The camera draws what it needs so this is perfectly okay.
It should pose no problem, as the device using the power, not the device
You must define what signal the alarm clock produces before you can
It sounds to me like he wants someone to explain how to do it, the problem
Why not just buy a timer? You can get them with plug outputs, that can
We'll see if he ever gets back to the group...
No. The batteries set the upper bound on how much current a load can
an.2005.04.05.21.51.55.857477@example.net...Geez! I never needed such noise to awaken! What the heck knocked you out
Learn just one formula such as V=IR and rearrange terms mathematically
Pedantic mode on.
i took out what i think is the capacitor. it has two wires going into
it and one other place where a wire can be connected. the back of it
(opposite the wires) is open and there is a metal disk that has a screw
righ tin the middle (variable?). like i said i have no idea what im
talking about lol. im not seeing any units of capacitence or anything
really. it does say pat 2585704 & others. also says klixon and
mee26rx-368 each terminal is numberd 1-3 thanks for your help! id hate
to see an expensive pump (that i got for $20
Woops, wrong search function. That patent appearsdewdrops@2die4.com> wrote in message
news:1112767705.668848.249400@l41g2000cwc.googlegroups.com...
i took out what i think is the capacitor. it has two wires going into
it and one other place where a wire can be connected. the back of it
(opposite the wires) is open and there is a metal disk that has a screw
righ tin the middle (variable?). like i said i have no idea what im
talking about lol. im not seeing any units of capacitence or anything
really. it does say pat 2585704 & others. also says klixon and
mee26rx-368 each terminal is numberd 1-3 thanks for your help! id hate
to see an expensive pump (that i got for $20
I could not find that patent # at http://www.uspto.gov/ .
Please attempt to describe (or better yet, if you've got the gear to do
Don,In article <1112766603.662249.49500@g14g2000cwa.googlegroups.com>,
dewdrops@2die4.com wrote:
thank you very much for the replys! im goign to go open up what i
think
may be the capacitor casing. how can i tell if the capacitor is any
good? the only electrical meter i have is for measuring resistence.
im
thinking the plug may be weird because of the motors age??? thanks
again
Please attempt to describe (or better yet, if you've got the gear to
do
so, snap a good clear picture of it, jpeg it down to a reasonable size
and put it online someplace, then post a pointer to it here) the plug
a
bit better than "weird"?
Two vertical blades? One vertical, one horizontal? Both horizontal?
Three blades in a roughly circular pattern? Two round pins? Four
blades
in a circle? Two *LARGE* blades plus a largish round pin? One blade
vertical, the other blade an "L" shape, plus a round pin? Something
else?
Knowing *EXACTLY* what the plug looks like *CAN* (unfortunately, not
"will") give a good idea of what kind of juice it wants, but
describing
it as "weird" is just about as useful as a screen door on a submarine
in
terms of figuring out what it expects to be fed.
And as a bit of advice: If you need to ask "How can I tell if a
capacitor is any good?", then quite frankly (and with apologies in
advance for what's probably going to sound like a slam or a putdown,
but
isn't intended to be one) you aren't qualified to consider opening the
beast up, and you should make no attempt to do so unless you don't
care
that you'll probably screw something up (possibly including
yourself...
Electricity's "bite" is MUCH worse than its "bark") beyond repair.
You've already gone FAR beyond what I'd recommend for anyone with what
your level of expertise seems to be by dinking around trying to run it
from an wires jury-rigged to an incompatible socket. Hell, to lay it
all
out there for everyone to see, you've already gone further than *I*
would have, and I'm far from being a beginner at the electricity and
electronics game. If you're lucky, you didn't damage it. If you're
not,
you may well have "smoke-tested" it, and it's never going to run no
matter what you do. The very fact that you need to ask "how do I..."
says that you're tinkering with something you don't have the proper
knowledge to be messing around with without standing a very good
chance
of trashing it.
As I said, this isn't intended as a slam or putdown - It's simple
statement of fact. The need to ask "how do I..." about such a basic
concept as determining whether a capacitor is good or bad is almost
always an indicator of someone who is "in way over his head". Don't
despair, though... Stupidity might be forever, but simple ignorance
such
as you're displaying can be easily cured with a dose of edgyookayshun
Here's your first dose:
Set up your meter for high ohms - 20K or thereabouts should be fine.
Once you've gotten to the leads of the capacitor, use a screwdriver
with
a well-imsulated handle to short the leads of the capacitor together.
Get yourself mentally prepared for a miniature lightning bolt to go
off
when the screwdriver makes contact. Eyeball the capacitor - Does it
have
a polarity marking? (Usually a plus or a minus sign, possibly a line
of
them, situated near one of the leads) If so, observe polarity -
red/plus
to plus, black/minus to either minus or the other lead from the
capacitor - and touch the probes to the capacitor lead. If the cap
isn't
totally dead, you should see almost zero ohms at first, then rising to
a
higher value.
If you see that, leave the probes in place for a few seconds, then
remove them, switch your meter to read volts (20 will probably be
overkill, but if you've got a higher range, start there, and work down
until you get a reading) and touch the probes to the capacitor leads
again. You should see an initial high value that slowly drops to zero.
If either of these tests doesn't do what I told you would happen, then
the cap is probably toast, and you'll need a new one. If they both
work
as described, put the lid back on. It's at least acting as a capacitor
should, regardless of whether age has sent it "off value" or not, and
it
probably isn't the source of your problem.
--
Don Bruder - dakidd@sonic.net - New Email policy in effect as of Feb.
21, 2004.
Short form: I'm trashing EVERY E-mail that doesn't contain a password
in the
subject unless it comes from a "whitelisted" (pre-approved by me)
address.
See <http://www.sonic.net/~dakidd/main/contact.html> for full details.
Is it possible that the OP wants to use an old-fashioned mechanical
I think he was a troll Terry, there has been no return of the OP to the
If the method is:PeteS schrieb:
Let's do the actual mathematics
Assume we charge a 10uF and 1uF to 10V. The 10uF will have 100uC
(Q=CV) and the 1uF 10uC of charge. The total charge is 110uC.
Now lets stick them in series.
The first thing is to calculate the new effective capacitance =
C1*C2/(C1 + C2) = 0.909uF. A thing about series caps is the total
capacitance is always less than the smallest cap.
As Q = CV, then V = Q/C so 110uC/0.909uF = 121V (near enough).
Hi Pete
I did it by calculating the total energy stored in both capacitors which
is 5.5 mJ. The corresponding voltage with both caps in series is V = V0
* (C1+C2)/sqrt(C1*C2) after charging each cap to V0.
So, what's correct? Conservation of charge or energy?
Let's test your calculation with two equal capacitors, say 10 uF at 10
volts. If you were right voltage would be 200 uC/5 uF = 40 V ! That's
obviously wrong. Of course the correct answer is 20 V. That's the same
voltage you get using my formula.
Regards
--
Michael Redmann
"It's life, Jim, but not as we know it." (Spock)