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## Main Question or Discussion Point

References are available to calculate skin effect, R

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 the

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

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

P = 0.07 Ohms * 10

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%.

_{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 the

**total I**due to both the AC and DC currents.^{2}R lossesHow 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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