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For a source represented by its Thevenin equivalent (Vth in series with Rth), the load draws the maximum possible power when the load resistance equals the source resistance: RL=Rth. At this matched condition exactly half the source voltage appears across the load, half across Rth, and only 50% efficiency is achieved — the other half is dissipated inside the source itself. This trade-off matters: matched loading maximizes delivered power but wastes half the energy as heat in the source, so it is used for signal/RF matching, not for efficient power distribution.
| Quantity | Formula |
|---|---|
| Optimal load | RL(opt) = Rth |
| Maximum power | Pmax = Vth2/(4Rth) |
| Load current at match | IL = Vth/(2Rth) |
| Load voltage at match | VL = Vth/2 |
| Power for any RL | PL = Vth2RL/(Rth+RL)2 |
| Efficiency for any RL | η = RL/(Rth+RL) |
It states that a source delivers maximum power to a load when the load resistance equals the source's (Thevenin) internal resistance, RL=Rth.
Pmax = Vth2/(4Rth), obtained by setting RL=Rth in the general power formula PL=Vth2RL/(Rth+RL)2.
At RL=Rth, the source voltage splits equally between Rth and RL, so equal power is dissipated in each — only half the total power reaches the load.
No. For efficient bulk power transfer (like a power grid or battery charger) you want RL≫Rth so efficiency approaches 100%, even though delivered power is then below the theoretical maximum.
When maximizing signal or power delivery matters more than efficiency — e.g. RF/antenna systems, audio amplifiers, sensor signal paths, and other low-power signal-transfer applications.
Compute the Thevenin equivalent resistance seen from the load terminals (deactivate independent sources and find the resulting resistance), using our Thevenin Equivalent Circuit Calculator.
Yes, generalized: maximum power transfers when ZL equals the complex conjugate of Zth (ZL=Zth*), which reduces to RL=Rth for purely resistive circuits.
Delivered power decreases from the maximum, but efficiency increases toward 100% since less power is wasted in Rth — the classic power-vs-efficiency trade-off.
Both delivered power and efficiency drop — most of the source voltage is dropped across Rth and little power reaches the load; the extreme case RL=0 delivers zero power (a short circuit).
No — load current is maximum as RL→0 (short circuit, I=Vth/Rth). At the matched point the current is only half that: IL=Vth/(2Rth).
The theorem is usually stated using the Thevenin equivalent (Vth, Rth) of the source; the same RL=RN condition applies identically using the Norton equivalent, since RN=Rth.
Modern audio amplifiers are usually voltage sources with very low output impedance, designed to deliver a wide voltage swing efficiently into a much higher speaker impedance, rather than a strict impedance match — the true matched-power theorem is more relevant to older RF/telephone-style low-impedance sources.
At RL=Rth, the source itself also dissipates Pmax, which can mean significant internal heating if Rth is a physical resistance rather than a modeling abstraction — always check the source's real power rating.
Pmax is inversely proportional to Rth for a fixed Vth — halving Rth doubles the maximum deliverable power, which is why low-impedance sources can supply much more peak power.
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