Too much for 200A

[Tim Voss]

Ok, I re-read you post, and I think you are saying I CAN put 400 Nameplate rated amps of 120v equipment on the 200amp 240v service. Is that correct?

[Tim Voss]

Ok, so now I have all this 120v equipment plugged-in.

I'm not measuring or talking about amps in any wire. I'm reading the rating printed on the equipment. You are saying I can't add? Actually it's quite easy. Toaster 1 says "20 amps", toaster 2 says "20 amps", etc... 20+20+...

When I add that up, it adds to 400.

[Jerry Peck]

When I add that up, it adds to 400.

Tim,

I can go (legally - speed limits):

70 on I-95
70 on I-4
45 on SR 40
185 is what that adds up to.

*I* can add numbers too. Just means nothing.

[Tim Voss]

Actually, means I can plug-in 400 Nameplate rated amps of 120v equipment on my 200a 240v service.

It's a good way to know if it's "Too much for 200A" when I don't have a meter handy.

[Tim Voss]

You see, it actually does mean something. When I plug-in 401, I have overloaded the breaker, but when I plug in 400 I haven't.

[Tim Voss]

There must be something special about that number, 400.

[Jim Zborowski]

Tim, What I'm getting out of your last post is you need 120 volts and 1 amp to light a bulb. Then, if you have 2 bulbs you need another 120 volts and another amp to light that bulb, and so on. So the logic then is that each bulb " consumes " the entire 120v/1 amp to light.
Unfortunately you have to consider Volts, Amps, Watts, and Resistance in a circuit. You also have to consider parallel, series, and series/parallel loads
( resistance ). If you have 2 - 100 ohm resistors in series the resistance is added. If you place them in series, the resistance is divided. This then changes the power ( load ) requirement between a series and parallel circuit.
Also, for your scenario to work, the first bulb would need a filament the size of ooo service wire to keep from burning out.

[Tim Voss]

Hi Jim, did you read the entire thread? We've been through all that.

No, what I'm getting at is quite simple and doesn't require much calculation or detailed knowledge of electric theory.

I have 400 standard 120watt, 120volt, 1amp light bulbs, or 20 standard 2400watt, 120volt, 20amp toasters, doesn't matter. The volt and amp rating is stamped right on the nameplate.

For 120volt loads, I can add up 400amps of Nameplate rating and put that on a 200amp 240v service. But if I try to put 401amps of Nameplate rating, that will overload the breaker.

And actually, the light bulb does "consume" the entire 120v 1amp that goes through it. P = va = 120v * 1a = 120watts

[Bruce King]

Lets use twenty, 20 amp 120V circuit breakers each feeding a 20 amp load device as an example, with all turned on.

With 10 breakers on each leg of a regular panel there are only 10 current paths in the panel, each leg has a load in series with the other leg.

Each load is in series with a load on the other leg so ONLY 10 current paths are present in the panel.

Each current path has 20 amps flowing: 10 paths x 20 amps = 200 amps.

Its simply because you have to figure two loads on each path because that is the way the current travels. Each current path travels through one 20 amp breaker on opposite legs with a total of two breakers per path.

Yes, if you add up the loads by looking at it seperately you get 400amps but you can not use this method with series loads.
You have to use VA or watts as the total load so that it does not appear to be implying 400 amps is flowing in any one wire.

[Tim Voss]

Yes. I have always said there is not 400a in one wire. It is the amps of all the 120v loads added together that give you 400amps.

[Bruce King]

You have to use VA or watts when adding up series loads, never use the term 400A, its just not technically correct when discussing this example.

When you state something in "amps", you should only be referring to a current that can be actually measured.

Its one of the reasons that the term "watts" is available for use, this example we have been tossing around is a perfect example of why using the proper terminology at the right time is important.

[Tim Voss]

Actually, it is technically correct in this example.

When you have 120v loads, it does not matter if they are in a 120v circuit or a 240v circuit. The v in the va will always be 120v, because that is the voltage drop across the load. For example, the 120watt, 120volt, 1amp light bulb. Even when you put 2 in series on 240volts, there is a 120v drop across each bulb and 1amp flowing through each bulb. So the va for each bulb is 120v * 1a = 120watts. The Total va for the circuit is (120v * 1a) + (120v * 1a) = 240watts.

