Parallel resonant-style network with losses: A 9 mH inductor (with internal series resistance R_W = 60 Ω) is in parallel with a 0.015 µF capacitor across an 18 kHz AC source. What is the circuit impedance magnitude?

Introduction / Context:Parallel L–C networks near resonance can exhibit large impedances. When the inductor has copper loss (represented by a series R_W), the total impedance is finite and must be computed via branch admittances. This question reinforces complex arithmetic for parallel branches with loss and reactance.

Given Data / Assumptions:

• L = 9 mH, internal series resistance R_W = 60 Ω.
• C = 0.015 µF = 15 × 10^-9 F.
• Frequency f = 18 kHz.
• Ideal capacitor; inductor modeled as R_W in series with jX_L.

Concept / Approach:Compute branch impedances, convert to admittances, sum admittances, then invert to get Z_total. Specifically, Z_L = R_W + jX_L with X_L = 2π f L, and Z_C = −jX_C with X_C = 1/(2π f C). Total Y = 1/Z_L + 1/Z_C and Z_total = 1/Y.

Step-by-Step Solution:

Verification / Alternative check:Limiting insights: If R_W were zero, the L and C would nearly antiresonate, producing a very large impedance. The finite R_W limits the impedance to a moderate kilohm value, consistent with the selected option.

Why Other Options Are Wrong:

• 1,018 Ω: This is essentially the inductor’s reactance magnitude alone, not the network impedance.
• 17,340 Ω: Unrealistically high for the given R_W; would require much lower loss.
• 290 Ω: Far too low; parallel near-resonant L and C should not collapse to a few hundred ohms with these values.

Common Pitfalls:

• Treating the inductor as lossless or forgetting to combine via admittances for parallel networks.

Final Answer:1,734 Ω