Your derivation is correct. It also match the energy "law" of a capacitor, E=0.5CV^2.

But you are saying that you get the same answer with the resistor. In earlier post you say you get CV^2, if the voltage is a ramp.

I think you have mixed up the energy provided from the supply when it applies a step voltage into the RC network. In that case the energi provided will be CV^2 and independed of R. You can not use this relationship with a ramp with zero transition time. Then you can not solve the i=C*dv/dt relationship of the capacitor and we are back where we started with the two capacitors.

as for the original post - the energy remains the same - remember that 2 variables are changings - voltage and capacitance.

As for the Ideal voltage source and ideal capacitor - sure why not? Just need infinite current for a brief period of time.

BTW - if you can give me a true ideal voltage source, we can solve the worlds energy problems.

Rakesh wrote on Feb 5^th, 2011, 2:36pm:


The equations you are starting with is ok for me since you include the resistor. If you solving the differential system, please not forget that the voltage ramp is flatting out which influence the current.

You claiming that the answer will not change. I do not understand which answer will not change? And you can not say anything about the answer unless you have sokved the differential system and have an expression of i(t) and the integrate v(t)*i(t).

Yes, I got the same equation when trying to solve the ramp problem with a resistor. It is valid up to t=T. I did not proceed, I'm to lazy I guess

But you are on the right track. After t=T the voltage is constant and there is a new differential equation to solve, with the initial condition set by the first equation.