References are available to calculate skin effect, R(adsbygoogle = window.adsbygoogle || []).push({}); _{AC}/R_{DC}given wire gauge and frequency. But my problem is a complication of this simple calculation.

Say I have a 1 Amp RMS AC current component and a selected wire size that gives me a skin effect of 7X the DC resistance. But I also have a hefty 10 Amp DC current component.

I'm interested in calculating thetotal Idue to both the AC and DC currents.^{2}R losses

How are these two combined?? For the sake or argument, the DC resistance of the wire is 0.01 Ohms.

1) I can treat them as separate and independent currents and get a small power loss:

P = 0.01 Ohms * 10^{2}Amps^{2}+ 0.07 Ohms * 1^{2}Amps^{2}= 1.07 Watts.

Or 2) I can imagine that the 7X skin effect also impedes the DC current flow.

P = 0.07 Ohms * 10^{2}Amps^{2}+ 0.07 Ohms * 1^{2}Amps^{2}= 7.07 Watts.

The results are wildly different between the two.

Which method--or a third method, is correct to first order? I mean, never mind the finer details. I just want to know what my power loss is to +/-20%.

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# Skin Effect calculation

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