Parallel wires

Thread Starter

@nd!ru7

Joined Mar 31, 2018
10
I have a question. If you have two parallel runs and one wire is significantly larger than the other will the small wire overheat? Example: You have a 6awg wire in parallel with a 18awg wire connected to a 50amp load @ 240V. What happens?
 

Reloadron

Joined Jan 15, 2015
6,527
You have a 6awg wire in parallel with a 18awg wire connected to a 50amp load @ 240V. What happens?
Nothing much. Look at wire charts and note the resistance of AWG 6 verse AWG 18 per foot. Kirchoff's voltage and current laws define how the current will split. So with a purely resistive 50 Amp load what do you come up with?

Ron
 

drjohsmith

Joined Dec 13, 2021
382
I have a question. If you have two parallel runs and one wire is significantly larger than the other will the small wire overheat? Example: You have a 6awg wire in parallel with a 18awg wire connected to a 50amp load @ 240V. What happens?

two parts to the question
50 amps in a wire, will it heat up,
yes, there are tables around that tell you how much, just google

second
will one wire get hotter than the other,
depends if the same current is flowing down both wires, i.e they are in series,
or if they share the current, dependent upon there resistance, i.e. they are in parallel

can you expand please, may be a drawing, as to how your wires are arranged,
 

Thread Starter

@nd!ru7

Joined Mar 31, 2018
10
Nothing much. Look at wire charts and note the resistance of AWG 6 verse AWG 18 per foot. Kirchoff's voltage and current laws define how the current will split. So with a purely resistive 50 Amp load what do you come up with?

Ron
Uncoated copper wire resistance:

6awg stranded copper .491/1000ft

18awg solid copper 7.77/1000ft

Just getting the info uploaded so I can reference it. It's been a while since I've done electrical theory calculations.
 

Thread Starter

@nd!ru7

Joined Mar 31, 2018
10
Parallel Circuit.jpg

So using the formula I x R1/(R1+R2). 7.77 ohm wire would carry 47.028 amps and the .491 ohm wire would carry 2.972 amps. So the smaller wire is limiting the current to 2.972, but can it overheat?
 

MrChips

Joined Oct 2, 2009
25,928
I have a question. If you have two parallel runs and one wire is significantly larger than the other will the small wire overheat? Example: You have a 6awg wire in parallel with a 18awg wire connected to a 50amp load @ 240V. What happens?
The answer is no.
The resistance of the small wire is higher and therefore will conduct a lower current.

(It might make a difference if you want to start looking into the physical details of the wire, such as surface area, thermal resistance of the insulator, etc.)
 

Thread Starter

@nd!ru7

Joined Mar 31, 2018
10
The answer is no.
The resistance of the small wire is higher and therefore will conduct a lower current.

(It might make a difference if you want to start looking into the physical details of the wire, such as surface area, thermal resistance of the insulator, etc.)
Thanks. Is there any difference if the load is a 50 amp motor?
 

sghioto

Joined Dec 31, 2017
3,058
7.77 ohm wire would carry 47.028 amps and the .491 ohm wire would carry 2.972 amps.
Just the opposite the #6 wire (.491) would carry appx 47 amps.
Your calculations are base on 1000ft of wire or 500ft each way.
So 2.97 amps X 7.77 ohms is appx 23 volts which is the loss through 1000ft of #18 wire and #6 wires.

but can it overheat?
No because:
23V X 3A = 69 watts dissipated in the #18 wire spread out over 1000ft or 69mw /ft
23V X 47A = 1081 watts dissipated in the #6 wire spread out over 1000ft or a little more then a 1 watt/ft.
 

Thread Starter

@nd!ru7

Joined Mar 31, 2018
10
The answer is no.
The resistance of the small wire is higher and therefore will conduct a lower current.

(It might make a difference if you want to start looking into the physical details of the wire, such as surface area, thermal resistance of the insulator, etc.)
Wait...if you had only the small wire hooked up to 50 amps the wire would burn up. I wanted to know if that changes if connected in parallel with a larger conductor?
 

