[AMPS] Power Handling of Resistors

[Jon Ogden]

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."


--
FAQ on WWW:               http://www.contesting.com/ampsfaq.html
Submissions:              amps@contesting.com
Administrative requests:  amps-REQUEST@contesting.com
Problems:                 owner-amps@contesting.com
Search:                   http://www.contesting.com/km9p/search.htm