Capacitor storage in watts

[capthook]

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]

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]

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.

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]

@ 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]

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