Total amperage in a service panel

[milesyoung]

I went back and reread your posts and I think this is the crux of the matter:

I believe 60 amps can be drawn from each leg at 120 volts for a total of 120 amps.

I have to ask you:
What if, on the transformer secondary, you instead had 240 separate 1 V windings with a 60 A breaker on each. That could supply the same amount of apparent power as a single 240 V winding with a 60 A breaker, right?

Would you then say you have a total of 14400 amps available or call that a 14400 amp service panel?

 

[256bits]

Okay, now I know where I went wrong. I was thinking current would flow through the neutral line, back to the transformer, and then split and into each leg of the transformer. But that's nonsense, right?

From the diagram you posted, Hot1 at the top, neutral, Hot2 at the bottom
You could measure voltages as 240v - 120v - 0v,
or as 120v - 0v - (-)120v
where currents and voltages are in phase.

If you measure the bottom leg as Hot2 to neutral ie as 120v - 0v, then that means it is just 180 degress out of phase from neutral to Hot2. ie 0 to (-)120v.

Current through the transfomer secondary would always be in one direction eiither Hot1 to Hot2 or revesred Hot2 to Hot1 depending upon the state of the cycle. so if you have loads on Hot1 and Hot2, when Hot1 current flows out of neutral, the current from Hot2 is flowing into the neutral and cancel out.

You could have a transformer where both currents from both legs flows through the neutral in the same direction. If the second leg is wiried in the counter direction as to the first leg, then any current from Hot1 or Hot2 would flow into or out of neutral and both currents would add. If both loads are equal the neutral would carry double that of Hot1 or Hot2 and the neutral wire would have to be of a greater thickness to compensate for I squared R heat.

 

[Averagesupernova]

Jim Hardy, something doesn't look right with your example of capacitive and inductive loads. Can't put my finger on your math though. My crude sketches show current canceling in the neutral with XC and XL loads. If I am correct, that means a service feeding lots of 120 V inductive loads will have a lot more current on the neutral than what it was designed for. And of course there is a much greater chance of this than there is with large capacitive loads on one leg. Also, the XC and XL load as you described form a series resonant circuit. Any more thoughts on this? I feel silly questioning a guy with your background.

 

[jim hardy]

Jim Hardy, something doesn't look right with your example of capacitive and inductive loads. Can't put my finger on your math though. My crude sketches show current canceling in the neutral with XC and XL loads. If I am correct, that means a service feeding lots of 120 V inductive loads will have a lot more current on the neutral than what it was designed for. And of course there is a much greater chance of this than there is with large capacitive loads on one leg. Also, the XC and XL load as you described form a series resonant circuit. Any more thoughts on this? I feel silly questioning a guy with your background.

Please do question me, I tend to reverse things often. Mild case of autism...

I never even thought about their being resonant. That might be a clue...
will have to chew on that a bit.
Were it not for the neutral connection we'd surely have a series resonant circuit with its substantial voltage gain. Probably the neutral connection prevents that.

Regarding phase, which was your question -
remember that wrt neutral the two driving voltages are 180deg out of phase, and that's a half turn apart... so currents a quarter turn behind one and a quarter turn ahead of the other could be in phase.
If I've invented some untrue math to justify that I haven't yet caught my mistake. I'll not be embarrassed or offended if you find it though, I'm here to learn things.. especially about myself.
The brain can accept things that are quite untrue and then invent justifications for them. Eternal vigilance...

If you find an error please post !

old jim

 

[milesyoung]

Jim Hardy, something doesn't look right with your example of capacitive and inductive loads. Can't put my finger on your math though. My crude sketches show current canceling in the neutral with XC and XL loads.

I'm not Jim, but I hope you won't mind my comments.

If you have equal impedances on both legs, the currents in both will be 180° out of phase (since the voltages from the center-tapped transformer are 180° out of phase), so you won't have any neutral current.

If you, for some odd reason, wanted to maximize the neutral current, you could advance the current in one leg by 90° and retard it by 90° in the other so the current phasors "line up" instead of being in opposite directions, which is what Jim did by specifying the load impedances as:

Z1 = 2∠+90
Z2 = 2∠-90

If I am correct, that means a service feeding lots of 120 V inductive loads will have a lot more current on the neutral than what it was designed for. And of course there is a much greater chance of this than there is with large capacitive loads on one leg.

As long as you keep the loads on both legs balanced, be they inductive or capacitive, there can't exist any neutral current.

In practice, in a typical residential home, any unbalance is probably not going to be of any real consequence. Jim's example was pathological in nature, i.e. extremes in terms of current amplitude and load unbalance.