You can use math on that equation to come up with some equivalent equations.

(120v * 1a) + (120v * 1a) = 240v * 1a = 240va
(120v * 1a) + (120v * 1a) = 120v * (1a + 1a) = 120v * 2a = 240va

BOTH are equally correct, and in fact, it is MORE appropriate to express it as 120v * 2a BECAUSE it is a 120v light bulb, NOT a 240v light bulb. You don't want to give the impression that you can put that bulb on a 240v circuit, it will blow up.

[Jerry Peck]

it is MORE appropriate to express it as 120v * 2a BECAUSE it is a 120v light bulb, NOT a 240v light bulb.

Tim,

Now THAT is *COMPLETELY* wrong.

You do not have 2 amps at 120 volts.

Think about what you have: you have 1 amp at 120 volts twice (each lamp has 1 amp through it and each lamp has 120 volts across it).

Thus, the better way to say it would be 1 amps at 240 volts (because this *IS* what you have ... whether or not the center point between the two lamps was, or was not, grounded.

You don't want to give the impression that you can put that bulb on a 240v circuit, it will blow up.

You are giving the impression that you are using a 240 watt lamp at 120 volts, and you are not. You are using 120 watt lamps at 120 volts.

[Tim Voss]

* SIGH *

I guess it's time for Electricity 002: Calculating power in a multi-load circuit

Refer to the included diagram for this discussion.

The total power of a circuit is equal to the sum of the power for each load.

P(total) = P(1) + P(2) + ...

In circuit A, there are 2 loads with an ammeter and voltmeter showing the measured values.
Total power in circuit A
P(total) = P(1) + P(2) = (120x200) + (120x200) = 120x400

In circuit B, there are 2 loads with an ammeter and voltmeter showing the measured values.
Total power in circuit B
P(total) = P(1) + P(2) = (120x200) + (120x200) = 120x400

From "outside the box" you CAN NOT determine any difference between the circuits. All measured values are the same. They both deliver 400amps of power at 120volts. They are both equivalent.

For the total power calculation, it does not matter if the circuit is parallel or series, they are both driving the same exact loads (load 1 and 2).

Another way to calculate the Total Power is to reduce the multiple loads into an equivalent theoretical load. Now if we look inside the box, for circuit A, the total equivalent power is 400amps at 120volts and for circuit B the total equivalent power is 200amps at 240volts, but the actual power is 400amps at 120volts, because that is what is actually measured at the loads.

Jerry, you suggest that you can express the power in terms of amps at 240v, and I agree.

But you are * COMPLETELY * wrong when you say that you cannot also express the power in terms of amps at 120v.

[Jerry Peck]

* SIGH *

I guess it's time for Electricity 002: Calculating power in a multi-load circuit

Let's try Electricity 000, which should have come before Electricity 001 and Electricity 002.

When loads are in series, the current flow is not additive for a total amperage current flow. There is the same current flowing through each load, hence "in series".

There is a voltage across each load with the voltages being additive for the total voltage.

Jerry, you suggest that you can express the power in terms of amps at 240v, and I agree.

Yes, that is what I've been saying.

But you are * COMPLETELY * wrong when you say that you cannot also express the power in terms of amps at 120v.

That is not what I've said anywhere. IN FACT *I* have been telling YOU to refer to those loads in power terms (watts for resistive loads such as lamps, VA for other ac loads) ... it has been YOU who has REFUSED to use those power terms and to instead insist on only using the term "amps" in trying to add up series loads.

You seem so close to getting it, then you back away and insist on adding up "amps" for series loads. Why do you refuse to use the power terms you stated above which should be used?

[Tim Voss]

No Jerry, YOU are wrong. Did you even look at the diagram?

In the series circuit, the voltages don't add. Look at the measured voltage for Load 1, it is 120 volts. Look at the measured voltage for Load 2, it is 120 volts.

You don't even seem close to getting it.

[Tim Voss]

Jerry, what part of P(total) = P(1) + P(2) don't you understand?

