Timeline for Battery charging a Capacitor
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 5, 2020 at 13:26 | comment | added | Michael Levy | @Alfred Centauri , to round out my initial comment, can you tell me if you think that the following link (en.wikipedia.org/wiki/Radiation_resistance) is what you and (aapt.scitation.org/doi/10.1119/1.1435344) are referring to? If so, perhaps it would be beneficial to include a reference in your answer for the gentle reader. | |
| Feb 4, 2020 at 13:44 | comment | added | Michael Levy | I suppose radiation resistance is a technical term, so there's leeway; but for me, it just seams like the wrong part of speech. Like I said, this is a very minor point to an excellent answer. | |
| Feb 4, 2020 at 1:23 | comment | added | Alfred Centauri | @MichaelLevy, I meant to say radiation resistance. Is that an incorrect use of the term? It's used in a similar context, for example, here: "Contrary to the lumped-parameter description without radiation, the charging/discharging is not instantaneous, but rather is limited by the radiation resistance." | |
| Feb 4, 2020 at 0:34 | comment | added | Michael Levy | Very minor point, but I think in your last sentence you intend to stay "radiative resistance" rather that "radiation resistance." | |
| Sep 21, 2018 at 22:21 | comment | added | Alfred Centauri | @freecharly, the general explanation is just energy conservation isn't it? In solving for the circuit current (not shown), KVL is invoked, e.g., $v_S(t) = v_R(t) + v_C(t)$. So the energy lost to the resistor must be the difference in the energy delivered by the source and the energy stored in the capacitor. | |
| Nov 1, 2017 at 15:08 | vote | accept | Marc Striebeck | ||
| Nov 27, 2016 at 20:11 | comment | added | freecharly | This is a perfect answer to the question. It is, however, puzzling that half the energy is dissipated in the resistor independently of the value of the resistance and only half the energy is stored in the capacitor. Also, the result is the same if you transfer the charge by other means than by the battery+resistor to the capacitor. There should be a general explanation for that. | |
| Aug 28, 2015 at 17:44 | comment | added | Ellie | OP couldn't ask for a clearer answer, well done +1. | |
| Aug 23, 2015 at 23:19 | history | answered | Alfred Centauri | CC BY-SA 3.0 |