Amateur Electronics question
jdarby at lplizard.com
Mon Jul 10 20:12:08 CEST 2000
Thank you for your response! I have crossed out the faulty formula and
written in nice, neat lettering the correct one, plus the nifty "Electron
Plumbing" analogy in the margin. It all makes much more sense when you have
the correct formula, and I thank you for stopping my brain from hurting.
Viva la DIY list!
Your eternally endebted student,
From: Christian Hofmann <chris at scp.de>
To: synth-diy at node12b53.a2000.nl <synth-diy at node12b53.a2000.nl>
Date: Monday, July 10, 2000 12:46 PM
Subject: Re: Amateur Electronics question
>On Mon, 10 Jul 2000 10:16:08 -0400
>"Jon Darby" <jdarby at lplizard.com> wrote:
>> I officially became fed up of not knowing EXACTLY how all the little
>> electronic projects I've built work and deceided to educate myself on
>> electronic theory. I have a seriously juvenile resistance question, so I
>> figured since everyone on the list must have had a baby-steps phase in
>> education I would ask and not worry too much about being humiliated. Here
>> the formula my book gives for finding total resistance with two resistors
>> wired in parallel:
>> Rtotal=1/R1 * 1/R2
>> Assuming this is the correct formula, shouldn't two resistors with a
>> of 1 each equal a total resistance of 1? The answer the book gave was .5,
>> where am I missing the logic? Grrrr! Day two and I'm already stumped.
>> a pantload for any help!
>The correct formula is 1/Rtotal = 1/R1 + 1/R2
>i.e. Rtotal = 1 / (1/R1 + 1/R2)
>Taking the value 1 as an example indeed gives 0.5, but not using their
>As a rule of thumb:
>resistors in series
>- add resistance values for total resistance
>- resistance increases
>resistors in parallel
>- add conductivity values (i.e. 1/R) for total conductivity
>- resistance decreases
>And for the public humiliation part - we'll leave that for next week.
>Be prepared... :-)
>P.S.: after some time of just reading, but not being able to post to the
>list, I'm glad to be back. Yippieee...
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