Engineering Measuring Current Flow Through Neutral Wire with Open Phase R

[lukus09]

https://www.physicsforums.com/attachment ...

how do i measure the current flow threw the neutral wire if phase R is open circuited? the impedance per load is 3+7j

 

[Kruum]

Your link gives a 404 error...

 

[lukus09]

https://www.physicsforums.com/attachment.php?attachmentid=18054&d=1237464777
that should work

 

[Kruum]

Do you need to physically measure it by using an ammeter or calculate it? And you'll have to show your attempt before we can give you a push in the right direction.

 

[lukus09]

both. i attached an ammeter to the neutral wire and i put an access wire from the input of the ac power suplly to the out R and i have got a reading, but i need to hand calculate it.

 

[Kruum]

Okay, I've never heard of access wire (I'm non-english) but I assume it open circuits the R phase, as you described in your first post.

But this calculation should be fairly simple to do with only the basic knowledge of AC circuits and KCL. Just think what happens to the current if a circuit is opened and what happens when currents meet in a node. I assume you have had theory on 3-phase and you know the phases of the power supplies in such a system.

 

[lukus09]

as far as i am aware it is now an unbalanced load meaning the current of R, Y and B is Ir+Ib+Ic?and I am sure the impedance 3+7j is needed to work it out to.

 

[Kruum]

You're right, but if the R-phase is opened, is there a current trough that wire? And you can substitute the one voltage supply with three different ones in each loop with the same voltage, but different phase since they are in star formation. Can you tell me what the phases are?

If you are having trouble figuring the phases out, here's a good reading to find them out.

 

[lukus09]

do i need to know the phases to work out the current in the nuetral wire?

 

[Kruum]

do i need to know the phases to work out the current in the nuetral wire?

Yes you do. The phases of the voltage supplies affect the phases of the currents. The phases of the voltage are standardized, though.

 

[lukus09]

sorry jus realized the impedance is 7+3j. i have worked out the phases to be Z1 is -23.2, Z2 is -143.2 and Z3 is 263.2

 

[Kruum]

Do you mean the phases of the currents? And is the impedance 3+j7 or 7+3j, you have given two different values?

 

[lukus09]

yes and it is 7+j3

 

[Kruum]

yes and it is 7+j3

Right. The phase of I2 is correct, but I guess there's a typo in I3. Shouldn't it be -263.2? And the current for I1 would be right if the wire wasn't open circuited. I have given you hints earlier, of what the value of I1 should be.

Once you find out the I1, you are all set to calculate as you have said, I1+I2+I3. You probably have the magnitude as well, if you have determined the phases.

 

[lukus09]

yes it is a typo it is -263.2. I only need the magnitude do i not? just to find the amps in the neutral wire though and the magnitude is the same for each phase i get it to be 7.62A?

 

[Kruum]

I only need the magnitude do i not? just to find the amps in the neutral wire though and the magnitude is the same for each phase i get it to be 7.62A?

That could be, but usually when they ask for current they want the phase as well. You are right, though, the magnitude is the same for each phase and its the same for the current in the neutral wire. But I get about double the magnitude you get. How did you calculate it.

 

[lukus09]

squareroot(14^2 + 6^2) = 15.23

 

[Kruum]

Um, now you have calculated the magnitude of the impedance and doubled it. To find out the current, you have to use Ohm's law.

 

[lukus09]

so its 115/squareroot(7^2+3^2) = 15.1? but that is the same as the line current :S:S

 

[Kruum]

In this case their magnitude are the same, since only two wires are carrying a current.

 

[lukus09]

so is 15.1 the correct answer, what explains the non zero value other than its unbalanced?

 

[Kruum]

If you used the correct values, then it should be. But I can't think of any other reason than unbalanced load.