So, if you connect the plus pole to the front legs on a cat ( a catacitor),
and the minus to the rear legs, and also place your sister (resister) in
series, and switch on the 600 VDC supply, then the cat will make a
tremendous jump (inrush) but will come nowhere because of your sister, who
got 50 times the weight of the cat. But the cat is charged with anger and
will take any opportunity to kick back. I understand that, but the rest is
hard to see. So, you paint one side of the cat with conductive silver epoxy,
let it cure at room, which will take time as it follows an asymptotic curve.
And after that you switch on again. And you say, that the cat can never pull
the cord again and the cord can never slacken. The only solution I can
find, is that the cat died because it disliked the silver epoxy and tried
to lick it off, and as the cat's gone, it can't pull the cord any more. So
far OK, but why can't the cord slacken? Your sister ought to give a shrill
and run to the cat and take it up and sob and pat the cat, and the cord will
slacken. Your insisted bilateral equipotential anticircumstance ....I can't
fix that.
Inge
PS. I'm still thinking that it's a 47 pF cap, because the end cap silver has
migrated across, silver has atomic number 47 and when the circuit is
switched on, you will here a tiny 'pFff' when the bridge burns off, and the
cap works again.
----- Original Message -----
From: "Brian Ellis" <[log in to unmask]>
To: <[log in to unmask]>
Sent: Monday, October 23, 2006 5:23 PM
Subject: Re: [TN] NTC: Friday enigma.
> OK, the monetary value is obviously zero, as it is short-circuited by
> the metal. But the enigma is its electrical value as a capacitor. If a
> capacitor is connected to a DC source via a resistor to limit the
> current, there will be an initial surge of current of a value determined
> by the source voltage and the resistance of the circuit, determined by
> Mr Ohm. The current will then decay exponentially as the capacitor
> charges, providing a back EMF to the source, heading towards an
> asymptote of zero (assuming a perfect capacitor), when it is fully
> charged. The higher the value of the capacitor, the higher will be the
> current after a given time, and the longer will be the decay curve. In
> this case, after the given time, the current will be the same as the
> initial value, indicating that the value of the capacitor will be
> infinity farads because it can never be charged to the source voltage
> and the charging current can never decay. :) :)
>
> Brian
>
> Brian Ellis wrote:
>> On http://stevezeva.homestead.com/files/c46.JPG there is an 'orrible
>> photo of a ceramic capacitor. What is its value? :-) :-)
>>
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