MrChips

Joined Oct 2, 2009
25,928
Do the Ohn's Law calculation followed by the power calculation:

V = 250V and I = 50A are not independent!

I = V / R

P = I x I x R
 

Thread Starter

@nd!ru7

Joined Mar 31, 2018
10
Do the Ohn's Law calculation followed by the power calculation:

V = 250V and I = 50A are not independent!

I = V / R

P = I x I x R
I don't understand. If the
Do the Ohn's Law calculation followed by the power calculation:

V = 250V and I = 50A are not independent!

I = V / R

P = I x I x R
But if you had a load drawing 50 amps and you connected a #18awg wire between the two it would burn up.
 

MrChips

Joined Oct 2, 2009
25,928
I don't understand. If the

But if you had a load drawing 50 amps and you connected a #18awg wire between the two it would burn up.
Depends on the length of the wire.

The resistance of the load itself is:

R = V / I = 250V / 50 A = 5Ω

Now you have to add the resistance of the wire.

Rtotal = Rwire + Rload

If the wire is long, the current will no longer be 50A.
In any case, you have to calculate the power dissipated by the wire = I x I x R

18AWG wire is 6.385Ω per 1000ft.
Assuming that the wire is short enough so that its resistance is much less than 5Ω we can assume that the current is always 50A.
Hence the power P = 2500 x R watts.
The power is proportional to R.
10 ft wire = 0.06385Ω
P = 2500 x 0.06385 = 160W

The next complication is that as the wire heats up, so does the resistance. Hence the current will be lower.
 

Thread Starter

@nd!ru7

Joined Mar 31, 2018
10
Depends on the length of the wire.

The resistance of the load itself is:

R = V / I = 250V / 50 A = 5Ω

Now you have to add the resistance of the wire.

Rtotal = Rwire + Rload

If the wire is long, the current will no longer be 50A.
In any case, you have to calculate the power dissipated by the wire = I x I x R

18AWG wire is 6.385Ω per 1000ft.
Assuming that the wire is short enough so that its resistance is much less than 5Ω we can assume that the current is always 50A.
Hence the power P = 2500 x R watts.
The power is proportional to R.
10 ft wire = 0.06385Ω
P = 2500 x 0.06385 = 160W

The next complication is that as the wire heats up, so does the resistance. Hence the current will be lower.
Where do you get 2500? and would the current go up if the wire resistance was significant?
 

Thread Starter

@nd!ru7

Joined Mar 31, 2018
10
Depends on the length of the wire.

The resistance of the load itself is:

R = V / I = 250V / 50 A = 5Ω

Now you have to add the resistance of the wire.

Rtotal = Rwire + Rload

If the wire is long, the current will no longer be 50A.
In any case, you have to calculate the power dissipated by the wire = I x I x R

18AWG wire is 6.385Ω per 1000ft.
Assuming that the wire is short enough so that its resistance is much less than 5Ω we can assume that the current is always 50A.
Hence the power P = 2500 x R watts.
The power is proportional to R.
10 ft wire = 0.06385Ω
P = 2500 x 0.06385 = 160W

The next complication is that as the wire heats up, so does the resistance. Hence the current will be lower.
#18awg THHN is only rated to carry 14 amps continually. Any high and the insulation starts to degrade.
 

MrChips

Joined Oct 2, 2009
25,928
#18awg THHN is only rated to carry 14 amps continually. Any high and the insulation starts to degrade.
We are discussing theory or in practice?

#18AWG is rated to carry 15A safely.

I don't know what happens at 50A.
I don't know what happens to the insulation at higher temperatures.
 

Thread Starter

@nd!ru7

Joined Mar 31, 2018
10
We are discussing theory or in practice?

#18AWG is rated to carry 15A safely.

I don't know what happens at 50A.
I don't know what happens to the insulation at higher temperatures.
I would never wire something up like this, but I'm just trying to understand the theory behind a real-world application. Although paralleling different size connectors would never be done. The thought occurred to me what if? I have not studied electrical theory in so long that I couldn't make sense of it.
 
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