Timeline for Problem graphing inductor current-time
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 21, 2021 at 9:19 | comment | added | user288518 | see my answer , this is not trivial .......... electronics.stackexchange.com/questions/572096/… | |
| May 9, 2013 at 6:46 | comment | added | JamesHoux | Great question and fantastic answers, but some food for thought for "Cluelessly": If you really want a ball of fun, do some engineering of what happens after the switch opens. :) | |
| May 8, 2013 at 13:56 | vote | accept | Cluelessly | ||
| May 8, 2013 at 4:12 | history | protected | W5VO | ||
| S May 8, 2013 at 3:22 | history | suggested | CommunityBot | CC BY-SA 3.0 |
Missing L in eqn
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| May 8, 2013 at 3:14 | review | Suggested edits | |||
| S May 8, 2013 at 3:22 | |||||
| May 8, 2013 at 2:36 | comment | added | Kaz | Also Arrow Paradox. A flying arrow has a position determined by time. At some time \$t = a\$, it has a definite, fixed position. If it has a definite, fixed position, how can it possibly be moving? You've fallen into an ancient, well-known philosophical trap. | |
| May 8, 2013 at 2:33 | comment | added | Kaz | You could also ask, what allows a stone to roll down a hill? Initially it has no kinetic energy. It does not move. If it does not move, how can it make downward progress? If it does not make downward progress, how can it trade potential energy for kinetic energy? If it cannot gain any kinetic energy. So a stone should not roll down hill. :) | |
| May 8, 2013 at 2:31 | comment | added | Kaz | The back EMF may be 1V at the first instant, but remember that back EMF requires a changing current. A steady current doesn't generate back EMF; don't confuse back EMF with the magnetic field in the inductor, which is steady with steady current. So, if the back EMF is 1V, it stops all current. But if all current stops, then there is no change in current, and so nothing sustains the back EMF. But any collapse in the back EMF allows a trickle of current which props up the EMF again. This chicken and egg situation resolves itself with the calculus (see Alfred Centauri's answer). | |
| May 8, 2013 at 2:05 | review | First posts | |||
| May 8, 2013 at 2:05 | |||||
| May 8, 2013 at 2:00 | answer | added | Alfred Centauri | timeline score: 5 | |
| May 8, 2013 at 1:46 | history | asked | Cluelessly | CC BY-SA 3.0 |