# Capacitor discharge math discrepancy

I'm trying to cross check mathematically the mA being discharged from a .33uF capacitor which is charged to about 1200V by a circuit which is provided 37W of power to it. I do it three ways.

1) The circuit draws 35W total. 13V @ 5A = 65W. But the power supply drops the 13V to 7V to maintain the 5A so the effective wattage is 7V x 5A = 35W. I just assumed this 35W carries through the entire circuit and out T2 because energy cannot be created and destroyed and there is little to no resistance from what I can see. So the wattage coming out of the capacitor is still roughly < 35W. We should use Vavg I think, not Vpeak, so 35W / 600Vavg = 58mA.

2) Now (1) makes sense, but I was hoping there was a cross check. I noticed on a page like this: http://cnx.org/content/m42427/latest/?collection=col11406/latest (see Example 2) you can calculate the mA from a capacitor via it's reactance. X = 1 / (2 x Pi x freq x C). As shown there I do the same. I put the gas discharge tube on the scope $X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{6.28318 × 42 × 330\cdot 10^{-9}} = 11483 \Omega$

720Vrms / 11483 = 0.06270A or 62mA Great, this works. Essentially the same as (1). So (1) and (2) validate each other. This also means the wattage then is 720Vrms x .0627 = 45W. 45W is a little too high and not valid I think. After all Vrms is for AC, I have changing DC. So if we used Vavg its 600/11483 = 52mA, 52mA x 600Vavg = 31W. This makes more sense and probably also accounts for the wattage loss from stepping up the voltage @ T1. (35W - 31W = 4W loss).

3) But then I found E = (V² x C) / 2 which is used to calculate the energy per pulse from a capacitor discharge. Also conveniently calculated at: http://www.vishay.com/resistors/pulse-energy-calculator/ E = (1200Vpeak² x 0.00000033) / 2 E = .2376 Joules per pulse Joules per second = .2376 x frequency Joules per second = .2376 x 42.4Hz Joules per second = 10.07. Joules per second is the same as watts. Watts = 10W. This is shown easily and proven with the Vishay pulse calculator at that URL above. 10W? Why the sudden massive drop in wattage? So this is where I am right now, I want to know why (3) is only giving me 10W. How can 1 and 2 work out, but 3 be so far off. Is (3) not an applicable formula to use in this case?

• Please check mathjax to polish up your question, I inserted an example. The text is very hard to read. Also insert the images in the text where you refer to them, not at the end. – jippie Jan 12 '14 at 11:41
• Also, please try to narrow this down to a specific question. You wander all over the place and you seem to have many implied questions here. If your question is about charging a capacitor, then don't throw in every other part of your system. – Joe Hass Jan 12 '14 at 13:03