[Jerry Peck]

When loads are in series, the current flow is not additive for a total amperage current flow. There is the same current flowing through each load, hence "in series".

There is a voltage across each load with the voltages being additive for the total voltage.

No Jerry, YOU are wrong. Did you even look at the diagram?

In the series circuit, the voltages don't add. Look at the measured voltage for Load 1, it is 120 volts. Look at the measured voltage for Load 2, it is 120 volts.

Tim,

I'll make an "assumption" here: YOU drew the diagram, right?

Now (presuming that my "assumption" is correct) you need to learn to understand what you drew.

Circuit A is a parallel circuit, the current flow is additive, the voltages are the same.

Circuit B is a series circuit, the current flow is the same, the voltages are additive.

Don't believe me when I say :

When loads are in series, ...

There is a voltage across each load with the voltages being additive for the total voltage.

?

Look at Circuit B: You are showing Load 1 with 120 volts across it, you are also showing Load 2 with 120 volts across it, you are also showing the additive total of those voltages across each load as being 240 volts.

That is correct.

Thus YOU DREW it right (if you drew it), but apparently you do not understand it (did you copy it from someone else?) because you are now stating:

In the series circuit, the voltages don't add.

and

Look at the measured voltage for Load 1, it is 120 volts. Look at the measured voltage for Load 2, it is 120 volts.

Yeah, I have looked at them, and in Circuit B (the series circuit) Load 1 is connected to Load 2, with the voltage across Load 1 (120 volts) added to the voltage across Load 2 (120 volts) to gives the total voltage (240 volts), which, in Circuit B is 120 volts + 120 volts = 240 volts. Did you catch the "+" sign? That's for "addition".

Now, if that is not additive ... I don't know what kind of "new math" you are using.

*IF* *YOU* are referring to Circuit A, then you are even more confused than I thought as Circuit A is a parallel circuit, not a series circuit.

[Tim Voss]

In the series circuit, what is the power in Load 1?

120w = 120v * 1a

So what is the voltage across Load 1?

[Tim Voss]

Ok Jerry, let me try to explain this another way.

Let's say you cannot see "inside the box" because it's a locked panel and you have no idea what is in there.

What would you say is the total power for loads 1 and 2 on circuit B. Just look at the diagram and give me an honest answer.

[Jerry Peck]

In the series circuit, what is the power in Load 1?

120w = 120v * 1a

So what is the voltage across Load 1?

Ok Jerry, let me try to explain this another way.

Good, another way would be better as you've answered you own question in the above.

Let's say you cannot see "inside the box" because it's a locked panel and you have no idea what is in there.

What would you say is the total power for loads 1 and 2 on circuit B. Just look at the diagram and give me an honest answer.

First thing you do is use a VOM and check the voltage across each load and across both loads, getting 120 volts, 120 volts, and 240 volts (talking about Circuit B here).

The second thing you do is use a clamp-on ammeter to read the current through Load 1 and through Load 2.

Now, though, even doing that would not tell you whether or not they were in series unless you also knew the supply voltage (which we do not know because it is hidden inside the box).

Based on your drawing, BOTH Circuit A and Circuit B would show 120 volts at each load, BOTH Circuit A and Circuit B would show the same 200 amps current through each load.

The only way to find out if one was in series or not would be to disconnect one load, in the series Circuit B (as drawn with no common grounded connection), the other load would also 'shut off', whereas in the parallel Circuit A, the other load would 'stay on' ... HOWEVER ... if Circuit B had a common grounded neutral point, then the other load would also 'stay on' - ruining your ability to determine what was what based on those tests.

Now, back to your question:

What would you say is the total power for loads 1 and 2 on circuit B. Just look at the diagram and give me an honest answer.

For Circuit B *AND FOR CIRCUIT A*, the "total power for loads 1 and 2" would be 24,000 VA + 24,000 VA = 48,000 VA. If the loads were purely resistive loads, that would also be 24,000 watts + 24,000 watts = 48,000 watts ... REGARDLESS which circuit was being referred to, Circuit A or Circuit B - the "total power" (which is what you asked for) would be the same.

I've been saying that all along, so I'm not sure why you are asking that question.

