# Why doesn't the voltage contribute to heat in a conductor?

1. Nov 20, 2015

### hereforthebeer

This seems like such a simple thing so maybe everyone can help me talk this through...
So in a simple circuit with two conductors mated by physical means (think a plug into a socket) there will be heat because there is effectively arcing between the two conductors. So everyone keeps saying it is the current that is driving the heat. Why do they not consider the voltage as a contributing factor? I mean you need all three components to complete a circuit, am I missing something...?

P=I^2 * R
P=I*V
Heat = Power * time

2. Nov 20, 2015

### Buzz Bloom

Hi hereforthebeer:

All of the equations are OK. I think the issue is the use of language rather than the equations and the names of what the variables in the equations stand for.

Thinking about the units may help.

Current is measured in amperes. An ampere is coulombs per second. A coulomb is a specific number of electrons. Therefore a current consists of electrons in motion.

When electrons move through a resitsance, the resistance acts like friction on the electrons. I assume you understand the moving something against friction produces heat.

Voltage is what moves the electrons.

You are right that all three things, voltage, current, and resistance, are the constituents of the making of heat in a circuit. Each of the three constituents has its own role.

I hope this helps.

Regards,
Buzz
.

3. Nov 20, 2015

### hereforthebeer

Thank you Buzz. That absolutely makes sense and is a great explanation

I guess I was just confused by the fact that in this conversation I was having I was being told (very confidently) that current is creating the heat. It sounds like I should have just put my ear plugs in...

They are also stating the current is the main factor in the amount of heat created. This is how I assume that is not true. Increasing the voltage or current by the same amount produces the same wattage.
V=2 I=2 then P=4
V=3 I=2 then P=6
V=2 I=3 then P=6

4. Nov 20, 2015

### Buzz Bloom

Hi hereforthebeer:

If the resistance remains the same, the only way to change the current is to change the voltage, and the current is the proportional to the voltage. If the voltage is fixed, changing the resistance changes the current inversely.

Regards,
Buzz

5. Nov 20, 2015

### meBigGuy

The power dissipated in a component is the current times the voltage drop across the component.
Since the current through a component is dependent on both the resistance and the voltage, there are many ways to express it and misinterpret the results.

P = IV (power depends on voltage and current) <- Says increasing V increases power --- But generally it increases I also, which increases power even more
P = I^2 R (power depends on current and resistance) <-- but this sort of says increasing R increases power -- True only if I stayed the same which takes more voltage
P = V^2 /R (power depends on voltage and resistance)

It is important to distinguish carefully how power depends on resistance, current, and voltage, since these are all interdependent.

Physically, it is caused by the drifting electrons bumping into wire atoms, so it that sense the current is the direct cause.

6. Nov 21, 2015

### tech99

May I just mention that I notice that in the schools in the UK they tend to tell the kids that the voltage is the energy carried by each charge, based on the definition Energy = Charge x Potential. This causes utter confusion and devastation in young minds, and loss of yet another engineer!!

7. Nov 21, 2015

### Averagesupernova

I suspect that the current being the 'direct cause' is why it is so often said that current is what causes the heat and not the voltage. In other words, you can have voltage across something but no current, so no heat. But you cannot have current through something (except a superconductor) with no voltage lost across it, so you will have heat.

8. Nov 21, 2015

### sophiecentaur

If PD doesn't equal the Energy per Coulomb then what does it equal? It is likely that the young minds can be confused and devastated by a half baked presentation of that fact. Blame the standard of teaching but not the Physics.

9. Nov 21, 2015

### sophiecentaur

That doesn't constitute a circuit, A circuit will consist of a Voltage source, a load (say a Resistance), some wires and then your plug and socket. The heat in your socket will be (could be) generated by arcing or just by some low series contact resistance. The voltage across that arc / contact resistance will be very low and the current will be largely determined by the resistance of the load - which is a lot higher than the resistance in the connector. So the Power dissipated in the plug/socket will be I2Rplug. The statement about the current controlling the heating is based on that. A constant mains supply voltage and a variable current as the load is varied. It's a bit of sloppy Engineer speak that is fine between Engineers but can confuse the newcomer.

10. Nov 21, 2015

### meBigGuy

I think the concept of "a simple circuit with two connectors" includes the elements you mentioned.

The arc is more complex than ohmic heating.
Arcing does not, in it self, create heat except to the extent there are effects that heat the surrounding air. The arc does, however, also release additional energy as light and other EM frequencies and ionization effects in the surrounding environment. That in itself will cause voltage drop across the arc and an effective resistance which is not dissipated directly as heat.

