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?You have a 6awg wire in parallel with a 18awg wire connected to a 50amp load @ 240V. What happens?
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?
Uncoated copper wire resistance: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
The answer is no.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?
Thanks. Is there any difference if the load is a 50 amp motor?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.)
Makes no difference.Thanks. Is there any difference if the load is a 50 amp motor?
Just the opposite the #6 wire (.491) would carry appx 47 amps.7.77 ohm wire would carry 47.028 amps and the .491 ohm wire would carry 2.972 amps.
No because:but can it overheat?
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?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.)
Yes as explained in post #9Wait...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?
Well worked outView attachment 257407
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?
I don't understand. If theDo 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.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
Depends on the length of the wire.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.
Where do you get 2500? and would the current go up if the wire resistance was significant?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.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.
P = I x I x RWhere do you get 2500? and would the current go up if the wire resistance was significant?
We are discussing theory or in practice?#18awg THHN is only rated to carry 14 amps continually. Any high and the insulation starts to degrade.
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.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.