Now it is *my turn* to ask questions (the cover to *the box* has been removed to allow you to see and check what is inside *the box*):

a) In Circuit A, what is total current flow through Load 1? Through Load 2? Through Loads 1 and 2?

b) In Circuit B, what is the total current flow through Load 1? Through Load 2?, Through Loads 1 and 2?

[Corn Walker]

What a fun thread to reply to for my first post.

I think you need to get away from measuring the voltages between each load and common. You will always measure 120V between each leg and common and 240V between the two legs.

What you need to do is follow the electrons, and here is where phase is important and what it means to "balance" the panel.

If you have a 1A load attached from leg A (0 degree phase) and common 1A worth of electrons will travel from leg A to common. If you place a second 1A load on leg A you will now have 2A worth of electrons traveling from leg A to common.

Okay, now lets put a 1A load attached from leg B (180 degree phase) to common. Now what you have is 1A worth of electrons traveling from leg A to common and 1A worth of electrons traveling from leg A to leg B. You do NOT have 3A traveling from A and B onto common.

In a balanced panel all of the electrons are flowing from leg A to leg B and none are flowing back on common. Of course, in an AC system, these voltages flip direction every few milliseconds (so you have electrons occasionally flowing from B back to A) but the principle is not changed because of it.

Corn Walker
(former Physics Teacher)

[Tim Voss]

a) total current flow through load 1 = 200a, total current flow through load 2 = 200a, total current in circuit A = 400a
b) total current flow through load 1 = 200a, total current flow through load 2 = 200a, total current in circuit B = 200a

But you didn't really answer my question.

First thing you do is use a VOM and check the voltage across each load and across both loads, getting 120 volts, 120 volts, and 240 volts (talking about Circuit B here).

I don't think you are really going to try and measure the voltage across both loads, nor start disconnecting them to see if they are in series or parallel. That is completely impractical in a real situation. In my example there are only 2 loads, but if there where 30 circuits in a panel that is 2.6525286 × 10^32 combinations you would have to measure.

No, I think you would use a VOM and check the voltage across each load (only).

And you have measured the current with your clamp-on ammeter.

Then what would you do?

[Jerry Peck]

But you didn't really answer my question.

I did.

You asked: "What would you say is the total power for loads 1 and 2 on circuit B. Just look at the diagram and give me an honest answer."

I answered (I'll add bold and underlining for you): "For Circuit B *AND FOR CIRCUIT A*, the "total power for loads 1 and 2" would be 24,000 VA + 24,000 VA = 48,000 VA. If the loads were purely resistive loads, that would also be 24,000 watts + 24,000 watts = 48,000 watts ... REGARDLESS which circuit was being referred to, Circuit A or Circuit B - the "total power" (which is what you asked for) would be the same."

[Tim Voss]

Ok, that is correct, but show your work please.

How did you calculate 24,000 VA for Load 1 and 2?

[Jerry Peck]

Ok, that is correct,

I know it is correct.

but show your work please.

How did you calculate 24,000 VA for Load 1 and 2?

Your drawing shows 200 amps flowing through a load with 120 volts across it.

120 volts times 200 amps equals 24,000 volt-amps

The rest I already showed you.

But here it is AGAIN.

24,000 VA + 24,000 VA = 48,000 VA

I know you are trying to eventually show that you get "400 amps", but you don't.

You can keep playing these games, the outcome will never be "400 amps".

Because you never have "400 amps" on that circuit we were discussing.

Are you trying to use up another 100 posts having me show you that?

[Tim Voss]

Yes, I'm trying to help you "see" the 400amps, as you requested.

Ok, so now you have said:

(120v * 200a) + (120v * 200a) is the total power for circuit B.

So we have two 120v loads on circuit B using 200amps each. Now if someone said, "I have two 120v loads here. How much amps at 120v do I need to power these loads? Hmmm, looks like I need 200amps for the first one and another 200amps for the second one."

Would you not say, "You need 400amps at 120v to power those loads"?

[Corn Walker]

The way I would explain (incorrectly - but it helped to get the point across) power phasing to my students is that you have 120V pushing 200A through the circuit and -120V pulling that same 200A through the circuit. The total amount of Power available in the circuit to perform any work is 48000VA. That is NOT to say you have 400A total current flowing though the circuit - you don't.

