Difficulty: Medium
Correct Answer: 1,734 Ω
Explanation:
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:
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:
Common Pitfalls:
Final Answer:
1,734 Ω
Discussion & Comments