0
\$\begingroup\$

Why is it that a stranded wire has higher resistance than a solid wire?

Although the area of the stranded is more which means more skin?

""The resistance of the stranded conductor is slightly more than the solid conductor of equivalent cross sectional area. Since a stranded conductor is spiraled, each strand is longer than the finished conductor.""


Source: National Electrical Code NEC, 2017

NFPA 70, National Fire Protection Association

ISBN: 978-145591280-3

enter image description here

\$\endgroup\$
1
  • \$\begingroup\$ Skin effect only applies for high frequencies. Stranded has less copper for the same wire cross sectional area than solid (circular strands are not tessellating). \$\endgroup\$
    – DKNguyen
    Aug 16 at 5:23
1
\$\begingroup\$

Skin effect applies only to AC so we can forget about that here (and the conductors are in contact- we don't have a Litz(endraht) wire).

Now, on DC only- you have one particular chart that shows stranded wires have higher resistance but there are other charts. For example:

enter image description here

In this case it appears to be related to the type of stranding, with "C" concentric stranding showing a lower resistance than solid. There is not a strong correlation with the actual areas either, rather the stranding method appears to be the main factor.

So presumably there is some interaction between the contact between the strands, the strand configuration etc. that causes an increase in resistance in the twisted styles.

enter image description here

Here are the actual areas in mm^2 calculated from the total strands/strand size using the formula for AWG area which should be good to about 6 significant digits.

enter image description here

The formula for resistance also yields somewhat different results than the chart above as well. There are a number of assumptions (temperature, copper conductivity, which is a function of annealed vs. work-hardened etc) as can be seen in the references 7. from Wiki.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.