Circuit Network Theorem Power Transfer

Maximum Power Transfer Theorem: Choose the Load That Takes the Most Power

Author: Abloom Instruction Date: November 10, 2026

Every real source has some internal resistance — a battery, an amplifier output, or a whole network reduced to its Thévenin equivalent. If you connect a load to it, how much power the load gets depends on the load's resistance. Too small a load and most of the power burns inside the source; too large a load and hardly any current flows at all. Somewhere in between, the load takes the most power possible. The maximum power transfer theorem tells you exactly where that point is.

Thevenin source of 12 volts with 6 ohm internal resistance feeding a variable load resistor, beside a graph of load power against load resistance peaking at 6 watts when the load is 6 ohms, with an efficiency curve rising to 50 percent at that point

Figure 1: A source with internal resistance Rth feeding a variable load RL (a), and the load power and efficiency plotted against RL (b). The power peaks exactly where RL = Rth — but the efficiency there is only 50%. (Schematic, not to scale.)

The theorem

For a source with open-circuit voltage VS and internal (Thévenin) resistance Rth, the load receives maximum power when the load resistance equals the source resistance:

RL = Rth

At that point the load current is IL = VS / (2 × Rth), the load voltage is half the source voltage, and the maximum power is:

Pmax = VS² / (4 × Rth)

Why? The load power is PL = IL² × RL = VS² × RL / (Rth + RL)². If you differentiate that with respect to RL and set the result to zero, the maximum falls out at RL = Rth. The same result holds for AC circuits, with resistances replaced by impedances: the load impedance must be the complex conjugate of the source impedance.

Worked example — a 12 V source with 6 Ω inside

Take a source with VS = 12 V and Rth = 6 Ω, and try several loads.

Step 1 — the matched load: RL = Rth = 6 Ω. The current is IL = 12 / (6 + 6) = 1 A. The load power is PL = 1² × 6 = 6 W. Check with the formula: Pmax = 12² / (4 × 6) = 144 / 24 = 6 W ✓. The load voltage is 1 × 6 = 6 V — exactly half the source voltage. The other 6 W burns in Rth, so the total drawn from the source is 12 W and the efficiency is 6 / 12 = 50%.

Step 2 — a smaller load, 3 Ω: IL = 12 / (6 + 3) = 1.333 A. The load takes PL = 1.333² × 3 = 5.33 W — less than the maximum, even though the current is higher. The internal resistance burns 1.333² × 6 = 10.67 W, so the efficiency is only 5.33 / 16 = 33.3%.

Step 3 — a larger load, 12 Ω: IL = 12 / (6 + 12) = 0.667 A. The load takes PL = 0.667² × 12 = 5.33 W — exactly the same as the 3 Ω load. Halving or doubling the matched resistance costs the same amount of power; the curve is symmetric in that sense. But the efficiency is better: Rth burns only 0.667² × 6 = 2.67 W, so the efficiency is 5.33 / 8 = 66.7%.

Step 4 — the extremes: With RL = 1 Ω the current is 1.714 A but the load takes only 1.714² × 1 = 2.94 W (efficiency 14.3%). With RL = 24 Ω the current is only 0.4 A and the load takes 0.4² × 24 = 3.84 W — but the efficiency is 24 / 30 = 80%. As RL grows, the efficiency keeps rising toward 100% while the power falls toward zero.

Maximum power is not maximum efficiency

That is the catch, and it decides when the theorem matters. At maximum power transfer the efficiency is always exactly 50% — half the energy is wasted in the source. So:

Use matching where the signal or the power delivered matters most: audio amplifiers driving speakers (an 8 Ω speaker on an 8 Ω output), radio transmitters driving antennas, and sensor circuits where you want the largest possible reading. Use a much larger load where efficiency matters most: power distribution, battery-powered equipment, and anything where wasted heat costs money or battery life.

To apply the theorem to a real network, first reduce it to its Thévenin equivalent — find Vth and Rth exactly as in the Thévenin post — then set RL = Rth.

Key points to remember

  • Maximum power transfers to the load when RL = Rth, and then Pmax = VS² / (4 × Rth).
  • Worked example: 12 V source, Rth = 6 Ω → matched load 6 Ω, IL = 1 A, PL = 6 W, load voltage 6 V.
  • Halving or doubling the matched load gives the same power: 3 Ω and 12 Ω both give 5.33 W here — but the 12 Ω load is twice as efficient (66.7% vs 33.3%).
  • At maximum power the efficiency is always 50% — match for power in audio and radio circuits, but choose a larger load where efficiency matters.
  • For AC, match impedances: the load impedance must be the complex conjugate of the source impedance.