Thank you Paul, I knew you would come through for me.

I have seen a diagram explaining that the two busses shared the load and were only allowing 100 amps per side. That just did not seem right to me as I was under the impression that both legs were rated at 200 amps. I do reaiize that you cannot add up the two legs and get 400 amps.

Another way of looking at it is 200 amp in and thus 200 amps out and if a little more or less is being used it goes up and down the neutral.

(lets not get into power factor)

What say Paul? Does this sound like it helps one to understand?

rlb

Hey Richo,

Sounds good to me…basically think of it as true 200A capacity per LINE…if 50A DRAW is on one LINE…and 30A DRAW is on another line…quess what the unbalanced load is on the Neutral…???..Any Takers…

P.S. Fellow Electricians…DO NOT speak on Cont. Amps…this is WAY to excessive for this example…

I am going to guess the difference between the two loads or 20 amps

Larry

So Paul,

Are you suggesting that a “perfectly balanced” load would have ZERO amps on the neutral…

Go for it Paul

Where do you want to measure the neutral current? It makes a difference

lol…oh dear…not THIS discussion again…we are talking calculations for GENERAL knowledge in computing…lol…I dont believe in a true balanced system…so regardless of OTHER examples…their is always some load on a Neutral…lets NOT go their…brains are smoking already.

lol…I am talking about the example loads…lol…50A on one line and 30A on the other…the unbalanced load of the example…lol

I should have used smaller loads…lol… ( Thanks Joe- I got the example from an old PDF…it is a Sallcup Reference…Thanks ! )

The Code considers the neutral conductor a current-carrying conductor only when it carries the unbalanced current from other ungrounded phase conductors. When circuits are properly balanced, the neutral carries very little current. When sizing the load for a 2-wire circuit, the grounded neutral conductor carries the same amount of current as the ungrounded phase conductor. This type of installation has no unbalanced load; therefore, the neutral conductor carries full current.

Example: What is the neutral load for a single-phase, 120V, 2-wire circuit supplying a load of 14A?

Step 1: Find amperage per Sec. 220-22 and Sec. 310-15(b)(4)(a).

Ungrounded conductor = 14A

Grounded neutral conductor = 14A

Solution: Size the neutral conductor to carry a load of 14A.

When sizing the load for a 3-wire circuit, the grounded neutral conductor must carry the unbalanced load of the two ungrounded phase conductors. This type of installation has an unbalanced load - unless both ungrounded conductors pull the same amount of current on each ungrounded phase conductor.

Example: What is the unbalanced neutral load for a 3-wire circuit carrying 64A and 52A on the ungrounded phase conductors?

Step 1: Find amperage per Sec. 220-22 and Sec. 310-15(b)(4)(a).

Ungrounded phase conductor: Phase A = 64A

Ungrounded phase conductor: Phase B = 52A

Unbalanced load = 12A

Solution: The grounded neutral conductor load is 12A for the unbalanced condition.

I kind of knew that I would be opening a large can of worms here but for my own peace of mind I had to ask. So now I am assuming that the answer to your first question re the unbalanced load was correct and the unbalanced load is the difference between the two loads, in the example given, 20 amps.

Larry

By the way Paul, thanks for the reference, it has proved to be very informative.

Larry

If that was ever possible, here is the link for an article: "Characteristics of the Neutral Conductor, written by James Stallcup.

Courtesy: EC&M

SNAP…thanks JOE…I totally FORGOT where that PDF post came from…do you have a link to it…I can’t find the original…

Thanks Joe, Paul,

I had it in my brain from a electronics circuit analysis class with a cube or delta “type” circuit based on multiple similar value of resistors.

A source applied to cube and points of voltage / current tested / calculated to see what is happening… In a “perfect” world zero current across a resistor at center of cube…

I’m surprised all of you are ignoring how volts and amps work mathematically.

For a standard single-phase 120/240V supply at 200A, what you have is 200A @ 240V. That is 200Ax240V = 48kW. You can also measure it as 400A@120V if you want to becuase 400Ax120V = 48kW as well. That would just be a highly unconventional way of representing the service, but both are equivilant mathematically. You could feed 200A worth of 240V equipment from that service, 400A worth of 120V equipment, or anywhere in between. The latter being the common means, of course.

For three phase, you have a factor of ~1.73 (sqrt(3)) introduced. So a 120/208V three-phase supply at, say, 200A, has 200A*208V*1.73 = 71.9kW.

Ultimately, when power sources are expressed using watts, it is easier to compare them.

Joey

The cable size (gage and conductor type) is based on amps, how it is installed and the length of the run computed with the allowable voltage drop - the insulation is based on voltage.

And yes this is part of the math to comput the size of a neutral conductor

As you can see it is all simple math – how much to you need and how big is the required pipe

In your example if the limiting factor was the 48Kw rating you would exced the curent rating of the conductors at 400 amps if they were rated at 200 amps. Thus your statement of 400 amp of equipment at 120 volts is not correct and will not work. Real simple the cable is only rated for 200 amps for the length that it is being used. Remember this is also part of the design. One can not run a # 14 extention cord from NY to Tx and light a 100 watt light bulb with 120 volts - too much resistance in the wire

rlb

rlb

I admit that referring to my example service as 400A @ 120V is unconventional, but to understand why I am not wrong, work out the following example.

Lets suppose you have the following two circuits available to you:

Circuit #1 offers 20A at 120v from two wires (hot/neutral).

Circuit #2 offers 10A at 120/240v from three wires (hot/hot/neutral).

Assume that the 80% rule does not apply, and assume that all wires are adequately sized for their amperage such that circuit #1 has a minimum of 12 guage wire and circuit #2 has a minimum of 16 gauge wire. Also assume that voltage drop is not a concern. None of these assumptions hurt the validity of the examples.

Work out the math to show that both supplies can power the same amount of 100 watt light bulbs without exceeding the capacity of any wires . Then you should understand what I was saying.

Or think about how twice as much power is available between a 30A, 120V and 30A, 240V air conditioner circuits. Or how a transformer on any given circuit could be used to provide twice the amperage at half the voltage, or half the voltage at twice the amperage. (or any other similar ratio such as 1/3 the amperage at 3x the voltage).

Work some of those examples out and you will see my point.

I think what I am saying is quite relevant to the original poster’s question.

Joe

Yes, power is equal to voltage times amperage. One of the points that is a good one to make is that we rate for current and voltage - Not wattage on a 120/240 three wire ckt. Yes, as we all know, it is all directly related.

The main condition that will mess the system up and cause current ratings to exceed design is LOW VOLTAGE. Low voltage will cause higher current and higher power loses in all areas of the electrical system

Paul A — help me out here – maybe we need a new thread – we are sort of off the original question

Wattage is work – Voltage and current rating is potential work (KVA rating)

rlb

Mathematically it may work but we are talking about a non-controlled, non-laboratory environment here Joey. Sure you can use assumptions all you want but we have so many variables to deal with you cannot use this approach in a real world practical application.

If this where the case then high power transmission stations, lines, equipment would not exist in the form they do. Think of all the loss and load issues, power fluctuations not to mention how the consumer changes the load on super hot days when all those A/C units are running and the factories are running all those motors…hmmmm Coils…hmmmm

Wouldn’t it be great if we could forget about resistance altogether.

How about the Loss of energy in heat!! You talk of watts but what a watt…

Well that is the conversion of one type of energy to another with loss…

Heat from a light bulb, motor …gone…lost…

So living in a imperfect world with all those variables leads me to believe that we can only get so much out of all that “stuff” …

lol…Why would you be surprised…you yourself said it was highly unconventional…lol