How to derive the maximum power transfer theorem using the relation p=v²/R?

Question

In most derivations of the Maximum power transfer theorem the derivation is done using the relation of power :

P = i²R And on further derivation it comes out to be Pmax = Vth²/4Rth

If I derive the power from the relation p= vth²/Rth

It's clearly wrong My logic is as follows : The voltage across the load resistance is Thevenin voltage Vth , and it's resistance is RL but for Maximum power to be transferred, RL must be equal to Rth therefore the formula. Can anyone explain what am I missing here? (Actually I derived it intuitively in the exam, it turned out to be a blunder).

Answer 1

My logic is as follows : The voltage across the load resistance is Thevenin voltage Vth

Your logic is flawed...

• \$V_{TH}\$ is the equivalent open circuit voltage before you add any load.
• When you add a load equivalent to the Thevenin resistance of the source, the terminal voltage across that load drops to half the original Thevenin voltage --> \$V_{TH}\$/2
• Hence squaring \$V_{TH}\$/2 produces \$V_{TH}^2\$/4