1
\$\begingroup\$

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).

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

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
\$\endgroup\$
1
  • \$\begingroup\$ Ah I get it, since the load resistance is equal to the thevenin resistance, the voltage drops half of it across the load and that squared gives the result. Thanks \$\endgroup\$ Oct 13, 2023 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.