Maximum Power Transfer to a Matched Load

Sweep a load resistance from zero to infinity across a source and the power it receives rises, peaks, and falls again. Too small and the voltage collapses; too large and the current does. The peak sits exactly where $R_L = R_{Th}$, and there the load takes $V_{Th}^2/4R_{Th}$. Twenty volts behind 5 Ω, matched by a 5 Ω load: the loop carries 2 A, and the load absorbs $2^2 \times 5 = 20$ W.

Notice what else happened in that example — the source dissipated 20 W too. Matched transfer is 50% efficient, always, by construction. That is the classic misreading of this theorem: it maximises power delivered, not power saved. No utility matches its generators to the grid; a power system deliberately runs with source impedance far below load impedance so efficiency approaches 100% and voltage stays stiff. Matching belongs where the signal matters and the watts do not.

So it lives in RF and audio and instrumentation: antennas into 50 Ω feedlines, transmission lines terminated to stop reflections, a solar panel held at its maximum-power point by a tracking converter. And the AC version wants the load impedance to be the complex conjugate of the source, so a source that looks inductive must be met with capacitance — reactance cancelled first, resistance matched second.