I know when we analyze a circuit's Thevenin equivalent, the best we can ever get is 50% power transfer to the load. I would assume that this holds true for power plants (source) and the rest of the grid (load). I realize this is a fairly simplified view.

Anyway, I'm taking a thermodynamics class in mechanical engineering this summer and we toured a cogeneration plant on campus. I asked about this at the plant, but neither the plant manager nor my thermo professor seemed to know what I was talking about.

This graphic below was produced from data collected by the DOE at LBL. The 62 units lost in the power plant could mean imply the efficiency of a typical power plant, or be half of the current maximum of about 60% efficiency divided in half due to Thevenin. Any way, this is just speculation. I'm hoping a power engineer can weigh in?

Here's a silly graphic.

  • 1
    \$\begingroup\$ "I know when we analyze a circuit's Thevenin equivalent, the best we can ever get is 50% power transfer to the load." Untrue. This comes from matching the source impedance to the load or line impedance (RT = RL), but that is not always desirable. Your 62 units lost in the plan would be your Carnot inefficiency, friction, electrical & magnetic heating, etc. \$\endgroup\$ Aug 5, 2014 at 1:02
  • \$\begingroup\$ Chris, I'm aware of the limits imposed by the carnot cycle. To be more precise, the cycles used in a power plant are combined-cycles which are not quite the same. In real life, plants are happy to see around 30%. 60% is the highest efficiency i've heard claimed by a plant. So let's take our 60% max theoretical carnot from just the heat engine, and hook it up to our tubine to spin the gen. I was merely speculating about the figures. Seems I have temporarily mixed up 'efficiency' with power transfer. \$\endgroup\$
    – RYS
    Aug 5, 2014 at 3:34

2 Answers 2


I know when we analyze a circuit's Thevenin equivalent, the best we can ever get is 50% power transfer to the load.

That's a misunderstanding of the maximum power transfer theorem according to which the maximum power that can be delivered to the load is 50% of the maximum power available from the source.

But that doesn't mean that the best power ratio is 50%.

For a Thevenin voltage \$V_{th}\$, Thevenin resistance \$R_{th}\$ and load resistance \$R_L\$, the power delivered to the load is

$$P_L = \frac{V^2_{th}}{\left(R_{th} + R_L \right)}\frac{R_L}{\left(R_{th} + R_L \right)}$$

which is indeed maximum when \$R_L = R_{th}\$.

However, the power developed by the Thevenin source is

$$P_{th} = \frac{V^2_{th}}{\left(R_{th} + R_L \right)}$$

and so the fraction of the source power delivered to the load is thus

$$\frac{P_L}{P_{th}} = \frac{R_L}{\left(R_{th} + R_L \right)}$$

So, for \$R_L >> R_{th}\$, almost all of the power supplied by the source is delivered to the load (though this power is much less than the maximum power available from the source).

I don't have the expertise to address the reasons for 62 units of energy lost in the power plant so this answer is just to address the proper interpretation of the maximum power transfer theorem.

  • \$\begingroup\$ The lost 62 units were addressed by Chris Stratton's comment: heat engine losses. \$\endgroup\$ Aug 5, 2014 at 1:11
  • \$\begingroup\$ whatsisname, power plants don't run on carnot. Anyway, thanks for this post Alfred. This is the proof that was coming to mind as I toured the plant today. I mistakenly thought that power transfer was one to one with efficiency! \$\endgroup\$
    – RYS
    Aug 5, 2014 at 3:37

No not at all. For one, the Thevenin model is inappropriate. Thevenin's theorem applies only to purely resistive loads. The power grid has a great deal of reactive elements to it.

Secondly, your claim about 50% power delivery is untrue. You may be thinking of the Maximum power transfer theorem, which states that a source can deliver max power when the resistance of the load matches the internal resistance of the source. But that doesn't mean the power station is at half efficiency.

If you hook a bunch of light bulbs up in parallel, the resistance seen by their power supply reduces. If you hook them up in series, it increases. Therefore, you could wire up lightbulbs with any number you wanted, and increase it until eventually the power supply can't deliver enough watts, while still matching its internal resistance.

  • \$\begingroup\$ Not saying that efficiency and power transfer are the same at all. I'm asking: on top of the theoretical limits of a gas, vapor or combined cycle power plant, does thevenein come into play at all? As I stated, I know that this is a simplified model, and that pains are taken to also match the reactance. Seems like I made the same mistake as Joule - from wikipedia "The theorem was originally misunderstood (notably by Joule) to imply that a system consisting of an electric motor driven by a battery could not be more than 50% efficient..." \$\endgroup\$
    – RYS
    Aug 5, 2014 at 3:31
  • \$\begingroup\$ So, seeing that efficiency is a function of the load at some point, are there any estimates as to what this efficiency loss might be? \$\endgroup\$
    – RYS
    Aug 5, 2014 at 3:36

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