[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: LC III
- To: tesla@xxxxxxxxxx
- Subject: Re: LC III
- From: "Tesla list" <tesla@xxxxxxxxxx>
- Date: Fri, 01 Apr 2005 07:42:07 -0700
- Delivered-to: email@example.com
- Delivered-to: firstname.lastname@example.org
- Old-return-path: <email@example.com>
- Resent-date: Fri, 1 Apr 2005 07:42:37 -0700 (MST)
- Resent-from: tesla@xxxxxxxxxx
- Resent-message-id: <_2dWt.A.OZF.b3VTCB@poodle>
- Resent-sender: tesla-request@xxxxxxxxxx
Original poster: "acmdq" <acmdq@xxxxxxxxxx>
> Original poster: "Bob (R.A.) Jones"
> > For the coils that I have measured, or calculated, the
> > Medhurst capacitance is close to one half of the
> > of a hollow cylinder to "infinity".
> I think the ration depends if you use Medhurst C from his
table or calculate
> it from fr.
> If I remember correctly theoretically there can be 38%
difference in value
> between the two, possible more for real coils.
> I factor of four surprises me. I would have expected it to
be closer to two
> but it my be dependent on the test conditions and the above.
I don't see a reason for a difference in the capacitance
obtained by either method. The Medhurst capacitance is
supposed to generate the right resonance frequency with the
DC inductance (and it really works), and so if the resonance
frequency is measured and the capacitance calculated
something very similar to the Medhurst capacitance will be
A possible reason for the Medhurst capacitance being
somewhat higher that one half of the capacitance of a
cylinder is the equivalent capacitance from one end of the
coil to the other end, that adds to one half of the body
capacitance of the cylinder to form the Medhurst capacitance.
Antonio Carlos M. de Queiroz
Acabe com aquelas janelinhas que pulam na sua tela.
AntiPop-up UOL - É grátis!