Skin Effect - Resistance

Resistance

The effective resistance due to a current confined near the surface of a large conductor (much thicker than δ) can be solved as if the current flowed uniformly through a layer of thickness δ based on the DC resistivity of that material. We can therefore assume a cross-sectional area approximately equal to δ times the conductor's circumference. Thus a long cylindrical conductor such as a wire, having a diameter D large compared to δ, has a resistance approximately that of a hollow tube with wall thickness δ carrying direct current. Using a material of resistivity we then find the AC resistance of a wire of length L to be:

R\approx {{L \rho} \over {\pi (D-\delta) \delta}} \approx {{L \rho} \over {\pi D \delta}}

The final approximation above assumes .

A convenient formula (attributed to F.E. Terman) for the diameter DW of a wire of circular cross-section whose resistance will increase by 10% at frequency f is:

The increase in AC resistance described above is accurate only for an isolated wire. For a wire close to other wires, e.g. in a cable or a coil, the ac resistance is also affected by proximity effect, which often causes a much more severe increase in ac resistance.

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