on 3/7/00 10:01 PM, Wt8r@aol.com at Wt8r@aol.com wrote:
> What I am saying here is, that if the capacitor bank in a 5KW, 4000 VDC power
> supply, with 16 or 32 uF of capacitance discharges through an arc
The amount of energy stored in the capacitor banks of a power supply that
has a voltage of 4000 Volts and 32 uF of capacitance is:
J = (C*E^2)/2 source: Radio Handbook by Bill Orr 23rd ed. pg 2-7
C is capacitance in Farads and E is voltage.
So using plugging values of C = 32E-6, and E = 4000 into my handy-dandy Boy
Scout calculator, we get the following:
J = (32E-6 * 16,000,000)/2
J = 512/2
J = 256 Joules!
So let's see:
We've established the fact that 320 Joules into my resistor network will not
damage the resistors.
Yet 256 Joules will destroy a tank circuit according to what you say.
Glad I don't live in your universe.
73,
Jon
KE9NA
-------------------------------------
Jon Ogden
KE9NA
Member: ARRL, AMSAT, DXCC, NRA
http://www.qsl.net/ke9na
"A life lived in fear is a life half lived."
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