11. Nov 21, 2015

### sophiecentaur

It could constitute a circuit but not the sort of circuit that would normally involve a plug and socket. To explain the paradox the OP implies, there must be something added to the question. Introducing typical components takes us to a familiar situation and a familiar explanation. A hot plug, due to poor contact gets hotter as it carries more current. Whatever the supply volts, 120 or 240V, a 10A load will cause the same heating effect. As I pointed out, Engineers make assumptions and approximations and will say the sort of thing that confused the OP.
How is that unreasonable?

12. Nov 21, 2015

### meBigGuy

The enemy of understanding is classification.

13. Nov 22, 2015

### sophiecentaur

Meaning ??

There is no arcing between the conductors in a normal plug and socket and under normal circumstances. If one were to be studying arcing then would one use a plug and socket as the testbed?
Perhaps it would have been better if that statement had been challenged at the start and some context had been asked for. As it stands, it is just not true. A 'simple circuit' requires a source of power and, depending on whether the source is basically a voltage source or basically a current source, the circuit will behave differently.
A plug and socket connected across a voltage source will vaporise. Connected across a Current source, it will get hot (according to the current passing). Classifying power sources broadly as Voltage or Current sources is a very valid approximation and well understood by Engineers. Nowhere else but my post seems to have introduced this aspect of the problem. imo, there is no answer until the question has been specified properly and I was going for the most likely scenario. I ask again: is that unreasonable?

14. Nov 25, 2015

Mr Beer...

One point to make ... the heat generated is still V * I - but it is the V drop in that element. As the current changed - the V drop changes, it is not constant, it is not the full V of the source.

In a small arc when plugging in a device may only be 0.1 A but for an instant the full 120VAC is across the arc - so for an instant you may have 12W of heat generated, may not sound like much but since the area of the arc is so small - that end up being a pretty high power event relative to the size of the arc. But arc are a special - not ideal case and really not a great place to start this discussion.

But the regular conductor case is this... The P (heat) = V * I , and the V is the drop as I mentioned. There are two ways to do the same math the longer way should illustrate what is going on. Assume a conductor has 1 Ohm resistance, and the circuit has 5 Amps.... The V drop = I * R = 5A * 1 Ohm = 5 Volts -- So the VOLTAGE DROP is dependent on the current. -- if there is no current there is no voltage Drop along the length of the conductor.

Then the P (heat loss) = V drop * I = (I * R) * I = ( 5 * 1) * 5 = 25 Watts of heat ... So the common expression is typically reduced to P (heat) = I^2 * R. ( 5^2 * 1 )

Also - then note that the Voltage available at the desired load ( assuming the losses in the conductor are not what you want) -- will be decreased by that same amount.

In the case above if the Source was 50 V, and the Load will use 5 A, the load will "see" 50 - 5 V - 45V -- again not ONLY when there is 5 A flowing. This is one reason why bad connections can be difficult to troubleshoot,,,, with no load (or load turned off) -- there looks like 50V available - when the 5A load is on - there is only 45V .... - if you need 2 x the conductor ( one length to the load and one to return to the source ) -- we now have 10V Drop and only 40V available to the load - if the current increases or the resistance of the conductor the V drop is worse.

15. Nov 25, 2015

### sophiecentaur

Yes. You are re-making my point (with an example) that it needs another component in the circuit to define the current that flows. If the resistance of the component is much higher than that of the contact then it's a current source as far as the connector is concerned. I really don't think arcing is an issue with mains plugs and sockets because there is always going to be some metal-metal contact due to the spring finger so it's usually, basically down to resistance of a small area of contact, involving corrosion, perhaps. If you have arcing then you've got trouble and a runaway melting metal situation. That's not a situation where you'd say "Oh my, that plug feels a bit warm" you'd probably be having smoke and flashes coming out.

16. Nov 25, 2015

### Averagesupernova

Been there and done that. Nothing says something is wrong like an arcing joint on a conductor at night that lights up the side of a building as if it were daylight.

17. Nov 25, 2015

### CWatters

But Is that the power dissipated in the load or the connector :-)

18. Nov 25, 2015

### sophiecentaur

Time to start on an experiment and do some measurements.

19. Nov 25, 2015

### sophiecentaur

He is confused because the question (or the description of the scenario) was incomplete.