If you start with a 120V 200A load placed on one leg the load will be referenced to common (we'll say this leg pushes 200A onto common). If you then add a 120V 200A load to the second leg then this second circuit doesn't push an additional 200A onto common; instead it pulls the same 200A through the second load that was pushed through the first. The combined push/pull force is additive (240V) allowing you to do more work (48000VA) with the same 200A (assuming ideal resistive loads).

You can say you have two 120V 200A loads on the circuit but you can not say you have 400A on the circuit because the amperages on the load side of the circuit can't be added, only the force or the power.

Corn Walker
(former Math teacher too)

[Jerry Peck]

Yes, I'm trying to help you "see" the 400amps, as you requested.

There is nothing to "see", because there are no "400 amps" anywhere there.

Ok, so now you have said:

(120v * 200a) + (120v * 200a) is the total power for circuit B.

So we have two 120v loads on circuit B using 200amps each. Now if someone said, "I have two 120v loads here. How much amps at 120v do I need to power these loads? Hmmm, looks like I need 200amps for the first one and another 200amps for the second one."

That's where you become incorrect, ending up totally incorrect here ...

Would you not say, "You need 400amps at 120v to power those loads"?

No. (This is the part where you "ending up totally incorrect".)

See, first, I would have to know how the circuits were wired (see "That's where you become incorrect" above and your quote above it), then I would ...

... AS I HAVE BEEN for the past 100 posts or so ...

... explain that THERE ARE ONLY 200 amps in that circuit.

Do you understand yet? Or do we need to try for 200 posts?

[Tim Voss]

I'm not saying I have 400amps on the circuit, I'm saying I have two 120v 200amp loads on the circuit and that is equal to 400amps of 120v loads.

[Jerry Peck]

The way I would explain (incorrectly - but it helped to get the point across) power phasing to my students is that you have 120V pushing 200A through the circuit and -120V pulling that same 200A through the circuit. The total amount of Power available in the circuit to perform any work is 48000VA. That is NOT to say you have 400A total current flowing though the circuit - you don't.

If you start with a 120V 200A load placed on one leg the load will be referenced to common (we'll say this leg pushes 200A onto common). If you then add a 120V 200A load to the second leg then this second circuit doesn't push an additional 200A onto common; instead it pulls the same 200A through the second load that was pushed through the first. The combined push/pull force is additive (240V) allowing you to do more work (48000VA) with the same 200A (assuming ideal resistive loads).

You can say you have two 120V 200A loads on the circuit but you can not say you have 400A on the circuit because the amperages on the load side of the circuit can't be added, only the force or the power.

Corn Walker
(former Math teacher too)

Maybe you can do a better job of explaining that to Tim than I have been able to ... I've never been a Physics Teacher nor a Math Teacher - maybe that's my problem?

[Carl Eisen]

Jerry,

You must have the patients of a Saint. I would have thrown in the towel a hundred post or so ago.

[Tim Voss]

Jerry,

How do you explain the fact that generator manufacturers agree with me and list the capacity of their generators in BOTH 120v and 240v amps?

Baldor Generator 125amps at 240v OR 250amps at 120v
Honda Generator 20.8amps at 240v OR 41.6amps at 120v
Grainger Generator 41.6amps at 240v OR 83.2amps at 120v

[Jerry Peck]

I'm not saying I have 400amps on the circuit, I'm saying I have two 120v 200amp loads on the circuit and that is equal to 400amps of 120v loads.

You don't, though,

You have two 120 volt loads (you agree with that) at 200 amps (you agree with that), and you thus have two 120 volt loads on a 240 volt system with 200 amps.

As as as ... and this is where you veer off into netherland ... you want to start comparing those two 200 amp 120 volt loads on the 240 volt circuit to two 200 amps 120 volt loads on two 102 volt circuits?

You need to switch to ... and we've been here before, even you have ... power ratings.

I.e., each 200 amp 120 volt load has a power rating of 24,000 VA, two of them have a power rating of 48,000 VA. That does not change regardless if they are connected to two 120 volt circuits or are connected across one 240 volt circuit - THE POWER does not change.

