how do i calculate the load on each line in a transformer?

[Kmansker]

hello all,

i recently took the masters electrical exam. i missed it by two points and am going to take it again in a couple weeks. my question is, on the exam i took there were two questions that gave you the current draw on the neutral then asked to solve for the current draw on each leg of the transformer. how would i go about this??

thank you, keith

 

[Besoeker]

hello all,

i recently took the masters electrical exam. i missed it by two points and am going to take it again in a couple weeks. my question is, on the exam i took there were two questions that gave you the current draw on the neutral then asked to solve for the current draw on each leg of the transformer. how would i go about this??

thank you, keith

Not enough information.

 

[Kmansker]

if i gave you the load on each line of a wye transformer and asked you to calculate the load on the neutral we would take the
square root of (A²+B²+C²)-(AB-BC-CA). Correct??? is there a way this can be done in reverse? for example being given the load on the neutral and then figuring the line?

thank you, keith

 

[GoldDigger]

if i gave you the load on each line of a wye transformer and asked you to calculate the load on the neutral we would take the
square root of (A²+B²+C²)-(AB-BC-CA). Correct??? is there a way this can be done in reverse? for example being given the load on the neutral and then figuring the line?

thank you, keith

There are four wires. Given any three you can calculate the fourth.
With special knowledge (such as one balanced 3 phase and one line to neutral load) you can calculate the unbalanced part of the load knowing only the neutral current.
For single phase 120/240 you have three wires, so you need to know any two currents to find the third.

 

[david luchini]

if i gave you the load on each line of a wye transformer and asked you to calculate the load on the neutral we would take the
square root of (A²+B²+C²)-(AB-BC-CA). Correct??? is there a way this can be done in reverse? for example being given the load on the neutral and then figuring the line?

thank you, keith

That formula should be... square root of (A²+B²+C²)-(AB+BC+CA)

It's only accurate when each line has the same power factor, but can be a good approximation otherwise.