Message-ID: <000701c28713$e15270c0$0504210a@c.scope>
From: "g6tmk" <g6tmk@thersgb.net>
To: rsgb_lf_group@blacksheep.org
References: <7D653C9C42F5D411A27C00508BF8803DCB9CBB@mail.dstl.gov.uk>
Date: Fri, 8 Nov 2002 10:43:04 -0000
MIME-Version: 1.0
Subject: LF: Re: To Ponder over the weekend
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Assuming the charge doesn't drain out with the oil, then, for the
relationship Q=CV to be maintained, the voltage must rise to compensate for
the loss of capacitance.
My guess is the 10uF capacitor becomes 1uF and the voltage rises to 100V.

I looked for the answer, upside down, at the bottom of the page ... but it
wasn't there...

73
Hugh M0WYE


----- Original Message -----
From: "Talbot Andrew" <ACTALBOT@mail.dstl.gov.uk>
To: "LF Group (E-mail)" <rsgb_lf_group@blacksheep.org>
Sent: Friday, November 08, 2002 9:38 AM
Subject: LF: To Ponder over the weekend


> This simple puzzle caused more discussion in my office this morning than
it
> had any right to.   Have a ponder while waiting for next QRSS symbol to
come
> through.
>
> Andy G4JNT
>
> --------------------------------------------
> >From Circuit Cellar magazine, Nov 2002
>
> An oil filled capacitor consists of two plates separated by a dielectric
> made of oil with a dielectric constant of 10.  The capacitance is 10uF and
> it is charged to 10V.  The oil is drained, leaving air between the plates.
> the capacitance falls to 1uF.  Does the stored energy level change?.  If
so,
> where did the extra energy come from or go to?
>
> --------------------------------------------
>
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