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Old 7th May 2009, 07:55 PM   #31
luka is offline luka  Slovenia
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cap is 2 plates close together, there is no current going through it, only if damaged! LOL, hard to picture this right?
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Old 7th May 2009, 08:36 PM   #32
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Quote:
Originally posted by theAnonymous1
I have no clue what I'm saying.
Like I said.

OK, I took a few minutes out of my lame day to read about AC and capacitors; I'm fairly sure I understand now. It's kind of embarrassing to admit this is something I didn't already know after being a member here for so long.
 
Old 7th May 2009, 08:45 PM   #33
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Click the image to open in full size. Like I said, 2 plates, "air" between them, only charge can go on this 2 plates, + and -, but no electron [current] goes from one onto other plate[can't because of "air" in between plates]. Only charge goes on and off this plates, in AC conditions, so like I said before, no heating, coz there is no current

Hope you understand this better now
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Old 7th May 2009, 08:52 PM   #34
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Thanks Luka, I had just got done reading about this and was editing my last post before I saw yours.
 
Old 7th May 2009, 09:03 PM   #35
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we are all here to learn, well most of us
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Old 7th May 2009, 09:21 PM   #36
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Quote:
shouldn't the cap be dissipating around ~32W? It didn't feel like it was dissipating any power at all.
and

Quote:
so like I said before, no heating, coz there is no current

Current certainly flows through capacitors; but only alternating current. Capacitors block DC.
When (AC) current flows though a capacitor, the current is phase shifted by 90 relative to the voltage.
Power dissipation is defined V.I.cos(phi) - i.e., the scaler product of current and voltage.
With a resistor, current and voltage are in phase, so phi=0, and
power =V.I
With the capacitor, phi=90, so Power dissipated = zero!

Inductive loads shift the phase the 'other way' by 90, a pure inductor will also dissipate zero power when carrying alternating current, though real inductors have a largish resistance too, so there are resistive heat losses involved as well, which is why inductors get warm.

The current flowing that dissipates no real (heat) power is called reactive power, and is a big issue for power distribution companies, because they have to install high current cabling, but get no return in revenue because the reactive power doesn't show up on the energy consumption meter. So industrial consumers are additionally charged for their reactive power usage.
 
Old 7th May 2009, 09:37 PM   #37
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So, anyone care to explain to a laymen like myself the function of power being transfered here?

It seems with the 7.5uF capacitor the max current flow after the rectifier can be no more than 288mA (measured into a complete short). With a 100R resistor in series the voltage is 28.5V @ 280mA. With a 5R resistor it's 1.45V @ 287mA. The measurements were taken without any capacitance after the rectifier.
 
Old 8th May 2009, 10:07 PM   #38
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The reactance of the capacitor at 60Hz is
1/(120.pi.C) = 353.6 ohms

The battery you say starts charging at around 17V DC. There is going to be a bit of an anomoly in the current measurements because the voltage difference between 17V DC and 120v AC means the waveform isn't a true sinusoid anymore. Most meters are only calibrated / accurate for true sinusoidal current.
But 17 is a fairly small fraction of 120, so the error won't be enormous.
120v - 17v gives 103v dropped over the capacitor reactance of 356 Ohm.
103/356 = 290mA.
Taking into account the error of the meter, the fact that the capacitor is only accurate to within about 10% of its marked value, your measured 270mA is within the expected range.

With a dead short, you get a measured 288mA. Here there is no DC error. 120V is dropped directly across the capacitor:
120/353.6 = 339mA.
So, either your AC line voltage is lower than 120, or the capacitor is out by 15% (6.37uF or 416 ohms at 60Hz).

120v/416ohm = 288mA - exactly as you measure it.

I don't understand where the 100R resistor or 5R resistor are put and where those voltages are measured, so I can't offer any input on those figures.
 
Old 8th May 2009, 11:00 PM   #39
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So what is the theoretical and real world efficiency of such a capacitor-based charger? If it's very efficient, it could be a good charger for homemade EVs, provided the battery is isolated and a circuit used to verify isolation. Add some contactors and electronics to switch in and out different value capacitors and it can actually be a good smart charger.
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Old 8th May 2009, 11:45 PM   #40
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Add some contactors and electronics to switch in and out different value capacitors and it can actually be a good smart charger.
Certainly the capacitor wastes no energy. But safety isolation would require a transformer, and if you are putting in a transformer I think you can do better than a simple resistance charger (which is what the capacitor effectively is, as seen by the battery)

Nickel-cadmium and Nickel-metal-hydride like constant current.
Lead-acid batteries like to be charged with constant voltage; constant current charging them over a long time period shortens their life.

Constant voltage can be achieved very efficiently by means of a switched mode power supply - several ICs exist to do this with minimal support components.

An old AT computer power supply should work - with a small modification to set the voltage.
 

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