0
\$\begingroup\$

With regards to the Maximum Power Theorem (Princeton.edu);

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 since, when the impedances were matched, the power lost as heat in the battery would always be equal to the power delivered to the motor. In 1880 this assumption was shown to be false by either Edison or his colleague Francis Robbins Upton, who realized that maximum efficiency was not the same as maximum power transfer. To achieve maximum efficiency, the resistance of the source (whether a battery or a dynamo) could be made close to zero.

I don't know if I am making the same "mistake" that Joule made, however I still don't see how at maximum power transfer, the efficiency is anything but 50%. At maximum power transfer, the source impedance = load impedance, hence equal power dissipated in the source and the load. Am I missing something here?

\$\endgroup\$
5
\$\begingroup\$

The text doesn't deny that the efficiency is 50% at maximum power transfer conditions. It is stating that maximum power transfer conditions are not the same as maximum efficiency conditions, so there may be a different point (namely, when Rs is much less than RL) where efficiency is higher but power transfer is lower.

See Wikipedia for more details: Maximizing power transfer versus power efficiency

\$\endgroup\$
  • \$\begingroup\$ Ok. But I am still unsure how can the source impedance of something be changed? Especially something like a dynamo whose source impedance is largely the winding resistance (I believe)? \$\endgroup\$ – midnightBlue Jan 27 '15 at 19:46
  • \$\begingroup\$ In general the source impedance can't be changed. It's only done on the fly in some specialized cases (e.g., en.wikipedia.org/wiki/Maximum_power_point_tracking) \$\endgroup\$ – Justin Jan 27 '15 at 19:51
  • \$\begingroup\$ @midnightBlue With AC one can change the relative impedance of the source or load with a transformer. \$\endgroup\$ – George Herold Jan 27 '15 at 21:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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