A lesson in voltage and amperage, please

[Perplexed]

OK, as I understand Ohm's Law (I = V/R), there's an inverse relationship between voltage and amperage. That is, the more volts, the less amps, and vice versa. Correct?

And a plasma cutter needs more amps to cut through thicker metal - correct? Ditto for a welder, more amps are needed to weld up thicker pieces of metal... correct?

So why is it that the heavy-duty plasma cutters and welders pretty much require a 240V power supply, since those units would then have lower amperage outputs than a light-duty 120V unit?

I must be missing something here Please point out my mistakes!

 

[cdaigle430]

I think it's because if they did run at 120v, then they would require more amperage than a typical household circuit can supply. If a plasma cutter is rated for 240v at 30 amps input then most likely the amperage requirments would double if it were rated for 120V-120v at 60a amps and so on.

A commercial grade plasma cutter would run off of a 3 phase circuit.

 

[kd5kiz]

Welders and Plasma cutters all use transformers to lower the OUTPUT voltage which greatly increases the amperage at the rod. Welders may have 30 open circuit volts which will decrease under load.

 

[Cat_Driver]

Also has to do with wire size. 240 has bigger wire than 110. If you're running equipment the wires would heat up.

 

[Sublifer]

OK, as I understand Ohm's Law (I = V/R), there's an inverse relationship between voltage and amperage. That is, the more volts, the less amps, and vice versa. Correct?

And a plasma cutter needs more amps to cut through thicker metal - correct? Ditto for a welder, more amps are needed to weld up thicker pieces of metal... correct?

So why is it that the heavy-duty plasma cutters and welders pretty much require a 240V power supply, since those units would then have lower amperage outputs than a light-duty 120V unit?

I must be missing something here Please point out my mistakes!

The inverse relationship is between amperage and resistance. Voltage and amperage have a direct correlation. Resistance being equal, more voltage means more amps and more amps means more voltage. Take a look at the equation and try some numbers. Start with R=1 for an easy view

 

[Perplexed]

Hmmm... OK then, here's data off the UL label on the motor of a table saw:

12.8A @115V or 6.4A at 230V

I had thought that meant the motor produced 12.8 amps at 115 (or 120) volts, and 6.4 amps at 230 (0r 240) volts. An inverse relationship. What am I missing this time?

 

[Little_Grizzly]

Think power instead of ohms law. W = I * V

 

[Reg]

OK, as I understand Ohm's Law (I = V/R), there's an inverse relationship between voltage and amperage. That is, the more volts, the less amps, and vice versa. Correct?

And a plasma cutter needs more amps to cut through thicker metal - correct? Ditto for a welder, more amps are needed to weld up thicker pieces of metal... correct?

So why is it that the heavy-duty plasma cutters and welders pretty much require a 240V power supply, since those units would then have lower amperage outputs than a light-duty 120V unit?

I must be missing something here Please point out my mistakes!

No,
There is a DIRECT relationship between volts and amps, i.e. the higher the voltage the higher the current (through any given SAME resistance).

PART of the reasons for most plasma cutters being 220-240 volt is manufacturing costs and design simplicity.

Ahhh, maybe you were thinking that for any given level of POWER there is an inverse relationship between voltage and current ?
Yes, for whatever power is needed OUT of a device it will consume less current from a 220 volt source than from a 110 volt source.
Clearer ?

This is where current affects cost.
Say there is some design arithmetic that determines that the input power will need to be 6,600 Watts (arbitrary, but reasonable).
That could be "Sucked out of the wall" as 6600/220 = 30 amps or as 660/110 = 60 amps.

In one case you need only half the current rating for the primary side that you need for the other case, i.e. thinner wires, cheaper plugs, switches, etc.
Theoretically you need more/better insulation for higher voltages, but the difference between insulating 220 and 110 is negligible.

 

[bcp]

Hmmm... OK then, here's data off the UL label on the motor of a table saw:

12.8A @115V or 6.4A at 230V

I had thought that meant the motor produced 12.8 amps at 115 (or 120) volts, and 6.4 amps at 230 (0r 240) volts. An inverse relationship. What am I missing this time?

It is a 1472 Watt motor:

12.8A x 115V = 1472 Watts

6.4A x 230V = 1472 Watts

 

[Reg]

Hmmm... OK then, here's data off the UL label on the motor of a table saw:

12.8A @115V or 6.4A at 230V

I had thought that meant the motor produced 12.8 amps at 115 (or 120) volts, and 6.4 amps at 230 (0r 240) volts. An inverse relationship. What am I missing this time?

