Timeline for Modelling a capacitor whose dielectric has resistance as a circuit element
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 28, 2017 at 19:02 | vote | accept | Yashas | ||
| Feb 28, 2017 at 18:48 | comment | added | Olin Lathrop | @Yash: ESR models some real resistance in the leads and internal wiring of the capacitor. It also models dielectric effects that look like resistance to the outside circuit. That's why it's called the equivalent series resistance. | |
| Feb 28, 2017 at 18:29 | comment | added | Yashas | I derived a model from the fundamental laws of physics. It tells that the capacitor is exactly equivalent to an ideal capacitor with a resistor $r$ connected in parallel. For some reason, my derived model does not recommend the ESR resistor. What is the purpose of the ESR resistor? I have ignored inductive effects in my derivation. Does ESR has anything to do with compensating inductive effects? | |
| Feb 28, 2017 at 16:07 | comment | added | Olin Lathrop | @Yash: No, as I even stated explicitly in the third paragraph, the model with ideal capacitor and parallel resistor will draw a little current due to applied steady state voltage. | |
| Feb 28, 2017 at 16:05 | history | edited | Olin Lathrop | CC BY-SA 3.0 |
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| Feb 28, 2017 at 16:05 | comment | added | Yashas | I think my argument is wrong. The capacitor might indeed settle at a final voltage. The current which flows through the capacitor might be due to the resistance only. At the end, there would be conduction current through the capacitor but zero displacement current. | |
| Feb 28, 2017 at 16:04 | history | edited | Olin Lathrop | CC BY-SA 3.0 |
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| Feb 28, 2017 at 16:03 | comment | added | Yashas | If you use your model to represent the original capacitor, after sufficiently long time, the capacitor will act like an open circuit? It won't allow any current to flow through its branch. If you had the original capacitor in place instead, the charges from the plates of the capacitor could flow through the capacitor and recombine. There would be a current through the capacitor. The model won't agree with the actual case. | |
| Feb 28, 2017 at 16:00 | comment | added | Olin Lathrop | @Yash: I never said what you're quoting, and it doesn't make sense anyway. It is unclear what you are asking. | |
| Feb 28, 2017 at 15:58 | history | edited | Olin Lathrop | CC BY-SA 3.0 |
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| Feb 28, 2017 at 14:06 | comment | added | Yashas | "After a sufficiently long time, there would be a constant current in the circuit as the capacitor behaves like an open circuit. There won't be any current passing through the capacitor branch. But for same reasons mentioned in the previous case, the potential difference across the capacitor will never reach a specific value and pretend to be an open circuit." How would you go about this issue? | |
| Feb 28, 2017 at 13:53 | history | answered | Olin Lathrop | CC BY-SA 3.0 |