In one case (your parallel circuits) you are drawing 200 amps + 200 amps = 400 amps. You have two circuits with 24,000 VA on each, or 48,000 VA.

In the other case (the one where we started and are still discussing - the series 240 volt circuit) you have 200 amps ... period. You have one circuit with 48,000 VA on it.

The POWER does not change.

[Tim Voss]

Corn Walker,

All I'm trying to say is that the force of the power is equal to 400amps at 120volts. Is that not true?

[Tim Voss]

Jerry,

How do the loads "know" they are on a 240v circuit? They are 120v loads. ( Not to get sidetracked, please answer my question about the generators.)

[Jerry Peck]

Jerry,

How do you explain the fact that generator manufacturers agree with me and list the capacity of their generators in BOTH 120v and 240v amps?

Baldor Generator 125amps at 240v OR 250amps at 120v
Honda Generator 20.8amps at 240v OR 41.6amps at 120v
Grainger Generator 41.6amps at 240v OR 83.2amps at 120v

Tim,

They are not agreeing with you.

They are stating the MAXIMUM CAPACITY of the generator, how you hook it up is up to you.

They are saying that *IF* you hook it up as in your parallel circuit drawing, you will get twice the current at half the voltage than *IF* you hook it up in the series circuit, where you will get twice the voltage and half the current.

You will notice they are talking about the maximum POWER the generator is capable of, and POWER does not change (see my other post above), and then giving you the two options of how you might wire it up (parallel - 120 volt circuits, or series - 240 volt circuit).

[Jerry Peck]

All I'm trying to say is that the force of the power is equal to 400amps at 120volts. Is that not true?

Tim,

Now you are talking POWER again.

Make up your mind.

POWER does not change, the voltage doubles, the current drops by half, and vice versa.

You cannot, however, say you are getting 400 amps through two 200 amps loads at 120 volts each on a 240 volt circuit (which is where all of this started).

[Tim Voss]

No Jerry, these are service backup generators that are hooked up as split-phase 240v emergency power backup. And yet, for the Baldor generator you can get 250a of 120v out of that 240v service.

How do you explain that?

[Corn Walker]

All I'm trying to say is that the force of the power is equal to 400amps at 120volts. Is that not true?

When I taught I would always tell my students to check their units.

Volts is a measurement of force
Amperes is a measurement of charge/sec
Power is a measurement of force*charge/sec

If you're talking about abstract numbers outside of the system then you can say 400A at 120V (i.e. 48000VA) is the same as 200A at 240V and is the same as 4800A at 10V which is the same as 48000A at 1V (Quiz: what size Cu wire is required for this assuming a 100m length?). Whenever you say this, however, you aren't talking about force or volume but power (aka work). The force in the first example is 120V and the force in the last example is 1V. The power in both examples is the same: 48000VA.

*Within* a system, however, you can't abstract charge/sec from force. You can't talk about the amperes in the system without also talking about the volts required to move those amps. In the case of a main panel with 200A service, we're talking about 200A at 240V. If you attach a 200A*120V load to one leg and then balance it with another 200A*120V load on the other leg, you are moving 200A*240V. In your circuit when you attach the second load you change the phase angle of the force vector, giving you 240V force for the combined loads. The current stays the same - 200A.

It is convenient when we talk about circuits to refer only to amps because the voltage is implied in a residential panel (not an assumption I'd make in a commercial panel). It is nonetheless incorrect to refer to only amps when doing the math to calculate the total load of a panel. When adding loads you need to consider the phase angle of the load to determine how much current will travel over each leg and common. 130A*120V at 0 deg + 70A*120V at 180 deg results in 70A*240V phase balanced power + 60A*120V unbalanced power.

Theoretically the 4/0 supply cable could have an undersized common conductor (2AWG - much like they do with ground conductors) with the assumption that your panel will never be more than 50% out of balance at full load. In practice, however, it is a safer bet to assume a 100% out of balance panel and size the conductor the same as the mains.

As for the generators, they are talking about the amount of power they can generate (e.g. 4800VA) and explaining it in terms the user will likely understand (40A at 120V or 20A at 240V). That is, if the user connects a 40A*120V load to the generator it will reliably produce the required power. If the user connects a 25A*240V load to the generator they can expect a voltage drop (or the circuit to open if it has an overload protector).

