Total amperage in a service panel

[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.
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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