Nope,
It just means that for the POWER that it can DRAW (NOT produce) it can take it from the wall as 12.8 at 115 or as 6.4 at 230.
POWER is the PRODUCT (multiplication) of volts times amps.
1472 Watts in this case.

 

[Perplexed]

Ahhh, maybe you were thinking that for any given level of POWER there is an inverse relationship between voltage and current ?
Yes, for whatever power is needed OUT of a device it will consume less current from a 220 volt source than from a 110 volt source.
Clearer ?

This is where current affects cost.
Say there is some design arithmetic that determines that the input power will need to be 6,600 Watts (arbitrary, but reasonable).
That could be "Sucked out of the wall" as 6600/220 = 30 amps or as 660/110 = 60 amps.

In one case you need only half the current rating for the primary side that you need for the other case, i.e. thinner wires, cheaper plugs, switches, etc.
Theoretically you need more/better insulation for higher voltages, but the difference between insulating 220 and 110 is negligible.

Now this clears up my confusion! I think I understand better the relationship between voltage, amperage, and output when it comes to using a cutter or welder. Thanks, Reg! And everyone else who contributed :thumbsup:

 

[grsthegreat]

Hmmm... OK then, here's data off the UL label on the motor of a table saw:

12.8A @115V or 6.4A at 230V

I had thought that meant the motor produced 12.8 amps at 115 (or 120) volts, and 6.4 amps at 230 (0r 240) volts. An inverse relationship. What am I missing this time?

that means the motor is USING 12.8 amps at 115 and USING 6.4 amps at 230. So the 220 motor is CHEAPER to operate. remember, you pay for current (amps) not volts. This is why large commercial shops run at 480 volts. a similar motor like above at 480 volts would only need 3.2 amps of current to operate. also it would require smaller feed wires and the motor will heat up less and last alot longer.

 

[grsthegreat]

Also has to do with wire size. 240 has bigger wire than 110. If you're running equipment the wires would heat up.

actually a 220 motor of the same HP ratine would require smaller wires than a 110 of the same rating. less amps = less heat

 

[SPYDERLK]

that means the motor is USING 12.8 amps at 115 and USING 6.4 amps at 230. So the 220 motor is CHEAPER to operate. remember, you pay for current (amps) not volts. This is why large commercial shops run at 480 volts. a similar motor like above at 480 volts would only need 3.2 amps of current to operate. also it would require smaller feed wires and the motor will heat up less and last alot longer.

You pay for Energy. Power [V x A] for a given time .... Kilo-Watt Hours. A 110 motor drawing 12.8A is using the same amount of energy as a 220 drawing 6.4. The electric power [Watts] each uses would be the same -- 220x6.4=110x12.8.
larry

 

[crash325]

Hmmm... OK then, here's data off the UL label on the motor of a table saw:

12.8A @115V or 6.4A at 230V

I had thought that meant the motor produced 12.8 amps at 115 (or 120) volts, and 6.4 amps at 230 (0r 240) volts. An inverse relationship. What am I missing this time?

An electric motor is not producing amps. It is consuming amps. Same with a welder or any other electric user.

Welder putting out 30 Volts at 200 amps will overload a 110V power supply.

I think that is what you are missing.

 

[rankrank1]

Perplexed,

You would be better served to look at the equation: Power (in Watts) = Voltage x Amps.

Lets use a Lincoln AC-225 stick welder for an example since it is the best selling welder of all time. Your house has 230 volts (plus or minus 10 volt tolerance)at the breaker panel. However, if you tried to stick weld with the 230 volts at breaker panel it would not work well as the voltage is too high even if you could prevent your breaker from tripping. Stick welding is a constant current process requiring roughly 35 volts at a constant amp setting. A Lincoln AC-225 draws roughly 45 amps from your house panel at wide open setting of 225 amps.

So we have for input: (230 volts household power input)(45 amps input) = (10,350 watts of potential work based on inputs, but it is unsuitable for welding so we use a welder transformer to convert to something useful).

Welder output uses a transformer to convert inputs to a useable output of: (35 arc volts) (225 amps constant output at wide open setting) = (7875 Watts of theoretical output still in AC power but it is power that is suitable for welding).

Notice that you lost potential Watts since you can not get more out of something than you put into it so there is conversion loss. Just like your car has more horsepower at the motor than it does at the wheels. Power loss in the transmission unit and driveline components to convert that engine power into something that you can actually use to move the car.

