Timeline for Do a Thevenin source and its equivalent Norton source always supply equal power?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 1, 2018 at 11:49 | comment | added | Dave Tweed | @Chu: ... and exactly the opposite when the terminals are shorted. | |
| Jun 1, 2018 at 7:17 | comment | added | Chu | Clearly different cases because the Norton model will dissipate power with no load connected (current flows through \$\small R_{N}=\small R_{TH}\$); the Thevenin model will not (source and \$\small R_{TH}\$ are in series, hence open circuit). | |
| May 31, 2018 at 21:41 | comment | added | The Photon | The internal efficiency of the Thevenin and Norton circuits aren't equal, and neither one is likely to match the internal efficiency of whatever real source your Thevenin/Norton circuit is modeling. | |
| May 31, 2018 at 21:34 | comment | added | Dave Tweed | Yes, their internal efficiency is different, but that's a different question altogether. | |
| May 31, 2018 at 21:28 | comment | added | sushanta | So last time I calculated the power generated by the voltage source in Thevenin and the power generated by the voltage source maintaining a constant iNorton. Apparently, although they are "delivering" equal power to the load but are generating different amounts of power. | |
| May 31, 2018 at 20:33 | history | edited | Dave Tweed | CC BY-SA 4.0 |
added 41 characters in body
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| May 31, 2018 at 20:25 | history | answered | Dave Tweed | CC BY-SA 4.0 |