Also, the XC and XL load as you described form a series resonant circuit.

The two loads are on separate legs, they can't resonante. They share a common return, but they can't exchange energy.

 

[jim hardy]

Also, the XC and XL load as you described form a series resonant circuit.

I thought quite a while on this.

MilesYoung is right
but I had to dig clear back to 1963 to get at the nuts&volts explanation. Seek time on this old gray drive ain't what it used to be...

observe that each of my two ohm impedances is paralleled by a transformer winding of pretty low impedance. So it's not quite your typical series resonant circuit.
My L and C are driven individually by fixed voltage sources.
Contrary to that a series resonant circuit is driven by a single voltage source.

In a series resonant circuit the individual L and C voltages are equal and opposite, but not fixed by the source the way they are in our circuit.
In fact in a series resonant each will be (source volts) X Q_{of circuit} , where Q = X/R.
The voltage gain of a series resonant circuit is real and can wreck big equipment where low resistance makes for a small denominator ergo high Q.

Recall that in a series resonant circuit the voltages across the L and C become equal in amplitude but opposite in phase.
Hence each can become far larger than the source voltage,
but they cancel out,
So terminal voltage remains low while current goes sky high.
Ohm's law holds, voltage across the capacitor and voltage across the inductor is I X Z and is huge but because they're 180° out of phase they cancel. You'd have to measure them separately.

But in our circuit the voltage across each element is fixed by the transformer winding.
So voltage cannot rise. So neither can the current. So they're not resonating even though it looks like they should.

They would resonate if you removed the connection from their junction to transformer centertap .

Maybe MilesYoung can summarize this one as nicely as he did the previous one !

old jim

 

[Averagesupernova]

Jim and Miles, all you have posted is exactly what I have thought. However, thinking about directions of currents in both the L and the C in various parts of the cycle I am getting mixed up somewhere. I just can't figure out where. I am 100% sure you are correct and 100% sure I am missing something to cause the contradiction in my head.
-
The part that does make sense: Take an inductor and capacitor with 2 ohms reactance and give each a very high Q. Disconnect the neutral and we would have a very large AC voltage across each component due to resonance hence a very high voltage between the neutral and where it was previously connected. You can't have a voltage between two points in a low impedance circuit like this and not expect current to flow when you hook said points together.

 

[Integral]

Not sure why no one has said Power factor

 

[meBigGuy]

Again I learn something I thought I knew (thanks for that). The 60 amp service can supply 60 amps on either or both legs. The neutral conducts up to 60 amps as required by mismatched loads. If the loads are matched it conducts zero (since the matched loads are a perfect voltage divider and the middle point is 0V). Cool.

 

[milesyoung]

Jim and Miles, all you have posted is exactly what I have thought. However, thinking about directions of currents in both the L and the C in various parts of the cycle I am getting mixed up somewhere.

Maybe an equivalent circuit that obfuscates the transformer bit will help (you probably have this on paper already, but just in case):

Consider the two voltage waveforms in the time domain superimposed atop each other. Their positive peaks are the time equivalent of 180° apart, so you have:

Positive peak of voltage across inductor -> 180° separation -> positive peak of voltage across capacitor or:
VL -> 180° -> VC

The positive peak of the current through the capacitor will lead its voltage by 90°:
VL -> 90° -> IC -> 90° -> VC

The positive peak of the current through the inductor will lag its voltage by 90°:
VL -> 90° -> IL -> 90° -> VC

So the currents are in phase and you'll have constructive interference in the neutral.

The neutral conducts up to 60 amps as required by mismatched loads.

As Jim showed, if you've recently discovered that your true purpose in life is to test for code violations through trial by fire, you can get up to 120 amps in the neutral for severely unbalanced loads.

 

[Averagesupernova]

As Jim showed, if you've recently discovered that your true purpose in life is to test for code violations through trial by fire, you can get up to 120 amps in the neutral for severely unbalanced loads.

Something I have never considered because there are never capacitive loads that would be capable of overloading the neutral in this way. I never should have doubted it since I have measured the neutral with balanced inductive loads and found they do in fact cancel neutral current. Very interesting.

 

[jim hardy]

Thanks guys, I sure didn't expect this much interest in that absurd extrapolation to the logical extreme .
and your equivalent circuit is just right miles, 60 amps through each Z at zero power factor.

'Pathological' was a great adjective - only practical use is for a thought experiment.

You folks made me re-examine my fundamentals of resonance too.

This was fun. THANKS ALL !

old jim