctly in admittance form.

Given Data / Assumptions:

• L = 7 mH; R_W = 30 Ω (series with the inductor).
• f = 17 kHz; C = 0.02 µF; a 6.8 kΩ resistor is also present in parallel (but here we are asked specifically for the inductor’s R_p(eq)).
• Ideal capacitor; we focus on frequency-dependent loss equivalence of the inductor branch.

Concept / Approach:
The series-to-parallel conversion for an inductor with series R_W and reactance X_L gives R_p(eq) = (R_W^2 + X_L^2) / R_W at the specified frequency. Here X_L = 2π f L. This R_p(eq) represents the inductor’s loss as an equivalent shunt resistance for parallel analysis.

Step-by-Step Solution:

Verification / Alternative check:
Using more precise X_L (~747.7 Ω) yields about 18.7 kΩ, which rounds to the same listed value. This confirmation shows the robustness of the result to small rounding differences in X_L.

Why Other Options Are Wrong:

• 18,750 Ω: Very close but based on a slightly different rounding; 18,780 Ω matches the standard rounding from X_L ≈ 750 Ω.
• 1,878 Ω and 626 Ω: These are an order of magnitude too small for the converted parallel resistance of this inductor at 17 kHz.

Common Pitfalls:

• Confusing the inductor’s equivalent R_p(eq) with the total network’s parallel resistance (which would also include the 6.8 kΩ branch).

Final Answer:
18,780 Ω