Current carrying conductors

[Jpflex]

I was wandering if you would include the control circuits for motor starters as current carrying conductors within the same conduit containing power and lighting circuits. I’ve seen some starter magnet contacts carry as much as 1-2 amperes per starter with paralleled connections having as much current as lighting circuits.

There was one time I found the inductance in Henry’s of a starter contact coil mathematically by measuring the starter contact current

When determining which conductors to count as current carrying within a conduit for ampacity correction, my book says you would not include control and signal conductors that generally will carry small amounts of current (same as lighting circuits - but lighting circuits are counted?) and these circuits on for only a short period of time (not true for starter control circuits which hold contacts closed as long as the motor runs).

Since control wires for motors are in continuously as the motor runs and can indeed draw adequate current equal to power and lighting circuits within 20 amperes (several motor starter signal circuits all paralleled from 1 branch circuit) shouldn’t they be counted as current carrying conductors when determining conductor corrected ampacity?

 

[texie]

Adjustments for more than 3 CCCs only apply to power and lighting circuits-not a motor control circuit. See 310.15(C)(1).

 

[Jpflex]

Adjustments for more than 3 CCCs only apply to power and lighting circuits-not a motor control circuit. See 310.15(C)(1).

However the code/ book references control circuits “that are not on continuous and adds little current / heat”

This may not be the case with motor starter control circuits that supply / parallel multiple starters (20 amperes or more)

The total current can be equal to power circuits and exceed lighting circuit amperes (not little current and not little heat added).

 

[Jpflex]

When further studying when to count a neutral as a current carrying conductor it is understandable that for a 3 phase 110/208 volt system example problem 1-19 shown, the neutral is NOT counted because the pair of phases A and B become a “single phase” as they create a series circuit between two loads and the neutral junction which joins the two loads as I’ve pictured in my drawing

However the same would be expected in book for problem 1-21 pictured for the same 120/208 volt system. However, just because a third phase A with its corresponding load is omitted , the neutral is now counted?

How can this be since phase B and C creat the same series circuit 208 volts with 2 x the load resistance as the first example?

 

[LarryFine]

In a balanced 3ph circuit, there is no neutral current. For each amp removed from any one phase, the neutral current rises by the same amount.

Presuming a balanced 20a MWBC, subtract 20a from one phase, and the neutral current rises to 20a.

 

[wwhitney]

When further studying when to count a neutral as a current carrying conductor it is understandable that for a 3 phase 110/208 volt system example problem 1-19 shown, the neutral is NOT counted because the pair of phases A and B become a “single phase” as they create a series circuit between two loads and the neutral junction which joins the two loads as I’ve pictured in my drawing

Your handwritten diagram is not correct. If you disconnected the neutral in each diagram it would be correct, and the two series 120 ohm loads would form a voltage divider, with the potential of the midpoint being half-way between points A and B. Which midpoint is not the neutral potential, however.

Once you connect the neutral, each 120 ohm load will see a 120V supply, and its current will be 1A. In the neutral, those 1A current will be at a 120 degree phase angle, the phase angle of the A-N-B voltages. If you do the vector math (or just remember this special case that comes up alot), that means the neutral conductor will be carrying 1A of current.

This shows that with just A-B-N in the circuit, each one is a current carrying conductor. What is special about a circuit that is A-B-C-N is that if you add a 3rd 120 ohm load connected C-N to your diagram, then the 1A current it draws from N exactly cancel the 1A current that A-B jointly draw from N. So of the 4 conductors, A, B, and C will see 1A, while N will see 0A.

Cheers, Wayne

 

[Jpflex]

In a balanced 3ph circuit, there is no neutral current. For each amp removed from any one phase, the neutral current rises by the same amount.

Presuming a balanced 20a MWBC, subtract 20a from one phase, and the neutral current rises to 20a.

In a balanced 3ph circuit, there is no neutral current. For each amp removed from any one phase, the neutral current rises by the same amount.

Presuming a balanced 20a MWBC, subtract 20a from one phase, and the neutral current rises to 20a.

By removing 20 amperes from one phase you actually mean removing resistance or a load from one phase in order to do this?

I can see how removing a load such as in phase A can stop current travel from phase A to B which would cause the neutral to take on full current as the only current path

 

[Jpflex]

Your handwritten diagram is not correct. If you disconnected the neutral in each diagram it would be correct, and the two series 120 ohm loads would form a voltage divider, with the potential of the midpoint being half-way between points A and B. Which midpoint is not the neutral potential, however.

Once you connect the neutral, each 120 ohm load will see a 120V supply, and its current will be 1A. In the neutral, those 1A current will be at a 120 degree phase angle, the phase angle of the A-N-B voltages. If you do the vector math (or just remember this special case that comes up alot), that means the neutral conductor will be carrying 1A of current.

This shows that with just A-B-N in the circuit, each one is a current carrying conductor. What is special about a circuit that is A-B-C-N is that if you add a 3rd 120 ohm load connected C-N to your diagram, then the 1A current it draws from N exactly cancel the 1A current that A-B jointly draw from N. So of the 4 conductors, A, B, and C will see 1A, while N will see 0A.

Cheers, Wayne

 

[LarryFine]

By removing 20 amperes from one phase you actually mean removing resistance or a load from one phase in order to do this?

I can see how removing a load such as in phase A can stop current travel from phase A to B which would cause the neutral to take on full current as the only current path

Yes, and correct.

 

[Jpflex]

When you say that if I disconnected the neutral in each of my hand drawings, I assume you mean disconnect neutral buss connection such as at the panel only but not the “neutral” wire nutted connection or splice witch joins the two loads between two phases?

