neutral current calculation....again

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Rattus....the first example that you are requesting doesn't appear to be a functional 3 phase circuit. I think the angles are too far from the standard 0, 120, 240 degrees. Maximum lead / lag possibilities are 90 degrees.
No?
 
basicbill said:
Rattus....the first example that you are requesting doesn't appear to be a functional 3 phase circuit. I think the angles are too far from the standard 0, 120, 240 degrees. Maximum lead / lag possibilities are 90 degrees.
No?
Click to expand...

Bill, the first set of currents represents the basic single phase loads--no calculation is expected. In is then calculated for forward and reverse phase sequences.
 
Now that we've progressed this far, there is one question left unanswered...

So far we've been determining neutral current as though all three loads are connected to the same point on the neutral conductor. However, there are many instances where the loads are not connected at the same point. Referring to the diagram below, the neutral current In(3) will be the same. That of In(1) should be obvious. So what is the magnitude and phase angle of In(2)?

My deductive reasoning says it is equal in magnitude but opposite in phase angle (i.e. 180?) to In(3). I leave it to you guys to prove (or disprove)!

neutralcurrent4.gif
 
Smart $ said:
Now that we've progressed this far, there is one question left unanswered...

So far we've been determining neutral current as though all three loads are connected to the same point on the neutral conductor. However, there are many instances where the loads are not connected at the same point. Referring to the diagram below, the neutral current In(3) will be the same. That of In(1) should be obvious. So what is the magnitude and phase angle of In(2)?

My deductive reasoning says it is equal in magnitude but opposite in phase angle (i.e. 180?) to In(3). I leave it to you guys to prove (or disprove)!

neutralcurrent4.gif
Click to expand...

Depending on the sense (direction) of In2, it is +/- (Ib + Ic). Your answer is right only if Ia = 0.
 
rattus said:
Depending on the sense (direction) of In2, it is +/- (Ib + Ic). Your answer is right only if Ia = 0.
Click to expand...

I disagree.

Granted my deduction is wrong, but I believe yours is too. If Ia = 0, then In2 = Ib... not Ib + Ic.

I've re-evaluted the answer. It is in the attached Word file. However, I wanted to see if you (or others) can get the same, or otherwise correct answer before I reveal mine... so the file is password protected. I'll furnish the password if someone comes up with the same answer, or in time if no one does...
 
Smart $ said:
I disagree.

Granted my deduction is wrong, but I believe yours is too. If Ia = 0, then In2 = Ib... not Ib + Ic.

I've re-evaluted the answer. It is in the attached Word file. However, I wanted to see if you (or others) can get the same, or otherwise correct answer before I reveal mine... so the file is password protected. I'll furnish the password if someone comes up with the same answer, or in time if no one does...
Click to expand...

Yup. I misread the labels. I read Ia, Ib, Ic instead of Ic, Ib, Ia. Anyway, it is simply a matter of applying Kirchoff at each junction.
 
Well, since no one other than rattus has taken a stab at providing the answer, I'll follow through on my word...

Smart $ said:
I'll furnish the password if someone comes up with the same answer, or in time if no one does...
Click to expand...

The password to open the Word .doc is "ok" without the quotation marks.

The formula (answer) is:
In(2) = Ia + Ib​
There are at least three ways to arrive at this answer using KCL as the basis.
 
Smart $ said:
Well, since no one other than rattus has taken a stab at providing the answer, I'll follow through on my word...



The password to open the Word .doc is "ok" without the quotation marks.

The formula (answer) is:
In(2) = Ia + Ib​
There are at least three ways to arrive at this answer using KCL as the basis.
Click to expand...

I agree with this proof.

And obviously as was shown before, you cant simply just add the magnitues of currents Ia and Ib but must take their phase angle into consideration
 
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