Difficulty: Easy
Correct Answer: True
Explanation:
Introduction / Context:Maximum power transfer is a cornerstone in RF, audio, and power electronics. In AC systems with reactance, matching is not just about magnitudes; the phase (reactive part) must be addressed via conjugation. This question confirms the complex-conjugate matching condition for peak power delivery.
Given Data / Assumptions:
Concept / Approach:
The average power to the load is maximized when ZL = Zs* (the complex conjugate of Zs). This cancels reactive parts (XL = −Xs) and matches resistive parts (RL = Rs), maximizing the magnitude of load voltage/current product that contributes to real power without reactive circulation.
Step-by-Step Solution:
Let Zs = Rs + jXs and ZL = RL + jXL.Power to the load P = |Vth|^2 * RL / |Zs + ZL|^2 (Thevenin form).To maximize P with respect to RL and XL, set XL = −Xs (reactive cancellation).Differentiate P with respect to RL; optimum occurs at RL = Rs.Thus ZL,opt = Rs − jXs = Zs* (complex conjugate).Verification / Alternative check:
At the match: input seen by source is purely resistive; voltage standing wave ratio is minimized in transmission-line terms, confirming peak power transfer.
Why Other Options Are Wrong:
“False” and “must equal exactly (not conjugate)” ignore reactive cancellation. “True only for resistive sources” is unnecessary; the conjugate form explicitly covers reactive sources.
Common Pitfalls:
Matching magnitudes only; forgetting that reactive parts must cancel to avoid circulating reactive power and suboptimal real power absorption.
Final Answer:
True.
Discussion & Comments