Timeline for Why E = (Vc + Vr) in a circuit with a capacitor and a resistor in series?
Current License: CC BY-SA 3.0
13 events
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| Jun 14, 2017 at 19:56 | comment | added | Glenn W9IQ | @photon - But thanks for cleaning up the answer. | |
| Jun 14, 2017 at 19:53 | comment | added | Glenn W9IQ | @photon - a capacitor and open circuit are electrically equivalent at DC? Oh my! | |
| Jun 14, 2017 at 17:50 | comment | added | Alfred Centauri | @GlennW9IQ wrote "A condition of KVL is a closed loop circuit." - How then do we find the open circuit voltage of a Thevenin equivalent circuit? | |
| Jun 14, 2017 at 3:46 | comment | added | The Photon | and the "low frequency limit" they're talking about is just the limit where the electrostatic potential is well-defined. Which is already part of the lumped frequency approximation that's required for any form of KVL to be valid. | |
| Jun 14, 2017 at 3:20 | comment | added | The Photon | @GlennW9IQ, see here in Wikipedia: "In the low-frequency limit, the voltage drop around any loop is zero. This includes imaginary loops arranged arbitrarily in space – not limited to the loops delineated by the circuit elements and conductors." | |
| Jun 14, 2017 at 3:18 | comment | added | The Photon | @GlennW9IQ, you must choose a closed sequence of nodes (i.e., start and end on the same node). But there don't have to be current paths (circuit branches) connecting the nodes in the sequence you selected. (Again, it's more useful for circuit analysis to choose nodes connected by branches, but it's not a requirement for KVL). | |
| Jun 14, 2017 at 3:16 | comment | added | The Photon | @GlennW9IQ, simply not true. If it was, KVL wouldn't apply to a loop containing a capacitor at DC. | |
| Jun 14, 2017 at 3:08 | comment | added | Glenn W9IQ | Actually not. A condition of KVL is a closed loop circuit. A non conducting spark gap fails that condition. | |
| Jun 14, 2017 at 2:57 | comment | added | The Photon | KVL applies for any round trip choice of nodes in a circuit whether there's components connecting the nodes or not. We just usually choose a sequence of nodes connected by nodes because then the component constitutive equations actually let us solve the circuit. | |
| Jun 14, 2017 at 2:56 | comment | added | The Photon | @GlennW9IQ, even if it's not conducting, KVL says the voltages must add up. Even if U3 is a totally open circuit, KVL says the voltages must add up. (Obviously this is limited to lumped circuits, or KVL itself won't be valid) | |
| Jun 14, 2017 at 2:55 | history | edited | The Photon | CC BY-SA 3.0 |
added 1017 characters in body
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| Jun 14, 2017 at 2:43 | comment | added | Glenn W9IQ | Perhaps a little too much exuberance in the example of a spark gap if it is not actively conducting. | |
| Jun 14, 2017 at 2:09 | history | answered | The Photon | CC BY-SA 3.0 |