How to use the Maximum Power Transfer Calculator
- 1. Choose DC or ACResistive Rth, or a complex Zth such as 3+j4.
- 2. Enter VthUse RMS, or pick Peak for AC.
- 3. See the best loadPmax, efficiency and how your own load compares.
The maximum power transfer theorem
A source with Thevenin voltage Vth and resistance Rth delivers the most power to a resistive load when RL = Rth. The maximum power is Pmax = Vth²/(4Rth). At that point half the power is lost inside the source, so the efficiency is only 50%. In AC circuits with Zth = R + jX, the best load is the complex conjugate ZL = R − jX. The reactances cancel and P = |Vth|²/(4R) with an RMS voltage.
Worked example: Vth = 12 V and Rth = 4 Ω. The best load is 4 Ω, the current is 12/8 = 1.5 A and Pmax = 144/16 = 9 W. An 8 Ω load gets 144 × 8/144 = 8 W, which is 89% of the maximum at 67% efficiency.
Exam tips
- Find the Thevenin equivalent seen by the load first. The theorem only applies to that.
- If the AC load must be a pure resistor, use RL = |Zth|. It gets less than the conjugate-match power.
- With a peak voltage, P = Vpeak²/(8R).
- Maximum power transfer suits communication and audio circuits. Power systems aim for high efficiency instead, with RL much larger than Rth.
The table shows the load power and efficiency at 0.25, 0.5, 1, 2 and 4 times Rth.
Frequently asked questions
Why is the efficiency only 50% at maximum power?
When RL = Rth, the same current flows through both, so they dissipate equal power. Half of the source power is lost in Rth.
What if the load can only be resistive in an AC circuit?
Choose RL = |Zth|. The tool shows this case next to the conjugate match, so you can compare the powers.
Do I enter RMS or peak voltage?
RMS is the default. Choose Peak if your question gives the amplitude, and the tool uses P = V²/(8R).
Is this tool free?
Yes. You fill in a short form once to see your result, and the other Uncodemy tools then open without it. All the maths runs in your browser.