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

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

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>
> 
> 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

The analysis is correct, but what you get is the average value (100 mA).
The RMS current is the equivalent DC current that causes the same
average power dissipation in a linear resistor.
A 10 us pulse with 100 A over a resistor R dissipates:
E=R*100^2*10e-6=R*0.1 Joules.
Over a 10 ms period, the average power dissipated is:
P=R*0.1/10e-3=R*10 Watts
With DC, the power dissipated is R*I^2.
R*Irms^2=R*10 => Irms=3.16 A. 

Antonio Carlos M. de Queiroz