Any electricians handy?

[KenB]

This is a simple matter of exact mathematical definitions, not "somewhat of a combination".

Okay, I should have choose different wording in my statement.

It is the dissipated power in the wire that creates the heat; and, power IS defined as the product of both current squared AND resistance. To say that "It is the current that melts wires.." Is akin to saying that it is the ham (or the eggs) that makes a ham and egg breakfast. Your either having ham AND eggs, or your not.

I know and I am sure you will agree that without resistance there would be zero current flow.

power IS ALSO defined as Voltage x Current or Voltage squared divided by Current.

We have been discussing 12AWG wire. But suppose we were discussing a different material that had 100 times the resistance we could still achieve the same power dissipation by increasing the voltage but at a much lower current.

Example:
12AWG at 10ft and 1.6 ohms/1000ft = 0.016 ohms
Voltage required to produce 100A through 0.016ohms = I times R = 1.6 volts
P=I^2 x R = 100 x 100 x .016 = 160 watts

R x 100 = 0.016 x 100 = 1.6 ohms
V required to produce 160 watts through 1.6 ohms = square root of (P x R) = sqrt(160 x 1.6)=16 volts
I = V divided by R = 16/1.6=10 amps

The formula that you are using is if an exact resistance is known. However when a conductor is heated its resistance value changes.

So we have produced the same 160 watts power dissipation with 10 amps as we did with 100 amps by just varying the resistance and voltage.

;-) And the 10 amps at 120 volts will not melt the insulation, but the 100 amps at 12 volts will.

 

[Bama]

Either of ya'll remember when the biggest question was if power flowed from negative to positive or positive to negative.

 

[dun]

Either of ya'll remember when the biggest question was if power flowed from negative to positive or positive to negative.

Interesting you should mention that. According to the missle techs in the Navy (30+ years ago) it is just the opposite for missles but it's right for everything else.

dun

 

[dcara]

First, Thanks for this discussion. Now on with the matter at hand.

I know and I am sure you will agree that without resistance there would be zero current flow.

I would agree that with infinate resistance you would have zero current. With zero resistance you would have whatever current the source impedance supported. I=V/R.

power IS ALSO defined as Voltage x Current or Voltage squared divided by Current.

Your first equation (P=V*I) is correct. Your second one is not. (P does not equal V squared divided by I). However, P does = voltage squared divided by resistance.
BTW all the different forms can be derived from each other and ohms law. For instance, P=V*I and I=V over R. Therefore P=V*V over R which is the same as saying P=Vsquared over R.

The formula that you are using is if an exact resistance is known. However when a conductor is heated its resistance value changes.

The point is that we can still generate the same power of 160watts but with only 10 amps. It doesn't matter if the 160watts is being produced by 10amps through 1.6 ohms or 100 amps through 0.016 ohms. The only thing that matters is that 160watts of power is generating the same heat in both cases which may be enough to melt the plastic sheathing.

And the 10 amps at 120 volts will not melt the insulation, but the 100 amps at 12 volts will.

This goes back to what power is being dissipated in the wire. See previous explaination

Case 1
Source power = 120V x 10A=1200W
Power dissipated in wire = current squared x resistance = 10 x 10 x .016 = 1.6 watts
Power delivered to load = 1200-1.6=1199.4 watts

Case 2
Source power = 12V x 100A=1200W
Power dissipated in wire = current squared x resistance = 100 x 100 x .016 = 160 watts!
Power delivered to load = 1200-160=1040 watts

 

[Bama]

Either of ya'll remember when the biggest question was if power flowed from negative to positive or positive to negative.

Interesting you should mention that. According to the missle techs in the Navy (30+ years ago) it is just the opposite for missles but it's right for everything else.

dun

My point exactly, I can prove both of them right and both of them wrong as well. Same thing about power flow 30 years ago. My dogs to worn out to run in their race at hand though.

 

[dcara]

Either of ya'll remember when the biggest question was if power flowed from negative to positive or positive to negative.

Its both, and at the same time. Here's a simplified explaination of how it works.
Current flow is actually the movement of electrons from one atom to the next. An electron has a negative charge. An atom without an electron has a positive charge. This is sometimes called a hole because the atom has a hole in which to put the electron when it gets one. An atom with an electron has a neutral charge. Since opposite charges attract, any atoms without an electron (i.e positive charge) is attracting electrons and may even steel an electron from a neighboring weaker atom thus making the victim atom a positive charge. So as these electrons with negative charge move down the wire towards more stong positive atoms they jump from atom to atom leaving positively charged atoms (holes) behind. Hence, the negative electrons are moving towards the positive. But with each jump they leave a hole behind and so the hole flow as its called is from positive to negative.

