Calculating AMPs RMS across split wires

Thread Starter

eduncan911

Joined Nov 14, 2011
29
Given all wires of equal resistance (and connectors), is there a formula to calculate the AMPs two or more wires would draw when wired in parallel?

IOW, if a single 16AWG is rated for 6A RMS @ 24VDC. How would I calculate how many 28AWG are needed to carry the same 6A in parallel?

Is it as simple as doubling the AMP RMS rating for the given 28 AWG until I find the capacity?

I'm trying to clean up a messy project with a bunch of 16AWG and 28AWG wires by using a single 35-pin 28AWG wiring loom/connector. The idea is to use two or three 28AWG wired to carry the load previously of the 16AWG wire where applicable.
 

dl324

Joined Mar 30, 2015
11,845
if a single 16AWG is rated for 6A RMS @ 24VDC. How would I calculate how many 28AWG are needed to carry the same 6A in parallel?
Using RMS for DC is meaningless. RMS is used to indicate equivalent DC voltage for calculating power dissipation.

To answer your question, you use the cross sectional area. #16 is about 2600 circular mills mils, #28 is about 160. You'd need 17.

The current capacity of wire depends on diameter, cladding, bundling, length, and ambient temperature. #16 can carry much more than 6A under most conditions.

EDIT: correct spelling of mils.
 
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Thread Starter

eduncan911

Joined Nov 14, 2011
29
Where would I read more about "circular mills"?

I will also evaluate the max AMP draw of the load to figure out a lower value too. I know it doesn't draw anywhere near that.
 

SamR

Joined Mar 19, 2019
2,761
cmills is cmills, add them up! When you get up to RF you might also want to consider surface area. Google AWG, American wire gage.
 
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