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
>
----------------------------------------------------------------------------
--
-------------------------------
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
--
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
--
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
|