Technically it's AC. The voltage is always positive relative to earth, but since a capacitor is present and the measured voltage waveform across it has both upwards and downwards slopes, the current must have a definite and repetitive zero-crossing and reversal. Therefore the current alternates. It's AC - just not in the traditional sense. The voltage and current waveforms are simply non-sinusoidal and the voltage entirely positive-biased going off of the waveform you provided.

I'll admit algebra and calculus are not my strong suits so someone more mathematically enlightened will have to help you with the formulas - but I do know that capacitance, voltage and frequency are going to be the principle determinants of the circuit's current.

I'm not sure if the capacitor reactance formula applies for non-sine waveforms, but if so, we might be able to cheat using:

Xc=1/(2πFC)
And
I=E/Z

So if we assume:
F=60Hz
C=0.000060F (60uF, a typical motor run capacitor)
E=120Vrms
PF=0 (for the sake of simplifying Xc = Z, i.e. disregarding DC resistance and inductance)

1/(2*3.14159*60*0.000060)=44.2097Ω
120Vrms/44.2097Ω=2.7143A

2.7143A*120V=325.716VAR

So someone correct me if I'm wrong, but you could hypothetically take the RMS voltage of your continuous waveform, the capacitance of your cap and the frequency of the pulses to determine current. With a grain of salt.

 

I remember someone mentioning that you couldn't use the Reactive capacitance (Xc=1/(2πFC)) with pulsed DC as it was only for AC waves but as you mention it is technically AC.

Thanks for the input, we will have to go with the above until we can find a way to measure the current as you mentioned in a previous post.

 

Could you perhaps utilize some sort of hall-effect sensor to measure the current? Build some sort of a crude hall-effect ammeter and calibrate it, then insert it into your circuit?

Something to the tune of this: https://www.keysight.com/en/pc-1659326/oscilloscope-current-probes?cc=US&lc=eng

 

I remember someone mentioning that you couldn't use the Reactive capacitance (Xc=1/(2πFC)) with pulsed DC as it was only for AC waves but as you mention it is technically AC.

Thanks for the input, we will have to go with the above until we can find a way to measure the current as you mentioned in a previous post.

Well here's the thing. A square pulse (idealized, of course) can be decomposed into a sum of sine waves, The capactative reactance applies to all the frequency components contained in a square wave, As you are no doubt aware the reactance is very small for harmonics so they blow right through normal values of capacitance. They do less well with the flat tops of the waveform.

 

Why am I so sure that the forum and the TS are completely talking past each other?

I suspect that we still don't know what he is asking for. For one thing, power makes no sense here, I believe he means energy. But, the way the waveform is drawn, energy flows both directions with a net transfer of zero, i.e. before and after the pulse, the voltage across the capacitor is zero.

Bob

 

For one thing, power makes no sense here, I believe he means energy. But, the way the waveform is drawn, energy flows both directions with a net transfer of zero, i.e. before and after the pulse, the voltage across the capacitor is zero.

Reactive power. VARs, not Watts. No useful work is performed, yet measurable power still exists. Power in an electrical sense is simply the flow of energy, not necessarily the utilization of it.

E.g. A 1HP motor with a nameplate rating of 120V, 12A appears to consume 1440VA of power (apparent power), but only produces 746W of useful work plus 20-40% in thermal losses (true power). The rest is returned to the grid every half-cycle (reactive power).