Re: [femm] Power Losses

[David Meeker <dmeeker@xxxxxxxx>]

brian_hawes wrote:

> I have been using FEMM 3.1 to calculate r.f. losses in
> solenoids. I made each turn a separate circuit, and defined
> a current of 1-Ampere. I started by assuming that with I=1,
> I2*R=R, and hence P=R. I found I was getting resistance
> values ~ 1/2 what I expected. It finally dawned that with
> a.c. analyses, the defined current must be treated as an
> r.m.s value. Can someone confirm I am right, or have I
> dropped a factor two somewhere else?

The program is written so that all values are defined as peak, rather 
than RMS.

The trick is that power = i*i*R at DC, and it equals (1/2) i*i*R for AC, 
which is consistent with what you were seeing. The (1/2) in the AC 
comes because the current is sinusoidally varying. To see this, we 
could write current as an explicit function of time:

i(t) = ipk cos(w*t)

Since the /instantaneous/ power loss is i*i*R,

P = R*i(t)^2 = R * ipk^2 * (cos(w*t))^2

Now, one of those trig identities that is easy to forget is (cos(w*t))^2 
= (1/2) + (1/2) cos(2*w*t)
If we want the average power loss over time, we can just neglect the 
cos(2*w*t) part, because it averages out to zero over time. Considering 
just the constant part of (cos(w*t)^2) for the purposes of the average 
power yields:

Pavg =(1/2)* R*ipk^2

That's where the (1/2) comes from.

Dave.