# NCEES Power Problem 108



## BebeshKing PE (Jan 28, 2020)

How did they get the formula Ia+Ib= -In?

What I know is In=Ia+Ib+Ic. And if Ic=0, should it be Ia+Ib=In, right?

What am I missing here?

Thank you,


----------



## DLD PE (Jan 28, 2020)

Yes for unbalanced loads, Ia+Ib+In = 0

Maybe this will help:


----------



## DLD PE (Jan 28, 2020)

Oh no I'm having a brain fart:  How do we know Ic=0?  

I like how everyone posts these problems.  When I think I know something that I used to know, all of a sudden I realize it hasn't stuck yet.


----------



## BebeshKing PE (Jan 28, 2020)

MEtoEE said:


> Oh no I'm having a brain fart:  How do we know Ic=0?
> 
> I like how everyone posts these problems.  When I think I know something that I used to know, all of a sudden I realize it hasn't stuck yet.


I did the same solution Based on @supra33202 's solution per your attached comment above. I just don't know why the NCEES solution have the minus neutral current.

I know the problem was just asking about the magnitude, but I would also want to know if my concept is correct which is In=Ia+Ib+Ic. 'coz maybe in the future they will also asked for the vector of the neutral current.

In my thought, I assumed Ic=0 since this is a 3 phase system but there is no load in phase c and assumed to be open circuited. Hence, In=Ia+Ib


----------



## Orchid PE (Jan 28, 2020)

It doesn't really matter which side I neutral is on.

With Ia+In+Ic=In or Ia+Ib+Ic+In=0 they will work out to the same answer because of KCL.


----------



## BebeshKing PE (Jan 28, 2020)

Chattaneer PE said:


> It doesn't really matter which side I neutral is on.
> 
> With Ia+In+Ic=In or Ia+Ib+Ic+In=0 they will work out to the same answer because of KCL.


@Chattaneer PE, will the Neutral current displaced by 180 degrees if we use the other way around?

Ia+Ib+Ic+In=0

transpose In to the right of the equation:

Ia+Ib+Ic=-In  --&gt; 180 degrees displaced


----------



## Orchid PE (Jan 28, 2020)

BebeshKing said:


> @Chattaneer PE, will the Neutral current displaced by 180 degrees if we use the other way around?
> 
> Ia+Ib+Ic+In=0
> 
> ...


The angle will tell you the direction. All depends if it's included at summing into the node or leaving the node.

We know phase A and B will be entering the node, and because of that we know the neutral current will be leaving the node and of equal magnitude of A+B, but the angle will be 180 the opposite direction.


----------



## BebeshKing PE (Jan 28, 2020)

Chattaneer PE said:


> The angle will tell you the direction. All depends if it's included at summing into the node or leaving the node.


Thanks @Chattaneer PE.

So on this problem, the utility Y connected source is sending a current to the loads. And if the reference node is at the neutral, the 3 phase ungrounded currents are leaving the nodes and the  grounded neutral current is entering the node. Hence,

entering the node = leaving the node

In= Ia+Ib+Ic??

Please correct me if I'm wrong.

Thank you,


----------



## Orchid PE (Jan 28, 2020)

I would say phase A and B currents are entering the node, and neutral current is leaving the node.


----------



## Orchid PE (Jan 29, 2020)

Here's how I would draw the diagram:




From this we can make the equation _Ia + Ib = In_.

Ia = (200+100j) kVA / (7.62 ∠ 0 kV) = 29.3 ∠ 26.56

Ib = (200+100j) kVA / (7.62 ∠ 120 kV) = 29.3 ∠ 86.56

In = Ia + Ib = 29.3 ∠ 26.56 + 29.3 ∠ 86.56 = 29.3 ∠ -33.4

----

If we drew _In_ going into the node (-&gt;, towards the right), the equation would be _Ia + Ib + In = 0 _(summing all entering the node). Transform to _Ia + Ib = -In._

When we do all the calculations again, we get _In = -29.3∠ -33.4_. Since this is negative, we know _In _has to be *leaving *the node.

----

Just like in circuits when using KCL, the *sign *of the currents tells use the direction of flow *based on how we draw the diagram*.


----------



## BebeshKing PE (Jan 29, 2020)

Chattaneer PE said:


> Here's how I would draw the diagram:
> 
> View attachment 16225
> 
> ...


@Chattaneer PE, great explanation! Quick question though, is there gonna be a way in theory(or maybe in reality) that all the phase currents(Ia,Ib,Ic) including the neutral current(In) will all be entering a common node? If yes, in what scenario? 

Thanks,


----------



## Orchid PE (Jan 29, 2020)

BebeshKing said:


> @Chattaneer PE, great explanation! Quick question though, is there gonna be a way in theory(or maybe in reality) that all the phase currents(Ia,Ib,Ic) including the neutral current(In) will all be entering a common node? If yes, in what scenario?
> 
> Thanks,


The only time current flows on a neutral is when the load is not balanced.

Let's say phase A and B currents are the same, but greater than phase C current. In this scenario, the "excess" current from phases A and B that is not cancelled out by phase C will flow on the neutral, back towards the source.

----

Technically speaking, not to confuse you, since all this is AC the current alternates directions (60 times a second).


----------



## Zach Stone P.E. (Jan 29, 2020)

BebeshKing said:


> How did they get the formula Ia+Ib= -In?
> 
> What I know is In=Ia+Ib+Ic. And if Ic=0, should it be Ia+Ib=In, right?




*Here's the trick:*

Most books do not explain this well and gloss over it entirely. If you use a KCL equation to solve for the neutral current (In) from the three line currents (Ia, Ib, Ic) of a wye wye system, it looks like this:


In = Ia + Ib + Ic

*However, *the direction of the neutral current is typically shown from the *source to the load*, just like the line currents. *This is where the negative sign comes from*. How do we change the direction of the neutral current to express it mathematically as flowing from the source to the load? *We multiply it by negative one, or lead/lag the phase angle by 180 degrees (both have the same affect). *

The neutral current formula then becomes:


In = -(Ia + Ib + Ic)   or   In = -Ia - Ib - Ic    or   -In = Ia + Ib + Ic 

In this particular problem since there is no C line current, the following happens once we substitute Ic = 0 into the previous formula:


In = -(Ia + Ib + 0)

In = -(Ia + Ib)

However, since the problem *is only asking for the magnitude* of the neutral current, and not the complex neutral current with magntidue and angle, the negative sign does not matter. 

Multiplying a complex number by negative one is really just leading/lagging the phase angle by 180 degrees, *it has no affect on the magnitude.*

So even if you solved it without the negative sign for the neutral current, you'll still end up with the same magnitude of 29.3A that the author did.

Try it:


----------



## BebeshKing PE (Jan 29, 2020)

Zach Stone said:


> *Here's the trick:*
> 
> Most books do not explain this well and gloss over it entirely. If you use a KCL equation to solve for the neutral current (In) from the three line currents (Ia, Ib, Ic) of a wye wye system, it looks like this:
> 
> ...


Nice. This is what I thought, that typically the neutral current is also flowing from source to the load. and since it will be negative, the actual flow will be opposite.

Thank you @Zach Stone, P.E.


----------

