I have been studying RC circuits and noticed that even in a resistanceless circuit, when a capacitor is charged by a constant potential difference source, there is considerable heat generation, equal to the amount of energy stored in the capacitor, making the work done by the battery twice of the energy stored. Whereas, in the limiting case of an RC-circuit, as R tends to zero, the heat generated is finite and independant of R (which can be found by integration on the arbitrary current functions or otherwise), I was wondering if the heat generated could be explained as a result of the sudden change of kinematic state of the charges. Due to resistanceless wires, they gain quite a bit of kinetic energy as they rush towards the metallic plates of the capacitor and instantly come to rest due to the work function (I suppose). This seems to be an inherently elastic process and I would like to know if this idea can in fact be validated by suitable equations taking into account the kinetic energy of the charges and the force on them by the battery while in the external circuit.
1 Answer
...noticed that even in a resistance less circuit, when a capacitor is charged by a constant potential difference source, there is considerable heat generation...
This is not a correct statement.
You have seen derived the energy stored in a capacitor and the energy dissipated in any resistance in the circuit and the sum of these two being the energy supplied by the battery.
...in the limiting case of an RC-circuit, as R tends to zero, the heat generated is finite and independent of R...
The independence of $R$ is generally true. If the resistance is high, then the "significant" charging current is small, but that significant charging takes place over a longer period of time. If the resistance is low, then the "significant" charging current is large, but that significant charging takes place over a shorter period of time. Overall, the integral $\int I^2R\,dt$ stays the same.
Before you consider the resistance tending to zero, you should consider what happens when the resistance is very low. In this case, the inductance associated with the circuit (after all, it is a loop) becomes significant, and you now have to consider an underdamped LCR circuit with accompanying oscillation of the voltage, current and charge before reaching the final steady-state condition of the capacitor being fully charged and the current in the circuit being zero.
Although generally not significant from the point of view of energy loss from the circuit, that oscillation of the charges can become significant when the resistance is low because of the emission of electromagnetic radiation by the circuit. Unbound charged particles (mobile electrons in this case) that are accelerating emit electromagnetic radiation, corresponding to a loss of energy from the system. You could imagine this being the reason for the loss of energy from the (L)CR charging circuit when the circuit has no resistance.
Thus, the motion of the mobile charges is involved but not quite in the way that you explained it.
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$\begingroup$ That really clears up most of my misconception. Is it possible to support that idea quantitaively, with the equations and everything? $\endgroup$ Commented Aug 11, 2020 at 5:17
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$\begingroup$ Which equations are you asking for? $\endgroup$– FarcherCommented Aug 11, 2020 at 6:23
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$\begingroup$ Equations quantifying the energy carried away in EM radiation starting from the wave theory and the oscillations of the charges. (Assuming such equations can in fact be written.) $\endgroup$ Commented Aug 11, 2020 at 9:19
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$\begingroup$ A capacitor paradox is a paper which describes a near equivalent situation. $\endgroup$– FarcherCommented Aug 13, 2020 at 18:16
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$\begingroup$ I checked it out. Thanks for the search. Although my other teachers agreed with the idea of the energy being carried away in radiation, they also mentioned some of it going into the system's mechanical strain (Just thought you should know, or confirm?). $\endgroup$ Commented Aug 14, 2020 at 3:40