Are Electrons in Different Energy States Responsible for Variations in Power Output?

[abrogard]

Simple level question this:

How many kinds of electrons are there?

Like if 1 Amp = 6.24 x 10^18 electrons and Watts = Amps x Volts then in our local 240V system the watts = 240.

= 0.24 kW.

Now if we take that 1 Amp in an American 120Volt system then the Watts = 120W.

So the same stream of electrons - 1 coulomb per second - 6.24 x 10^18 electrons per second produces two different amounts of power.

If you're paying for it and you run the system for one hour you pay for 0.24kWh on one system and 0.12kWh on the other.

Where is the difference in power?

Is it that the electrons are in different energy states?

And what's the relevance of this to stick welding where rods are rated by current, Amps and welders are rated by current and thicknesses of metals are rated as requiring this or that current.

So if a stick works best at 100A and say 3/8"metal requires 100A if you go at it with a 240V system you are putting twice as much energy into it as a 120V system would. Right?

Actually of course the voltages in question with the welders would be the DC open circuit voltages they specify but the principle is the same. A 50V welder as against a 25V welder would be putting twice as much energy into the weld. I assume.

 

[russ_watters]

All electrons are the same. Kinda obviously, since you set it up that way, voltage is different in the two scenarios you gave. And since power is volts times amps, the power is different.

Voltage is like force. It determines how much current you can push through the same resistance. Your two welding systems are each designed for whatever voltage they need to get the amperage they need. You can't just change the voltage and expect the amperage to be the same.

 

[Integrand]

Like if 1 Amp = 6.24 x 10^18 electrons...

You are confusing quantity of charge with its rate of flow.

 

[abrogard]

All electrons are the same. Kinda obviously, since you set it up that way, voltage is different in the two scenarios you gave. And since power is volts times amps, the power is different.

Voltage is like force. It determines how much current you can push through the same resistance. Your two welding systems are each designed for whatever voltage they need to get the amperage they need. You can't just change the voltage and expect the amperage to be the same.

that's not an answer. that's my question.

 

[abrogard]

You are confusing quantity of charge with its rate of flow.

maybe I am but i need it spelled out for me, this looks pretty unambiguous (wikipedia):

The SI unit of charge, the coulomb, "is the quantity of electricity carried in 1 second by a current of 1 ampere". Conversely, a current of one ampere is one coulomb of charge going past a given point per second: In general, charge Q is determined by steady current I flowing for a time t as Q = It.

 

[russ_watters]

that's not an answer. that's my question.

What, exactly, is your question?

maybe I am but i need it spelled out for me, this looks pretty unambiguous (wikipedia):

The SI unit of charge, the coulomb, "is the quantity of electricity carried in 1 second by a current of 1 ampere". Conversely, a current of one ampere isone coulomb of charge going past a given point per second: In general, charge Q is determined by steady current I flowing for a time t as Q = It.

What do you need spelled out for you that isn't there?