Author Topic: Capacitor storage in watts (Read 31427 times)

capthook · « **on:** March 24, 2008, 10:16:15 PM »

How many Watt seconds will a capacitor store?

Is my math/process correct?

Capacitor storage:

WattSeconds (Ws) = 1/2(CV?)
Where:
C = capacitance in Farads
V = voltage

Example: 1000 uF capacitor @ 12v

1/2(.001x12x12)= .072 Ws or 72 mWs (milliwatt seconds)

1 AA battery: 1.5v x 4mA= 6 mW (*per second)

So the capacitor charged as above will provide:

72 mW / 6 mW = 12 seconds

The capacitor will provide the same amount of energy as the battery for 12 seconds?

Thanks!!!

CH

zerotensor · « **Reply #1 on:** March 24, 2008, 11:10:09 PM »

Well, you got the first part right:

Quote

Example: 1000 uF capacitor @ 12v

1/2(.001x12x12)= .072 Ws or 72 mWs (milliwatt seconds)

The unit mWs is also known as a millijoule. This is a unit of energy.
Your conclusion was right, but your presentation of the units was goofy.

Here is your calculation, in a little clearer format:

We draw 4mA at 1.5V, or 6 milliWatts = 6 milliJoules/second.
72 milliJoules / (6 milliJoules / second) = 12 seconds.

Hope that helps.

zerotensor · « **Reply #2 on:** March 24, 2008, 11:22:04 PM »

By the way, you should be able to draw way more than 4mA off your AA.

1 Amp is certainly possible. Where did you get the 4mA figure from?

capthook · « **Reply #3 on:** March 25, 2008, 01:48:02 AM »

Thank you for the replies!!

I was using mWs 'cause a Watt makes more sense to me than a joule - even though they are the same thing

I'm trying to collect output from my generator coils to a capacitor - and want to compare it to the input from my battery.

I got 4mA off the battery: AA 1.5 volts @ 4mA.

I haven't actually put a meter on the battery yet to calculate the input - I just assumed

it was as stated on the battery. I just got a potentiometer to adjust the input current - so I guess I can adjust it as such. I didn't realize you could draw 1 amp from a 4mA battery??

Anyway - I'm trying to accurately calculate power-in vs. power-out as well as figure how big a capacitor I might need....

I look forward to more suggestions/clarifications!

CH

zerotensor · « **Reply #4 on:** March 25, 2008, 02:22:27 AM »

Quote

I got 4mA off the battery: AA 1.5 volts @ 4mA.

I haven't actually put a meter on the battery yet to calculate the input - I just assumed

The inscription to which you refer means that the fresh battery will show a potential difference of 1.5 V across its terminals when a current of 4mA is drawn from it. It will read less than 1.5V as you pull more current.

The energy storage capacity of a battery is usually given in milliamp-hours (mAh). e.g. a fully-charged 1000mAh battery (like in one of my old mobile phones) should output an one Amp of current for one hour. (Actually, at such a high current draw, I doubt it would last that long. The rating is more of an average for "typical use", and will vary depending on the actual load conditions, temperature, etc...)

Are you reading the "12V" off the packaging on the capacitor, too? The actual voltage on the capacitor will change as you charge it up. If you were to charge your capacitor directly with the AA battery, when you measure the voltage across the capacitor you will see 1.5V (probably a bit less). The label on the cap is the voltage it is rated for. Above that voltage, the capacitor's dielectric will start to break down and if you're not careful it can explode!

So yes, by all means, read the label. Just make sure you understand the label. When you talk about the voltage on a capacitor, most people will assume you mean an actual voltage on the cap that has been measured, not the performance rating on the package.

This is basic stuff. Please be careful, Even a small cap like this can really hurt or even kill if you screw up.

P.S. A Watt is a Joule per second. They are not "the same thing" as you suggest.

capthook · « **Reply #5 on:** March 25, 2008, 07:14:37 AM »

Hi zerotensor - thanks for the reply!

The cap is rated @ 35v - my output coils will put out about 16v - so I'm good there as caps won't go above the 16v to them.