As a practical analogy, let's assume I can push 200 pounds up onto the 1st story roof using a pole and you can pull 200 pounds up onto the second story roof using a rope. If we were both on the ground pushing or both on the second story roof pulling we could move 400 lbs from the ground to the first story or from the first story to the second. But because I'm on the ground and you're on the roof (180 deg phase separation) we can only each move the 200 pounds one story or together move the 200 pounds from the ground to the second story.

Corn Walker

[David Banks]

Corn Walker. What a cool name.

[Tim Voss]

Corn Walker,

As for the generators, they are talking about the amount of power they can generate (e.g. 4800VA) and explaining it in terms the user will likely understand (40A at 120V or 20A at 240V). That is, if the user connects a 40A*120V load to the generator it will reliably produce the required power. If the user connects a 25A*240V load to the generator they can expect a voltage drop (or the circuit to open if it has an overload protector).

That's ALL I've been trying to say. Jerry says you CAN'T say that. Have you read the whole thread from the beginning? Are you saying I can connect 400amps of 120volt devices to a 200amp 240volt split-phase service? (Better watch out what you say, or Jerry will get you.)

[James Duffin]

The power company bills you kw you use per hour (khw). They don't care if the loads are 120 volt or 240 volt. The charge is the same for a 2-pole 20 amp breaker or two 1-pole 20 amp breakers if the breakers are all loaded to the max.

[Jerry Peck]

That's ALL I've been trying to say. Jerry says you CAN'T say that.

If I understand what Corn Walker said, he also said you can't say that.

(underlining is mine)

If you're talking about abstract numbers outside of the system then you can say 400A at 120V (i.e. 48000VA) is the same as 200A at 240V and is the same as 4800A at 10V which is the same as 48000A at 1V (Quiz: what size Cu wire is required for this assuming a 100m length?). Whenever you say this, however, you aren't talking about force or volume but power (aka work). The force in the first example is 120V and the force in the last example is 1V. The power in both examples is the same: 48000VA.

*Within* a system, however, you can't abstract charge/sec from force. You can't talk about the amperes in the system without also talking about the volts required to move those amps. In the case of a main panel with 200A service, we're talking about 200A at 240V. If you attach a 200A*120V load to one leg and then balance it with another 200A*120V load on the other leg, you are moving 200A*240V. In your circuit when you attach the second load you change the phase angle of the force vector, giving you 240V force for the combined loads. The current stays the same - 200A.

[James Duffin]

The volts don't matter in this application. You have the same wattage available at all times. In my area 48KW cost you $4.80 per hour no matter if you have a 200 amp 240 welder connected or 240 two hundred watt 120 volt light bulbs connected.

[Tim Voss]

Actually, I thought Corn Walker was agreeing with me. He said:

As for the generators, they are talking about the amount of power they can generate (e.g. 4800VA) and explaining it in terms the user will likely understand (40A at 120V or 20A at 240V). That is, if the user connects a 40A*120V load to the generator it will reliably produce the required power.

Seems like he is saying it is correct to say it both ways.

So how can Baldor Generators say 250amps at 120volts when they only have a double pole 125amp breaker on their 240v generator?

[Corn Walker]

So how can Baldor Generators say 250amps at 120volts when they only have a double pole 125amp breaker on their 240v generator?

Explaining how auto-switching circuits work is perhaps a bit beyond this topic but suffice it to say they operate differently than the utility feeds in a typical home. The Baldor generators have the ability to supply either 120V or 240V to a standard L1430R receptacle. Once you add voltage switching circuitry to the mix you're no longer comparing apples to apples.

Still they're not exactly being precise here - but they don't need to be. It would be exceedingly difficult to connect a total load of 250A to a single 120V leg.
The two 125A breakers are on independent power legs (fed by independent generator coils) that are kept 180 deg out of phase with each other to provide split-phase power. Baldor is answering a consumer question here: how much stuff can I run with this generator? The consumer will look at the name plates on the table saw, work lights, air compressor, and battery chargers (all of which operate at 120V) and add up the current draw of those devices. They get a number (62A) and see that it's less than the rating of the generator and they're good to go.

