Re: [EE]: considering charging capacitor losses. When a charged cap charges a discharged cap.
Dave Tweed <[email protected]> Thu, 2 Apr 2026 7:54:34 -0400
| Newsgroups | gmane.comp.hardware.microcontrollers.pic |
|---|---|
| Organization | Embedded Computer Engineering |
| Message-ID | <[email protected]> |
Justin Richards wrote: > When I was at TAFE (not sure what the equiv but its below Uni), a lecturer > made an off hand comment about when charging one capacitor from another > that the math does not work out and may have continued to say only half > the energy remains. That statement is based only on initial and final equilibrium states, given that charge is conserved and ends up distributed over twice the capacitance. Since this divides the voltage in two and total energy is proportional to voltage squared, half of the energy "disappears". That's all based on ideal components in a lumped-component model. The problem is that this ignores some real issues with how you get from one state to the other. When you connect two capacitors that have different voltages in parallel, it produces an infinitely high and infinitely narrow pulse of current (whose area is the net charge transferred) that the math of lumped-component models can't handle. In the real world, the capacitors and the connections between them are not infinitesmal in size. The transition from infinite resistance between them to zero resistance must take some amount of time and involve intermediate values of resistance. So, there are a number of ways to approach this. You can assume some small value of resistance in the circuit, in which case, the current pulse has finite values for magnitude and time, and you can easily show that the energy the resistor dissipates as heat is equal to half of the original energy in the system. On the other hand, if you want to assume zero resistance, but take into account the inductance of the connection, you once again avoid the infinities, but now the system never reaches a final equilibrium state -- it just oscillates forever. However, if you also consider the non-zero loop area of the inductance, then you'll find that the oscillations slowly radiate energy out of the system as electromagnetic waves. The system gradually approaches equilibrium after radiating half of the original energy away. -- Dave