With balanced loads, with no current on the neutral wire going back to the neutral bus within the panel, it is the same as there being no neutral or it being disconnected to the neutral buss within the breaker panel and should make no difference whether the neutral return line is present.

However I was taught in DC class that all voltage is “ consumed” after the final load in a circuit and since AC can be thaught of as DC reversing polarity quickly so this makes me wonder why wouldn’t voltage or current cannot be present on the neutral regardless if loads are balanced ?

 

[Jpflex]

Your handwritten diagram is not correct. If you disconnected the neutral in each diagram it would be correct, and the two series 120 ohm loads would form a voltage divider, with the potential of the midpoint being half-way between points A and B. Which midpoint is not the neutral potential, however.

Once you connect the neutral, each 120 ohm load will see a 120V supply, and its current will be 1A. In the neutral, those 1A current will be at a 120 degree phase angle, the phase angle of the A-N-B voltages. If you do the vector math (or just remember this special case that comes up alot), that means the neutral conductor will be carrying 1A of current.

This shows that with just A-B-N in the circuit, each one is a current carrying conductor. What is special about a circuit that is A-B-C-N is that if you add a 3rd 120 ohm load connected C-N to your diagram, then the 1A current it draws from N exactly cancel the 1A current that A-B jointly draw from N. So of the 4 conductors, A, B, and C will see 1A, while N will see 0A.

Cheers, Wayne

Ok I get that you meant splitting the neutral at the neutral junction between two loads in order to get a voltage divider as voltage is “consumed after the final load”

 

[roger]

Do this equation for three phase wye neutral current.
Say B is 10 amps and C is 10 amps
SQRT ( B^2 + C^2 – BC)

 

[LarryFine]

When you say that if I disconnected the neutral in each of my hand drawings, I assume you mean disconnect neutral buss connection such as at the panel only but not the “neutral” wire nutted connection or splice witch joins the two loads between two phases?

Yes, he means (I believe) leaving the two neutrals still tied together to the shared neutral.

With balanced loads, with no current on the neutral wire going back to the neutral bus within the panel, it is the same as there being no neutral or it being disconnected to the neutral buss within the breaker panel and should make no difference whether the neutral return line is present.

With a shared neutral that is disconnected at the panel, it will then behave just like a 1ph MWBC, with the same possible imbalance, except the L-L voltage would be 208v instead of 240v.

However I was taught in DC class that all voltage is “ consumed” after the final load in a circuit and since AC can be thaught of as DC reversing polarity quickly so this makes me wonder why wouldn’t voltage or current cannot be present on the neutral regardless if loads are balanced ?

There is no difference between AC and DC in this discussion.

Voltage (to neutral/ground) and current (between the two L-N halves) would still be present.

 

[wwhitney]

voltage is “consumed after the final load”

I have no idea what that statement is supposed to mean.

What you can say is that if you have a voltage source, and then you trace a path from one side of the voltage source through the circuit and load(s) back to the other side of the voltage source, the sum of the load voltage drops has to equal the voltage of the source.

So I guess you could say that "by the time you get back to the source in the circuit analysis, all the voltage must be consumed (dropped across the loads)." In that if you find that's not true, you've made a mistake in your analysis.

Cheers, Wayne

 

[wwhitney]

With a shared neutral that is disconnected at the panel, it will then behave just like a 1ph MWBC, with the same possible imbalance, except the L-L voltage would be 208v instead of 240v.

Just to amplify/clarify this, I think what you mean is "it will then (mis)behave just like a 1ph MWBC would (mis)behave with a disconnected neutral."

Cheers, Wayne

 

[Jpflex]

Do this equation for three phase wye neutral current.
Say B is 10 amps and C is 10 amps
SQRT ( B^2 + C^2 – BC)

200 - 208 close to zero but mathematically equals 8 or - 8 amperes but yes ok I’m going to have to keep at this concept because it’s not easy to take all this in at once

 

[LarryFine]

Just to amplify/clarify this, I think what you mean is "it will then (mis)behave just like a 1ph MWBC would (mis)behave with a disconnected neutral."

Roger dodger!

 

[wwhitney]

200 - 208 close to zero

No, B = C = 10A, so it would be SQRT(10² + 10² - 10*10) = 10A.

Which is a way of doing the math I mentioned in the middle paragraph of my first post, for the case of 120 degree phase shift.

Cheers, Wayne

 

[Jpflex]

Yes, he means (I believe) leaving the two neutrals still tied together to the shared neutral.


With a shared neutral that is disconnected at the panel, it will then behave just like a 1ph MWBC, with the same possible imbalance, except the L-L voltage would be 208v instead of 240v.

ok yes all these points I understood from the start and am in agreement with you all. If I said 240 volts I meant 208 volts between line to line phases

However I don’t yet get why in the second problem that how removing a load in a third phase will affect the remaining two phases to have a current in the neutral unless it’s as the last post stated that doing so causes the neutral to rise exactly how much amperes was removed from the line and how the third phase cancels the current on the adjacent phase. 120 degrees out ?

 

[jim dungar]

....I was taught in DC class that all voltage is “ consumed” after the final load in a circuit and since AC can be thaught of as DC reversing polarity quickly so this makes me wonder why wouldn’t voltage or current cannot be present on the neutral regardless if loads are balanced ?

You need to study 3-phase AC, as it is different than simple DC.
Poly-phase systems require some understanding of trigonometry, in particular the sine and cosine function. One common formula is the Pythagorean theory for right triangles which can be simplified to A² + B² = C² and its various permutations.