Maximum power transfer with reactive sources: For maximum power transfer from a capacitive source (negative reactance part), the load impedance must be the complex conjugate of the source impedance. True or false?

Introduction / Context:
In AC circuits with reactive elements, matching for maximum power transfer extends beyond equal resistances. Conjugate matching ensures that reactive parts cancel, optimizing power delivered to the load—crucial in RF, audio power stages, and impedance-matched sensors.

Given Data / Assumptions:

• The Thevenin equivalent seen by the load is Zs = Rs + jXs.
• A “capacitive source” means Xs < 0 (capacitive reactance).
• We can choose a load impedance ZL = RL + jXL.

Concept / Approach:

Maximum power transfer in AC occurs when ZL = Zs* (complex conjugate), i.e., RL = Rs and XL = −Xs. If Xs is negative (capacitive), then XL must be positive (inductive) and equal in magnitude, canceling the net reactance at the interface. This yields a purely resistive combined impedance, maximizing real power flow into the load.

Step-by-Step Solution:

Verification / Alternative check:

Differentiate load power with respect to RL and XL using phasor current I = Vth / (Zs + ZL). The optimum occurs at RL = Rs and XL = −Xs, confirming conjugate matching.

Why Other Options Are Wrong:

• “False” ignores the reactive cancellation requirement, which is standard in AC maximum power transfer derivations.

Common Pitfalls:

Maximizing voltage or current alone does not maximize power; both magnitude and phase must be considered. Also, conjugate matching is about power, not necessarily efficiency, which may require different criteria.

Final Answer:

True