Alt-BEAM Archive
Message #01597
To: beam@corp.sgi.com (mailing list)
From: Benjamin Edward Hitchcock beh01@uow.edu.au
Date: Fri, 19 Mar 1999 09:00:50 +1100 (EST)
Subject: [alt-beam] Re: caps
>1. lets say i have two 2.5V, 1F caps hooked up in series, this means
>they
>can hold a 5V charge with a 1F capacity right?
Not quite. It's a 5 volt charge with a 0.5 farad capacity.
The formula is the same as when you put resistors in parallel - you have
to add the inverses of the capacitances, then invert the result.
> 2. if i hook them up in parallel does this mean they can hold a 2.5V
> charge
> with a 2F capacity?
Yep. Got it in one.
The energy contained in the first setup = 0.5 * 5^2
= 0.5 * 25
= 12.5 Joules
The energy contained in the second setup = 2 * 2.5^2
= 2 * 6.25
= 12.5 Joules.
So both setups contain the same amount of energy (As you'd expect) but
their capacitance and maximum voltages are very different.
For more information, I have written some brief articles about the topic.
You can have a quick squiz at:
http://wollongong.apana.org.au/~ben/info/capacitors.html
or
http://wollongong.apana.org.au/~ben/info/laws.html
> 3. if i have a motor that requires 100mA and i hook it up to a large cap
> how
> much current will it allow the motor to take? is there a general rule
> that
> determines this?
At startup, the motor will take a fair bit of current. You can measure
this by holding the shaft still, and measuring the current drawn by the
motor.
Or you can divide the capacitor volts by the resistance of the motor when
it's not turning.
When the motor is turning, it will actually generate a small voltage which
opposes the voltage you apply. So when the motor is running, the
effective voltage across the armature decreases so the current decreases
as well. Normally pager motors will pull up to 250 mA on startup, and
this will decrease to about 30 mA once it gets up to speed.
For the same explanation written a bit differently, go to one of the links
above, and find the INFO article on DC motors.
Ben
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