[AMPS] Power Handling of Resistors

[Will, K6NDV]

Please stop the personal attacks!
Thank you.
Will,  K6NDV
AMPS reflector

-----Original Message-----
From: owner-amps@contesting.com [mailto:owner-amps@contesting.com]On
Behalf Of Wt8r@aol.com
Sent: Wednesday, March 08, 2000 6:04 AM
To: jono@enteract.com; amps@contesting.com
Subject: Re: [AMPS] Power Handling of Resistors



In a message dated 3/7/00 11:22:45 PM Eastern Standard Time,
jono@enteract.com writes:

>
>  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
>
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Anyone who has ever seen the results of such an arc discharge as I have
described knows the Dr. Jon is ANAL RETENTIVE.

This is enuf time wasted on this foolishness.

Dave in Dayton, WT8R

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