RE: RMS and Average Current (was - Why do primaries get hot?)

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "John H. Couture by way of Terry Fritz <twftesla-at-qwest-dot-net>" <couturejh-at-mgte-dot-com>


I used the coulomb solution.
   Coulombs = amps x time
   coulombs for 10 usec = 100 amps x 10 us = .001 coulombs
   10 ms/pulse =  1/.01 =  100 pulses/sec
   .001 x 100 = .1 coulomb/second = .1 amps RMS or DC

John Couture

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-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Sunday, March 03, 2002 11:41 AM
To: tesla-at-pupman-dot-com
Subject: Re: Why do primaries get hot?


Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br>

Tesla list wrote:
>
> Original poster: "John H. Couture by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <couturejh-at-mgte-dot-com>
>
> Antonio -
>
> You say average current is zero. How did you find the average current for
a
> 100 amp (square wave?) pulse lasting 10 us at each 10 ms (100 pulses per
> second) is 0.1 amps and the RMS current is 10 amps.
>
> I get (100 amps x .00001 secs) = .001 amp sec for each pulse
> and   (.001 x 100 pulses) = 0.1 amps for the RMS current.

I don't understand your calculation.
I was considering a series of pulses, starting from zero.
10 us in 10 ms is 1/1000 duty cycle.
The average value is then 100/1000=0.1.
The RMS value is sqrt(100^2/1000)=3.16.
Oops... I forgot the square root.

Antonio Carlos M. de Queiroz