Parallel wires

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tallgirl

Senior Member
Location
Glendale, WI
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Controls Systems firmware engineer
I do not think that the effect of conductor heating is significant enough to seriously affect current distribution and heating.
Just note that the ampacity of the pair of non-identical wires will probably be less than the sum of the two independent ampacities because of the difference ratios between resistance and heat dissipation.
Anything other than equal current distribution over equal conductors will end up with the current distribution NOT matching the relative ampacities. No opinion on which conductor will be more stressed at this point. Have to think more about it. Probably the larger one.
I was responding to Sea Nile's comment.

It's very unlikely that a simple analysis will reflect what's going to happen because each wire will change its behavior as the other changes its. The temperature coefficient of resistance for copper is about 0.4% / *C. That's a non-trivial change, but the question would be under what circumstances would the temperature of one conductor be significantly different from the other. Were I doing any kind of calculation to figure out how the currents are distributed I'd pick the extremes for both temperature and current and ensure that both conductors were within their rated ampacities over that entire range.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
In my opinion, by the time the current causes either conductor to overheat, the experiment is over.

I think the question is about them sharing current proportionately without either one overheating.
 

retirede

Senior Member
Location
Illinois
The heating (temp rise) in each branch should equal from a practical standpoint. I’m too lazy to calculate it.

The light bulb circuit is not a good analogy. The R vs T curves of the branches don’t match like they do for two copper branches.
 

tallgirl

Senior Member
Location
Glendale, WI
Occupation
Controls Systems firmware engineer
So long as the current is below the ampacity of the conductor (and that can be calculated is things like battery internal series resistance, and conductor resistance per foot, are known), "neither" is the correct answer.

If the battery is a stiff current source (lead-acid chemistry, for example), the bypass wire will be vaporized fairly quickly and then the light will illuminate.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
220130-2110 EST

If the light bulb has a voltage rating equal to the battery voltage, then it will burn out after its normal rated lifetime, or longer.

If the nominal lamp voltage equals the normal drop across the resistor with battery power applied, then bulb life is its nominal life.

If bulb voltage is greater than nominal voltage of the lamp when across the shunt, then the bulb will burn out first.

.
 

Sea Nile

Senior Member
Location
Georgia
Occupation
Electrician
If the light bulb has a voltage rating equal to the battery voltage, then it will burn out after its normal rated lifetime, or longer
I concede, the light bulb is a bad example unless you break the glass and release the gas that keeps the filament from burning out.
 
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