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# Re: [femm] Power Losses

- To: femm@yahoogroups.com
- Subject: Re: [femm] Power Losses
- From: David Meeker <dmeeker@ieee.org>
- Date: Fri, 09 Aug 2002 11:07:36 -0400
- References: <3D53C58E.30300@ieee.org>
- User-agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.0.0) Gecko/20020530

`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.`
``

- References:
  - Power Losses
    - From: "brian_hawes" <b.hawes1@physics.ox.ac.uk>

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