Amps on the neutral at panel?

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brian john said:
Actually you will seldom have leakage current on the neutral generally this is on the ground, fault current should be for a short duration for a phase to neutral fault, that is till the OCP operates to clear the fault.
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Seldom... but that is not the same as ipso facto never. And a high impedance line-to-neutral fault can exist without opening the OCPD.
 
don_resqcapt19 said:
But only on a three phase system. The harmonics do not cause increased neutral current on single phase systems.
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I often wonder about this. Harmonics can occur at any multiple of the base frequency. On three phase systems, the ones that add up are triplens, i.e. every third multiple of the system frequency.

Say you have a switching supply that triggers 3/4 of the way into the positive cycle on L1 and drops out at the zero crossing, and the same on on Line 2 (using neutral as the reference as to positive and negative), on a 120/240 system. Say each is drawing 1A rms. The loads are drawing current 180? apart, so all the current has to pass on the neutral. The current on the neutral would be 2A rms!!!
 
Allspark said:
Just hall can lites 3 tube ballast - hall recpts- and 4 bathroom fan lite heat combos using around 25 cks in panel but there are alot of neutral shareing going on
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If you have Excel (or an app that works with Excel's .xls format), downloading my Neutral Calculator may help. However, it does not take harmonic currents into account for the result.
 
Neutral load calculation is simple on a 3phase Y:
add the 2 lowest phase current readings, then subtract this from or to the highest phase reading.

lets say
Phase A= 19 amps
Phase B= 25 amps
Phase C= 10 amps
19+10=29-25=4amps on the neutral

Heres where it comes out reverse:

Phase A=5 amps
Phase B=10 amps
Phase C=20 amps
the two lower readings are the 5 and 15 so
5+10=15 reverse it to take the 15 from the 20 = 5 amps on the neutral

want to see an unbalanced panel figure out balanced?
A=5
B=10
C=15
5+10=15-15=0 amps on the neutral
This will only happen with a Y system
you said:
but there are alot of neutral shareing going on
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Check to make sure that each ungrounded (hot) conductor that shares a common neutral is on a different phase. we call these Multi-wire circuits, very common in commercial. If two or more ungrounded conductors share a neutral from the same phase, the neutral will be carrying the sum of all currents loaded on each circuit, this can overload the neutral, if on separate phases, then the neutral Carries the differential current, as in the above equation's
 
hurk27 said:
Neutral load calculation is simple on a 3phase Y:
add the 2 lowest phase current readings, then subtract this from or to the highest phase reading.
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This is an inaccurate method... perhaps very inaccurate.

For example:
hurk27 said:
lets say
Phase A= 19 amps
Phase B= 25 amps
Phase C= 10 amps
19+10=29-25=4amps on the neutral
Click to expand...
Without any account of non-linear load contributions and non-normal-operating currents, and assuming a unity power factor, the actual current on the neutral conductor is 13.08A.

Here's the basic formula to calculate neutral current.
NeutralCurrentFormula.gif

Note this formula assumes unity power factor. For unequal power factors, vector math or diagramming is necessary... or take the easy route and use my Excel Neutral Calculator.
 
Smart $ said:
... Say you have a switching supply that triggers 3/4 of the way into the positive cycle on L1 and drops out at the zero crossing, and the same on on Line 2 (using neutral as the reference as to positive and negative), on a 120/240 system. Say each is drawing 1A rms. The loads are drawing current 180? apart, so all the current has to pass on the neutral. The current on the neutral would be 2A rms!!!
Click to expand...
If I understand what you said, the two loads don't occur at the same time so they do not add on the neutral. Only one load is on at a time.
 
don_resqcapt19 said:
If I understand what you said, the two loads don't occur at the same time so they do not add on the neutral. Only one load is on at a time.
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Correct except that they do add... at least that is what I'm thinking... but I could be wrong :rolleyes:

It is actually the fact that current occurs at different times that make them add up. In essence there would be one "spike" on each line conductor 360? apart. However there would be one "spike" every 180? on the neutral. Each spike has an rms currrent of 1A per cycle.

Consider the following illustration using pulse waveforms...

1addingcurrents.gif
 
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