Do note, btw, that the primary unit by which generator power output is rated is watts, not VA.

[Tim Voss]

They are not supplying 250a to a single 120v leg, they are supplying 125a to one leg and 125a to the other leg. The unit doesn't have a receptacle, it is hardwired to the service panel and operates just like split-phase utility power.

Baldor is answering a consumer question here: how much stuff can I run with this generator? The consumer will look at the name plates on the table saw, work lights, air compressor, and battery chargers (all of which operate at 120V) and add up the current draw of those devices. They get a number (62A) and see that it's less than the rating of the generator and they're good to go.

That is EXACTLY the question I was answering. You can add up all the amp ratings on the name plates of your 120v equipment. And if it is less than 400A, then you can run that on your 200A 240V service, correct?

Jerry says I can't say it that way.

[Corn Walker]

That is EXACTLY the question I was answering. You can add up all the amp ratings on the name plates of your 120v equipment. And if it is less than 400A, then you can run that on your 200A 240V service, correct?

Jerry says I can't say it that way.

Well... it depends.

That's where the whole balancing thing comes into play. Once you reach 1/2 of the rated capacity you need to be concerned about where you're connecting those loads. If the loads were all 240V there'd be no problem since they would automatically balance across the two service legs and you'd have up to 200A of current draw to play with. 120V loads, on the other hand, need to be planned (hence why electrical inspectors require load calculations from the electrician laying out the panel).

If you were to ask me, "can I run 205A of 120V equipment on a 240V 200A service?" I'd still have to say it depends. There's no definite answer you can give to such a question - you need to know where each of the loads is to be connected. Less than the rated capacity (say 198A total of 120V equipment) and you can be unequivocal - above that you need to be cautious.

Now practically speaking, and bringing this back to home inspection, I would not be concerned if I saw 400A worth of breakers on a 200A Mains panel. However I would be concerned if I saw electric space heaters in every room since many homes share a single circuit for two or three bedrooms. In my own house every bedroom has 12/2 home run back to the panel for receptacles (on AFCI breakers at that) and an additional 12/2 home run for lighting serving no more than two rooms. Perhaps it was a overreaction to what was there before - two 20A and two 15A circuits feeding the entire house.

Corn Walker
tentatively unequivocal

[Tim Voss]

Right, I left out "Perfectly Balanced". We covered that before. So, yes, then, as long as it is balanced.

[Jerry Peck]

If you're talking about abstract numbers outside of the system ...

*Within* a system, however, you can't abstract charge/sec from force.

Tim,

You and Corn Walker can sort out the abstract numbers, it is getting to abstract for me, since we were initially talking about within the system.

Let me know if Corn Walker makes you feel good about your position ... I'm abstracting out of here ...

[Brandon Chew]

This thread is like a vampire in a bad horror film. No matter how many times you kill it, it just won't die.

[Tim Voss]

Ok, Jerry, but the part about running 400A of nameplate rated 120V equipment on the 200A 240V service is real. I guess since you already agreed that was possible (but to be fair, meaningless) it's a good time for you to "abstract" out.

[Jerry Peck]

it's a good time for you to "abstract" out.

*I* am "abstracting out", as I said, however, *YOU* have already "abstracted out".

As soon as you stepped out of talking about what we were talking about, i.e., "within the circuits" as Corn Walker put it, you "abstracted out" ... go for it man.

96,000 volts at 0.50 amps = 48,000 VA.

As does 1 volt at 48,000 amps.

[Tim Voss]

No, I'm still talking about the same 200A 240V split-phase service we started with. The same service you said you could put 400 name plate amps of 120v devices on.

[Corn Walker]

Only 14 more posts after this one and we can hit 200. Come on team, we can do it!

[Bruce King]

I thought you could only hit 200 posts on the subject of "how to test a garage door"

[Jim Zborowski]

#188
Corn, I like your annalogy.

[James Duffin]

I'll help reach the 200 mark.

I still say that the volts and amps don't matter. The PC transformer determines how much power is available. On a 200 amp service it is 48KW at 120 volt, 240 volt, or a combination of both.