Now lets say you are lucky enough to have an AC/DC Lincoln buzzbox. From here there is even more conversion loss in rectifying the AC output from the Welding transformer into DC power. Lincoln AC-225 are 225 amp on AC output, but only 125 amp on DC output, because even more power is lost in the conversion of the AC power output to DC power output in the amps department as the arc voltage will still be roughly 35 volts either way.

Hopefully you have seen that as the voltage goes up on the inputs, the amps can go down and you still can have the same watts of potential work. Smaller amps on the inputs means the wires supplying it can be smaller which is beneficial in long wiring runs in order to cut wiring expense.

Some older special Miller Thunderbolt buzzbox welders could be connected to 115 volt input or alternately to 230 volt input. Regardless of which input voltage source connected, the Miller Thunderbolt would still output 235 amps at the required 35 volts or so for welding. Do the math though as it reveals a very important fact.

(115 volt input) (90 amp input) = (10350 watts of potential work based on inputs)
(230 volt input) (45 amp input) = (10350 watts of potential work based on inputs)

Can you imagine the size of the wire that would have been needed if this Miller Thunderbolt was connected to 115 volts in order to handle a 90 amp input load? Plus many houses in the 1950's and 1960's only had 60 amp total service to the entire house, so the capability of connecting this particular welder to 115 volts was impractical in almost all cases. The old Forney and Marquette welders got their start back in the 1940's under the premise that they would never pull more than 40 amps thus still leaving 20 amps to run the rest of the house - LOL.

 

[KennyG]

OK, one more way to look at this. The purpose of a welder is to produce heat. The energy transferred to heat is the amperage squared times the resistance. So you want small voltage and high amperage from a given available amount of electrical power. There is a practical limit to how low the voltage can go since voltage is the "electrical pressure" to overcome surface resistance.

On the other hand, when you are transmitting the power over wires, you want small amperage to minimize the resistance heating losses. That's why 220 wiring and 220 appliances are inherently more efficient than 110 volt. Europe and Asia have an edge on North America in this respect.

The losses from heating are why the large transmission lines operate at 345,000 volts, 765,000 volts or even 1.2 million volts. The amperage is a lot less than most people would think.

 

[rankrank1]

...The losses from heating are why the large transmission lines operate at 345,000 volts, 765,000 volts or even 1.2 million volts. The amperage is a lot less than most people would think....

Actually not quite correct . Yes, the amperage is relatively low for the immense voltage, but the reason these lines operate at the immense voltages is that they supply entire neighborhoods/city blocks of houses on wiring that is actually smaller than the supply lines running to your house. There is only one way to supply the amp load of that many households on small wires and that is to dratically increase the voltage on the feeder lines and then use step down transformers to supply each house or a group of houses. Really the transformers on the utility poles are doing the same thing as a transformer based welding unit does (i.e. Decrease the volts and increase the amps).

 

[grsthegreat]

You pay for Energy. Power [V x A] for a given time .... Kilo-Watt Hours. A 110 motor drawing 12.8A is using the same amount of energy as a 220 drawing 6.4. The electric power [Watts] each uses would be the same -- 220x6.4=110x12.8.
larry

nope nope nope

remember, a 120 volt motor is only pulling on one leg , and this moves meter concurrent with the amperage used. Now if the same machine is pulling 220 and using 1/2 the amperage, only 1/2 the current draw is registered on the meter.

Ive discussed this with the utility company and with my load calculations instructor and they agree. why do yo think large companies use 3 phase..they spread the power over 3 poles. their cost is less.

believe me, Ive talked this over with many different people, as i didn't believe it at first. This is the reason we try to balance a load in a panel. don't want alot of the continuous loads on only one leg. those dang meters move when they sense current thru ANY leg. If current is going thru both legs, it only registers as one leg. Very odd, but it has been confirmed.

 

[rankrank1]

nope nope nope

remember, a 120 volt motor is only pulling on one leg , and this moves meter the same amount of time as 1/2 the load across 2 poles will using 1/2 the amperage.

Ive discussed this with the utility company and they agree. why do yo think large companies use 3 phase..they spread the power over 3 poles. their cost is less.

believe me, Ive talked this over with many different people, as i didn't believe it at first.

Spyderlk is correct and grs great is wrong on this. A kilowatt is a kilowatt regardless of the voltage used to get there. Yes, the amps are different in each case, but a kilowatt is still a kilowatt. 3-phase simply allows one to go even smaller with supply wiring than 230 volt does.

Industries use 3 phase for the same reason that a homeowner uses 230 volts over 115 volts for heavier load applications like a range or clothes dryer. Many Industries have big machines that require huge amp draws and there is a practical limit to the size wire that you can feed the machine with in order to meet ampacity.