Given 3 currents, find the unknown 4th current.

[mivey]

In this thread, there were some math formulas I thought I would consolidate or expand on. I hope you find this useful.

There is a quickie formula for finding the neutral currents but it only works for specific cases. I thought some might find the details of calculating the neutral currents helpful so here it goes:

Having three of four current vectors with the same sense (entering or leaving) a node, label them Ia@A?, Ib@B?, Ic@C?. Label the fourth as In, with the sense to be decided later. Add these using a calculator that can add vectors or convert to rectangular coordinates for summation (see below). If you do not have the current vector, but only have the power factor, you can convert the currents using the voltage vectors (see below).

If you want the sense of the fourth current to be opposite of the other currents, use the summation. If you want the sense of the fourth current to be the same as the other currents, use the negative of the summation (or add 180 degrees to the angle). Label the result as In@N?.

Now that is really it in a nutshell but there are more details. If you are interested, keep reading.

Using varying senses:
It would be worth noting that if all currents do not have the same sense, we can pick one as the reference for a positive sense. We add the currents that have the same sense, and subtract those that have the opposite sense. Alternatively, we could add 180 degrees to the angles for the currents with the opposing sense and then sum the currents.

You do not always have currents given to you using the same sense.

You can't always tell the sense of three-phase currents by looking at the given current angles alone as they may not be evenly dispersed around a circle (see post #145 in the other thread, for instance). For our current example, we have obtained values using the same sense, so we can just sum.

Converting to rectangular coordinates:
The current and angles use polar coordinates to represent a number located in two dimensions. For rectangular coordinates, Ia@A? = Xa+jYa where X is a scalar (or single dimensional) number on the horizontal axis and Y is a scalar number on the vertical axis. This gives you two one-dimensional numbers that can be added and subtracted separately.

The "j" is used to indicate the vertical axis (also know in the electrical engineering world as the imaginary axis). Now you know what an imaginary number is. "X+jY" is the complex number representation of a two-dimensional number.

Find X & Y:
Ia@A?: Xa = Ia*cos(A?), Ya = Ia*sin(A?)
Ib@B?: Xb = Ib*cos(B?), Yb = Ib*sin(B?)
Ic@C?: Xc = Ic*cos(C?), Yc = Ic*sin(C?)

Xsum = Xa+Xb+Xc
Ysum = Ya+Yb+Yc

Now you have the complex number "Xsum+jYsum", but we do not have the magnitude or angle of the current. Converting to polar coordinates will give you those values.

Converting to polar coordinates:
You can convert Xsum+jYsum to polar coordinates using a calculator or in several steps. The magnitude (denoted by using pairs of "|" symbols) is given by the square root of the sum of the squares: |Xsum+jYsum| = sqrt(Xsum^2 + Ysum^2).

The angle is given by the inverse tan (or arctan) of Ysum/Xsum. This is a little more complicated if your calculator does not directly convert to polar coordinates because you must account for what quadrant of the two-dimensional polar plane you are in. A calculator that does not handle complex math will usually give angles from +90 to -90 degrees and you will have to make the angle correction yourself using the following information:

The polar quadrants:
................90?(+Y)
.................|
.............II..|..I
.................|
(-X)180?---------|--------0?(+X)
.................|
............III..|..IV
.................|
.................|
................270?(-Y)

For different signs of X & Y, you will be in different quadrants:
+X & +Y = 1st quadrant
+X & -Y = 2nd quadrant
-X & -Y = 3rd quadrant
+X & -Y = 4th quadrant

Tan is positive in the 1st and 3rd quadrant and negative in the 2nd and 4th quadrant. When you take the arctan (arctangent) of a number, you will usually be given a positive or negative angle of less than 90 degrees (the 1st or 4th quadrant). That means for a +X & -Y (2nd quadrant), your angle result will be in the 4th quadrant so you will have to add 180 degrees to get the result to the 2nd quadrant. The same is true for -X & -Y (3rd quadrant). So to get the correct angle, you have:
1st Q with +X & +Y = arctan[(+Y)/(+X)] = XY?
2nd Q with -X & +Y = arctan[(+Y)/(-X)] + 180? = XY?
3rd Q with -X & -Y = arctan[(-Y)/(-X)] + 180? = XY?
4th Q with +X & -Y = arctan[(-Y)/(+X)] = XY?

