Capacitor storage in watts

[capthook]

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]

Well, you got the first part right:

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]

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]

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]

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.