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Electrical
Maximum power transfer from a Thevenin source
Largest power a resistive load can draw from a Thevenin source, reached when the load resistance equals the source resistance.
- Ref
- ELE-33
- Rev
- v1
- Field
- Electrical
- Result
- PL,max [W]
The maximum power transfer theorem says a source with Thevenin voltage Vth and resistance Rth delivers its greatest power to a load of RL = Rth. At that point PL,max = Vth²/(4Rth). The efficiency at maximum power is 50 %: the same power is wasted inside the source. For AC circuits the load impedance must be the complex conjugate of the source impedance.
Largest power a resistive load can draw from a Thevenin source, reached when the load resistance equals the source resistance. At that point half the power is lost in the source, so maximum power is not maximum efficiency.
Calculator
- Load resistance needed, RL = Rth4 Ω
- Current at maximum power, Vth/(2Rth)1.5 A
- Load voltage at maximum power, Vth/26 V
- Short-circuit current, Vth/Rth3 A
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A 12 V source with 4 Ω of internal resistance can deliver at most 9 W, when the load is also 4 Ω; at that point the load takes 6 V and 1.5 A.
- The maximum is reached by matching but the source also dissipates 9 W, so only half of the energy taken from the source does useful work.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (W) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Vth Thevenin (open-circuit) voltage | 1 to 100 V | 0.0625 | 625 | 624.9 | +10% | +21 % |
| Rth Thevenin resistance | 0.1 to 100 Ω | 0.36 | 360 | 359.6 | +10% | −9.09 % |
Sweep of Thevenin resistance
| Thevenin resistance (Ω) | Maximum load power (W) |
|---|---|
| 0.1 | 360 |
| 0.2371 | 151.8 |
| 0.5623 | 64.02 |
| 1.334 | 27 |
| 3.162 | 11.38 |
| 7.499 | 4.801 |
| 17.78 | 2.024 |
| 42.17 | 0.8537 |
| 100 | 0.36 |
Sweep one variable
Variables and how each one contributes
Maximum load power
What the equation gives you. Everything else feeds it.
Thevenin (open-circuit) voltage
The open-circuit voltage of the source, in volts. Raises the maximum power with the square of the voltage. Raising it by 1 V (to 13 V) moves the result by +17.4 %, from 9 to 10.56 W.
Thevenin resistance
The Thevenin resistance of the source, in ohms. Lowers the maximum power in inverse proportion: a source with a higher internal resistance can deliver less current. Raising it by 0.5 Ω (to 4.5 Ω) moves the result by −11.1 %, from 9 to 8 W.
Worked example with the values above
- Square the voltage: Vth² = 12² = 144 V².
- Four times the resistance: 4Rth = 4 × 4 = 16 Ω.
- PL,max = 144 / 16 = 9 W, at a load of 4 Ω.
Assumptions and limits
- A linear circuit that can be replaced by a Thevenin equivalent: a voltage source Vth in series with a resistance Rth.
- Steady DC conditions, or AC with purely resistive impedances.
Common mistakes
- Assuming it applies for maximum efficiency: it gives maximum power only.
- Using it for a source that cannot tolerate the dissipation at matching, such as a battery that would overheat.
Statistics and uncertainty
The peak is flat: a load 20 % away from Rth still delivers 99 % of the maximum power. The values Vth and Rth are found from an open-circuit and a loaded measurement, so errors in the meter affect the maximum.
Where it is used
- Matching radio, audio and sensor circuits to a source.
- Estimating the greatest power a source can deliver to a resistive load.
Maths you need
- Thevenin equivalent: a source and one resistance that replace the rest of the circuit
- Ohm’s law: V = IR