Capacitor storage in watts

[capthook]

So - whatever!

To heck with a resistor - the ideal solution is to properly size your coil in the first place.

P = V²/R
where P=power / V=voltage / R=resistance

so - if I provide a coil of 10 ohms with 1.5 volts it will consume:

(1.5x1.5)/10 = .225 watts

Target power consumption with no resistor loss - sweet!

Some examples for reference of what might get your target:

Feet        AWG        Ohms     Voltage       Current    # of turns   Ampere-turns
200           30             20              5             0.25        800        200
75            26              3                5             1.67        350        580
40            22            0.65              5             7.7         160        1230

CH

[capthook]

Why isn't Ohm's law providing the same results as actual meter testing??

Testing coils with 1 "C" battery: 1.5 volts

P = V²/R
where P=power / V=voltage / R=resistance

#1) 2.3 ohm coil: (1.5x1.5)/2.3= .98W
Meter: 1.03V x .43A = .44W

#2) .7 ohm coil: (1.5x1.5)/.7= 3.21W
Meter: .6V x .9A = .54 watts

#3) 2.8 ohm coil: (1.5x1.5)/2.8= .80W
Meter: 1.12V x .42A = .47W

#4) 30.7 ohm coil: (1.5x1.5)/30.7 = .073W
Meter: 1.42V x .044A = .062W

#4 is the only one even CLOSE!  #1 and #3 are about half projected, and #2 is off by a MILE!

I'm taking the digital mulimeter readings correctly: volts - across coil ends / amps - inserted into circuit

Why is the actual metered consumption way less than the projected consumption of Ohm's law?


Please - help!!!   

CH

[capthook]

Ahhh -

Got the answer @ http://www.physicsforums.com/forumdisplay.php?f=102

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"If in your first calculation you use the measured voltage drop your results are much better. The assumption that the battery can provide a full 1.5V to your low resistance loads is clearly bad. It looks like the missing voltage is dropped by the batteries internal resistance rather then the coil."

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I guess Ohm's law should have an asterisk stating circuit voltage for guys like me! (rather than battery voltage measured before hooking it up!)

And I guess I need to add battery resistance to my coil resistance....

What is the internal battery resistance of a "C" 1.5V/4mA battery? (tiny it appears)

CH

[teslonian]

Hi everybody! I'm new in posting at this forum even though I've been reading from it for probably several years here and there.

I need help on figuring out a good way to measure power discharging out of a capacitor. I don't have an 0-scope, only two radio shack meters, which btw are by far the best meters I've ever had, they are hard to bust, and that's about it.

I don't want any complicated math formulas, I can't deal with them, and some of them are 100 years worth of doctrine brainwashed into everybody's head. I think there are other ways besides formulas because some people have a hard time understanding.

One I idea I've seen on here so far is to connect a motor with a mechanical meter attached to it's shaft like a counter to count the number of turns. I think that would be a good way as it is true power flowing out of the capacitor and into the motor and that the amount of turns can be related to the power flowing through it in proportion.

But I only have a tiny little radio shack motor and I did try sticking some tape on it to make a sort of flag that could flap past something and then record it on my iphone and then upload it to my laptop and play it back on VLC Player in super super super slow motion and count the turns like that.

Any great ideas out there? It's all because of this site I found right here http://overunity.x10host.com/ and I am wanting to duplicate it and confirm this for myself, I believe there are great ramifications for it.