Timeline for What happens in circuits where the propagation time of the electric field is significant?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Nov 16, 2015 at 4:16 | history | bounty ended | CommunityBot | ||
| S Nov 16, 2015 at 4:16 | history | notice removed | CommunityBot | ||
| Nov 10, 2015 at 19:51 | comment | added | user93237 | What you need to do is find a good book on transmission line theory. At the time scales you're talking about, distributed capacitances and inductances become important. What happens after you close the switch is that a voltage and current pulse are sent through the wire and there is a lot of ringing and reflections of pulses back and forth until a steady-state situation equivalent to a conventional DC analysis solution of the circuit is reached. | |
| Nov 10, 2015 at 7:10 | comment | added | Keith McClary | Essentially you're asking "What if there is an electric current that does not create a magnetic field (with its concomitant energy)". | |
| Nov 9, 2015 at 15:24 | history | tweeted | twitter.com/StackPhysics/status/663738865379901441 | ||
| Nov 9, 2015 at 15:08 | answer | added | Floris | timeline score: 3 | |
| Nov 9, 2015 at 1:49 | comment | added | Scott Lawson | @DanielSank If what you are saying is true, the Telegrapher's equations would be invalid for steady state DC current, where the series inductances and parallel capacitances contribute nothing to the transmission. However, we know that the Telegrapher's equations remain valid when the series inductances and parallel capacitances are set to zero. If L=C=0 you would not realistically model a practical transmission line that does have parasitic elements. Fortunately, my original question is a thought experiment and not a practical system. | |
| Nov 9, 2015 at 0:59 | comment | added | DanielSank | The propagation of electromagnetic waves depends fundamentally on the same thing that causes parasitic capacitance and inductance. When you put a high frequency electrical signal on a wire, you get wave-like transmission, and that wave motion depends exactly on the inductance and capacitance per length of the wire. Energizer777's answer links to a Wikipedia article all about this. | |
| Nov 9, 2015 at 0:34 | comment | added | Scott Lawson | @DanielSank Could you please explain why that would make this question unanswerable? | |
| Nov 8, 2015 at 6:58 | comment | added | DanielSank | Assuming there's no parasitic inductance or capacitance makes this question un-answerable. | |
| Nov 8, 2015 at 3:23 | answer | added | Energizer777 | timeline score: 0 | |
| S Nov 8, 2015 at 2:51 | history | bounty started | Scott Lawson | ||
| S Nov 8, 2015 at 2:51 | history | notice added | Scott Lawson | Draw attention | |
| Oct 28, 2015 at 20:03 | comment | added | Scott Lawson | @brucesmitherson Because if the answer is the same then it would violate causality. It takes 1s for the information from S1 to reach D1, and 2s for the information from S2 to reach D1. Thus, D1 could not turn at t=1s in both cases. | |
| Oct 28, 2015 at 19:13 | history | asked | Scott Lawson | CC BY-SA 3.0 |