QuoteI said that circuit should be used to study some interesting effects.

Which ones, for example? For me everything about electricity is interesting - ("magnets, how they do work?")

Quote from: evostars on April 06, 2017, 05:44:38 PM
but Im interested in the power that puts the charge in the caps. its not linear,  at first there is a sudden great rise of voltage,  and then when it reaches its maximum voltage,  the voltage rise slows down.
so the current seems large at the start of charging,  when the voltage is still low.

Yes Evostars, the bogeyman is going to speak to you again.  I am going to make good sense, so the question for you is are you going to have enough character to respond to me?  Am I allowed to eat at the same lunch counter as you?

The power going into the capacitor is the product of the voltage times the current.  Both are functions of time.

p(t) = i(t) * v(t).

And of course the energy is the summation of the power over the charging cycle.  We can also say that the energy is the integral of the current times the voltage with respect to time from the start to the end of the cycle.

Let's go back to p(t) = i(t) * v(t).

We are charging a capacitor, and we know that the voltage on a capacitor v = q/C.

And we also know that q(t) is the integral of i(t) with respect to time.  So we can do some substitutions:

p(t) = i(t) * [integral i(t)dt/C]

And there you go.  If you have a DSO you can record the current waveform and the capacitor voltage waveform and have the DSO do the multiplication for you.

Alternatively you can record the current waveform only and then put the file on a USB drive.   Then you can load the current waveform into a spreadsheet and do a multiply-accumulate function and get your answer like that.  This is the more "profound" way to do this because as long as you know the current waveform and the capacitance you have enough information to get your answer.  It's like Spice running through your veins.

MileHigh

@evostars,

Discharging a capacitor is like decanting water from a five gallon jug; Slow starting, followed by a strong gush  at .67, tapering off to a slow flow: Charging exactly the reverse, max charge rate at .33 capacity:

No, the charge rate shown in your graph is the slope of the line. The maximum charge rate (steepest slope) is at zero capacity and slows down from there. The maximum discharge rate is at full capacity.

Perhaps you should review your calculus.

Milehigh and synchro one,
thank you for your input on the capacitor!
No DCO for me,  but I think i might be able to measure it with a larger capacitance, needing more time to charge. so i can make a graph.

and that picture of the discharge really makes sense.

thanks for the spice

and yes Tinselkoala,  its all about the steepness of the line, but I need some voltage change to get current. So 0.67 seems perfect