Solving for phase currents given line currents and line voltages in 3 phase unbalanced delta load

Question

I am given the line currents and the line voltages for an unbalanced 3-phase delta connected load but need to solve for the phase currents on the legs of the delta load.

I know that:

$$I_{L1} = I_{a} - I_{b}$$ $$I_{L2} = I_{b} - I_{c}$$ $$I_{L1} = I_{c} - I_{a}$$ $$I_{L1} + I_{L2} + I_{L3} = 0$$

where \$I_{a}, I_{b},\$ and \$I_{c}\$ are the phase currents, hence:

$$\pmatrix{I_{a}&I_{b}&I_{c}} \pmatrix{1&0&-1\\-1&1&0\\0&-1&1} = \pmatrix{I_{L1}&I_{L2}&I_{L3}}$$

So far, I have used a method where I right-multiplied both sides by the pseudoinverse of $$\pmatrix{1&0&-1\\-1&1&0\\0&-1&1}$$

Under the assumption that

$$ABB^{-1}=A=CB^{-1}$$

Solving for the pseudoinverse of the matrix gives: $$B^{-1} = \frac{1}{3}\pmatrix{1&-1&0\\0&1&-1\\-1&0&1}$$

But something about this seems really sketchy. Most of the examples and lectures I have found deal with solving for line currents given the phase currents or phase impedances, not the other way around.

How do I solve for the phase currents knowing only the line currents and the line to line voltages?

Answer 1

You can't invert the matrix because it's not an independent linear system, in another word, there is infinite solutions. You can try to find the phase currents using Kirchhoff laws, but I think the only way is to use superposition, as here: To ilustrate the idea, take this examples:

• Phase currents: ia=1, ib=2, ic=3 >>> you got line currents: i1=-1, i2=-1 and i3=2
• Phase currents: ia=0, ib=1, ic=2 >>> you got line currents: i1=-1, i2=-1 and i3=2
• Phase currents: ia=-1, ib=0, ic=1 >>> you got line currents: i1=-1, i2=-1 and i3=2

So, if you use the line currents i1=-1, i2=-1 and i3=2, you can get any of the phase currents above.

You can also note that the i3 is not used, because it has to be -(i1+i2) assuming the line currents are all entering the triangle.