Combining the magnitude and angle, you will now have |Xsum+jYsum|@XY?

Converting power factor data to vector data:
If you have been given current readings with a power factor, then you can use the voltage vectors to derive the current vectors. With Ia@pfA'', Ib@pfB'', and Ic@pfC'', the current angle is measured relative to the associated voltage angle. Cosine of this angle is the power factor.

The polarity of the voltage reference determines the sense of the current because we normally take positive current flow to be from a point of higher potential to lower potential (thanks to Ben Franklin, this is opposite of the electron movement). Sense is discussed in the other section, and the same rules apply. For this example, we will assume the same polarity for the voltage readings, thus the same sense for the current readings.

Now we need to convert the power factor value to an angle value. The power factor can be labeled as leading or lagging. Leading or lagging may also be indicated by a sign with a positive value for leading and a negative value for lagging. Lagging means the voltage waveform reaches a peak value before the current waveform (the current lags the voltage for inductive loads). Leading means the current waveform reaches a peak value before the voltage waveform (the current leads the voltage for capacitive loads). We mostly encounter lagging power factors as the majority of our loads are inductive, but don't assume that is always the case.

So, the power factor is a positive number from 0 to 1 with a lead/lag indicator. First we convert the power factor value to an angle value (do not include any negative indicator signs):
Va-Ia_angle = arccos(pfA'')
Vb-Ib_angle = arccos(pfB'')
Vc-Ic_angle = arccos(pfC'')

This gives us three positive angle values. We now need the direction of rotation of the reference voltages to find out if we add to or subtract from the voltage angle.

Find reference voltage rotation:
We normally take a positive angle direction to be counter-clockwise. We pick one of the voltages to be a reference ("a") and then measure "b" & "c" relative to this voltage. These choice of "a", "b", and "c" should match the choice of "a", "b", and "c" for the currents. We will make voltage "a" have a reference angle of zero so we will have Va@A'?, Vb@B'?, Vc@C'?, all with positive angles between zero and 360. Use these steps:
1) If A'? is not zero, then subtract its angle from the other two angles to get new angles for B'? and C'?.
2) If B'? or C'? is negative, add 360 degrees to get all positive angle values.
Now we can determine the rotation:
Positive rotation: angle B'? is greater than C'?
Negative rotation: angle C'? is greater than B'?

If we have a positive rotation, the current angle will be clockwise from the voltage angle for lagging and the current angle will be counterclockwise from the voltage angle for leading. The reverse is true for a negative rotation and is summarized here:

For positive rotation and lagging (or negative rotation and leading):
Ia_angle = Va_angle - arccos(pfA'')
Ib_angle = Vb_angle - arccos(pfB'')
Ic_angle = Vc_angle - arccos(pfC'')

For positive rotation and leading (or negative rotation and lagging):
Ia_angle = Va_angle + arccos(pfA'')
Ib_angle = Vb_angle + arccos(pfB'')
Ic_angle = Vc_angle + arccos(pfC'')

We now have three angles for our currents and can label them "A?", "B?", and "C?" to get Ia@A?, Ib@B?, and Ic@C?

Kirchhoff's Current Laws (KCL):
There are three ways you can state and apply KCL. All three are perfectly valid ways to analyze a node and your choice will depend on your particular situation. This is a summary of those three methods:

1) Picking senses where some currents enter the node and some currents leave the node is an application of this KCL statement: The sum of the currents entering a junction of conductors is equal to the sum of the currents leaving the junction of conductors.

2) Picking a sense where all four currents enter the node is an application of this KCL statement: The sum of the currents entering a junction of conductors is zero at all times.

3) Picking a sense where all four currents leave the node is an application of this KCL statement: The sum of the currents leaving a junction of conductors is zero at all times.

