Timeline for Is the energy loss in a voltage drop equal to heat/mechanical work/light emitted?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 14, 2013 at 20:03 | comment | added | The Photon | If the short has no potential across it, then it drops 0 V, not 100% of the voltage. But in reality there is no perfect short (nor any perfect voltage source)...so if you want to talk about short circuits your model has to include more detail than just perfect voltage sources and perfect wires. | |
| Feb 14, 2013 at 19:15 | history | tweeted | twitter.com/#!/StackElectronix/status/302133950007091200 | ||
| S Feb 14, 2013 at 17:46 | history | suggested | Chetan Bhargava | CC BY-SA 3.0 |
removed unnecessary tail
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| Feb 14, 2013 at 17:29 | review | Suggested edits | |||
| S Feb 14, 2013 at 17:46 | |||||
| S Feb 14, 2013 at 17:16 | history | edited | Phil Frost | CC BY-SA 3.0 |
corrected spelling
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| S Feb 14, 2013 at 17:16 | history | suggested | Jobin T Philip | CC BY-SA 3.0 |
corrected spelling
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| Feb 14, 2013 at 17:13 | answer | added | Phil Frost | timeline score: 3 | |
| Feb 14, 2013 at 17:09 | vote | accept | Vial | ||
| Feb 14, 2013 at 17:00 | review | Suggested edits | |||
| Feb 14, 2013 at 17:16 | |||||
| Feb 14, 2013 at 16:59 | comment | added | Shamtam | The voltage drop in a short circuit is ~100%. A short, by definition, has no potential across it. | |
| Feb 14, 2013 at 16:57 | answer | added | The Photon | timeline score: 9 | |
| Feb 14, 2013 at 16:53 | history | asked | Vial | CC BY-SA 3.0 |