The whole process (pun intended) is one of equalizing the surplus of holes at one end of the wire with the surplus of electrons from the other end of the wire.

positive atoms that don't As an electron jumps from the happy neutral atom to a atom that

 

[dcara]

positive atoms that don't As an electron jumps from the happy neutral atom to a atom that

Sorry about that last bit of geberish

 

[KenB]

First, Thanks for this discussion. Now on with the matter at hand.

I know and I am sure you will agree that without resistance there would be zero current flow.

I would agree that with infinate resistance you would have zero current. With zero resistance you would have whatever current the source impedance supported. I=V/R.

power IS ALSO defined as Voltage x Current or Voltage squared divided by Current.

Your first equation (P=V*I) is correct. Your second one is not. (P does not equal V squared divided by I). However, P does = voltage squared divided by resistance.
BTW all the different forms can be derived from each other and ohms law. For instance, P=V*I and I=V over R. Therefore P=V*V over R which is the same as saying P=Vsquared over R.

You are correct, That was a typo. on my part, I intended to say...P=V squared times I

The formula that you are using is if an exact resistance is known. However when a conductor is heated its resistance value changes.

The point is that we can still generate the same power of 160watts but with only 10 amps. It doesn't matter if the 160watts is being produced by 10amps through 1.6 ohms or 100 amps through 0.016 ohms. The only thing that matters is that 160watts of power is generating the same heat in both cases which may be enough to melt the plastic sheathing.

And the 10 amps at 120 volts will not melt the insulation, but the 100 amps at 12 volts will.

This goes back to what power is being dissipated in the wire. See previous explaination

Case 1
Source power = 120V x 10A=1200W
Power dissipated in wire = current squared x resistance = 10 x 10 x .016 = 1.6 watts
Power delivered to load = 1200-1.6=1199.4 watts

Case 2
Source power = 12V x 100A=1200W
Power dissipated in wire = current squared x resistance = 100 x 100 x .016 = 160 watts!
Power delivered to load = 1200-160=1040 watts

You keep wanting to keep a constant resistance in your equations.
The example that I gave was for a given load of 100 amps @ 12 volts, and 10 amps @ 120 volts in order for this to happen the load or device (read resistance ) would have to change in order for this to happen.
By the way I rather doubt 160 watts would melt the wire, maybe make it warm. ;-) 100 amps will make it VERY warm.

 

[dcara]

You are correct, That was a typo. on my part, I intended to say...P=V squared times I

I'm not try to be a smart ass here but, this statement is also wrong. You had P=V*I correct, P does not equal V^2*I as you stated here. The only form in which V^2 works is V^2 divided byR

You keep wanting to keep a constant resistance in your equations.

You'll note that my case1/case2 analysis never mentioned load resistance because it is irrelavent to the main subject of our discussion which is whether it is current or power that causes the heat build up in the wire. To analyze this all we care about is what is going on in the wire. The constant resistance I used was the resistance for 12AWG wire which is the same for both cases. Granted, the load resistances are different for these 2 cases, but what is going on in the load is a secondary issue here.

My case here is still the same which is that it is the power being dissipated in the wire itself which causes the wire to get warm, hot, or melt. And power dissipoation in the wire is defined by BOTH the current in the wire AND the resistance of the wire. The current in the wire is a function of both the load AND the wire resistance AND the source voltage.

Although I could calculate the exact temperature rise in the wire for these cases under given ambient temperature conditions I am sure you are correct in your statements regatrding the temperature as they are based on years of experiance.

 

[KenB]

You are correct, That was a typo. on my part, I intended to say...P=V squared times I

I'm not try to be a smart ass here but, this statement is also wrong. You had P=V*I correct, P does not equal V^2*I as you stated here. The only form in which V^2 works is V^2 divided byR

You keep wanting to keep a constant resistance in your equations.

You'll note that my case1/case2 analysis never mentioned load resistance because it is irrelavent to the main subject of our discussion which is whether it is current or power that causes the heat build up in the wire. To analyze this all we care about is what is going on in the wire. The constant resistance I used was the resistance for 12AWG wire which is the same for both cases. Granted, the load resistances are different for these 2 cases, but what is going on in the load is a secondary issue here.

My case here is still the same which is that it is the power being dissipated in the wire itself which causes the wire to get warm, hot, or melt. And power dissipoation in the wire is defined by BOTH the current in the wire AND the resistance of the wire. The current in the wire is a function of both the load AND the wire resistance AND the source voltage.

Although I could calculate the exact temperature rise in the wire for these cases under given ambient temperature conditions I am sure you are correct in your statements regatrding the temperature as they are based on years of experiance.

No, that is just plain common sense

Okay, I think we are both nit picking.
When I retired I thought I was done with arguing with engineers.
Well I am retiring again.

PS ;-) Keep your hands away from #12 wire carrying 100 amps.

 

[dcara]

I was going to say something similar but you beat me to it.

We can now return this thread to its righful owner, but I think they gave up on us and went somewhere else