It is important to realize the current sense you have picked or adopted.

 

[mivey]

In this thread,

Actually, this thread

 

[mivey]

typo

typo

The following paragraph in the "Converting to polar coordinates" section:

For different signs of X & Y, you will be in different quadrants:
+X & +Y = 1st quadrant
+X & -Y = 2nd quadrant
-X & -Y = 3rd quadrant
+X & -Y = 4th quadrant

has a typo and should read:

For different signs of X & Y, you will be in different quadrants:
+X & +Y = 1st quadrant
-X & +Y = 2nd quadrant
-X & -Y = 3rd quadrant
+X & -Y = 4th quadrant

It was noted correctly in the table after the following paragraph but the paragraph had the same typo and read:

"That means for a +X & -Y (2nd quadrant), your angle result will be in the 4th quadrant so you will have to add 180 degrees to get the result to the 2nd quadrant."

but should have read

"That means for a -X & +Y (2nd quadrant), your angle result will be in the 4th quadrant so you will have to add 180 degrees to get the result to the 2nd quadrant."

 

[Mayimbe]

You should have posted some references. [1]

Like a good published work. It will have given more soul to your post.

The rotation analisys, isnt quite clear for me.

positive rotation: counter clockwise. Example, at t=0, A?=0?, B?=-120?, C?=+120?
negative rotation: clockwise. Example, at t=0, A?=0?, B?=+120?, C?=-120?

 

[mivey]

You should have posted some references. [1]

Like a good published work. It will have given more soul to your post.

I could have looked up some specific reference material but this was just typed out as I thought it. It seemed to me that it was pretty common knowledge and I didn't think about including additional reading material.

I can see where it would help for a reader to be able to read through some more detailed texts containing more background information and example calculations.

A quick Google should turn up tons of reading material, although I doubt any of it would cut right to the chase for this specific calculation.

The rotation analisys, isnt quite clear for me.

positive rotation: counter clockwise. Example, at t=0, A?=0?, B?=-120?, C?=+120?
negative rotation: clockwise. Example, at t=0, A?=0?, B?=+120?, C?=-120?

Which agrees with my statement: "We normally take a positive angle direction to be counter-clockwise". I think your examples could definitely add clarity but I was trying to avoid using negative angles so I could use the "greater than" test.

Also, instead of saying: "If we have a positive rotation, the current angle will be clockwise from the voltage angle for lagging and the current angle will be counterclockwise from the voltage angle for leading. The reverse is true for a negative rotation ..."

I might could have said something like:
"If we have a positive rotation, the current vector will be located clockwise from the voltage vector for lagging and the current vector will be located counter-clockwise from the voltage vector for leading. The reverse is true for a negative rotation... "

The sentence still seems like it needs some work.

 

[mivey]

Also, instead of saying: "If we have a positive rotation, the current angle will be clockwise from the voltage angle for lagging and the current angle will be counterclockwise from the voltage angle for leading. The reverse is true for a negative rotation ..."

I might could have said something like:
"If we have a positive rotation, the current vector will be located clockwise from the voltage vector for lagging and the current vector will be located counter-clockwise from the voltage vector for leading. The reverse is true for a negative rotation... "

The sentence still seems like it needs some work.

How about:
"If we have a positive rotation, the current vector will lie in a direction that is clockwise relative to the voltage vector for lagging and the current vector will lie in a direction that is counter-clockwise relative to the voltage vector for leading. The reverse is true for a negative rotation... "

 

[Mayimbe]

How about:
"If we have a positive rotation, the current vector will lie in a direction that is clockwise relative to the voltage vector for lagging and the current vector will lie in a direction that is counter-clockwise relative to the voltage vector for leading. The reverse is true for a negative rotation... "

Im struggling to understand that statement, due to my low budget english. But since I know, how that thing goes, im really not concern about finding what you are triying to say.

What I can say, is that I have been there, I find sometimes diffficult to express my ideas in a way that all people can understand. Maybe thats why Nobel prizes in literature are not given to engineers very often.