Maximum power transfer from a capacitive source: What must the load impedance be relative to the source impedance to achieve maximum power transfer?

Introduction / Context:
The maximum power transfer (MPT) condition is crucial in communications, RF design, and audio interfaces. When reactive elements are present, merely matching magnitudes is insufficient; the phase (reactive sign) must be addressed to ensure full power delivery from source to load.

Given Data / Assumptions:

• Source represented by Thevenin or Norton equivalent with complex impedance Zs = Rs + jXs.
• Load impedance ZL is selectable.
• Sinusoidal steady-state operation in the frequency domain.

Concept / Approach:

For AC circuits with reactance, the MPT condition is ZL = Zs* (complex conjugate). This means RL = Rs (equal resistive parts) and XL = −Xs (reactive parts cancel). For a capacitive source (Xs < 0), the load must be inductive with equal magnitude reactance (XL > 0) to neutralize the reactive component at the interface.

Step-by-Step Solution:

Verification / Alternative check:

Differentiate the real power delivered to the load with respect to RL and XL; the optimum occurs at RL = Rs and XL = −Xs. RF matching networks implement this with L- or pi-networks that create the conjugate at the design frequency.

Why Other Options Are Wrong:

'Capacitive reactance equal to circuit resistance' mixes units and ignores the reactive sign. 'Be as capacitive as it is inductive' is meaningless without phase relation. 'None of the above' is false because a well-defined condition exists: conjugate matching.

Common Pitfalls:

Confusing DC (where XL = 0) with AC matching; attempting magnitude-only matching; forgetting that MPT can increase losses in some systems where efficiency, not power, is prioritized.

Final Answer:

have an impedance that is the complex conjugate of the source impedance