My goal is to be OU and disconnect the battery after a very,very short time and run the pulse coils off the energy collected/stored in the caps. I wanted to be sure my calculations were correct in figuring power in/out.

So I can read voltage at cap and get power out from coils after - say 60 seconds.

On power in from batteries - it's pulsed - so it's not (1.5v x 4mA) x 60 (seconds) but a fraction of that 60 seconds. How do I calculate the exact pulse duration? ie: 1 sec. per 360 degree rotation with 1 pulse per rotation for 10 degrees = 1/36 pulse time. 1/36 = .027 seconds.

Would this be the way to do this - or some other way?? I can't use a stopwatch for .027 seconds!

What is the best/proper way to get a precise calculation of Watts consumed by battery considering it is pulsed?

Thanks a ton!!

CH

capthook · « **Reply #6 on:** March 25, 2008, 10:01:25 AM »

Quote from: zerotensor on March 25, 2008, 02:22:27 AM

P.S. A Watt is a Joule per second. They are not "the same thing" as you suggest.

"Joule: The work done to produce power of one watt continuously for one second; or one watt second"

ie: 1 joule = 1 watt second
or 1 watt second = 1 joule

zerotensor · « **Reply #7 on:** March 25, 2008, 10:53:09 AM »

Quote

Would this be the way to do this - or some other way?? I can't use a stopwatch for .027 seconds!

What you really need to do is to measure it with an oscilloscope. If you don't have a scope, you could connect a small pickup coil to an audio plug and capture the waveform using your computer's soundcard. Use a software oscilloscope (just google around, you will find some free ones). position the pickup coil next to the pulse coil. You should be able to get at least 10kHz resolution with such a setup. It probably won't give you much useful amplitude information, but the frequency and duration of the pulses should show up quite well.

Quote

On power in from batteries - it's pulsed - so it's not (1.5v x 4mA) x 60 (seconds) but a fraction of that 60 seconds.

The 4mA figure written on the label of the battery has absolutely nothing to do with the current you actually draw. Put it out of your mind! The actual current you draw will not be 4mA (unless purely by chance). Calculating the exact current provided by the battery during a single pulse is very complicated. Measuring it is not much easier. You will have better luck if you look at time-averages, and even then, there will be anomalies, as the battery's chemistry reacts and changes over time.

capthook · « **Reply #8 on:** March 25, 2008, 02:06:56 PM »

@ zerotensor

The software scope is a really cool suggestion - especially since I don't have a "real" one. Here's a software version:
http://www.download.com/oscilloscope-lib/3260-20_4-6277260.html
(now to just drag those powerful magnets over next to my computer/hard drive for a good sample - lol - j/k)

The current I draw can be regulated - and thus accurately calculated - by a potentiometer or individual resistors - yes? At least until I get the power/airgap/coil size etc. specs finalized.....

Thanks for the excellent feedback!

==========

Question: how do you figure how much a resistor will "waste" as heat? The current on the backside of a resistor is reduced - both because of the "smaller pipe" and the dissipation of heat. How much is lost to heat? So say current on the backside is reduced by "10 units" - the discharge from the source battery is NOT reduced by 10 - because "x" is lost in heat - so battery discharge is still "x" ?

Is there a formula for this? I would guess so - I just haven't found it!

CH

capthook · « **Reply #9 on:** March 29, 2008, 06:57:43 PM »

Can anyone answer the above post? or put another way...

Power supply to pulse coil - 100 watts. Put a variable resistor in between dropping power to coil 50% (50 watts).

How much power is consumed by power supply now? 50 watts? 50 watts + "x" watts "waste" heat of resistor? Is there a formula for this?

TIA

CH

capthook · « **Reply #10 on:** March 31, 2008, 10:03:27 AM »

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 · « **Reply #11 on:** April 03, 2008, 09:15:57 AM »

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 · « **Reply #12 on:** April 03, 2008, 12:02:37 PM »

Ahhh -

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

-----

"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."

----

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 · « **Reply #13 on:** January 28, 2016, 05:20:02 AM »

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.

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Author Topic: Capacitor storage in watts (Read 31427 times)

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