# Unbalanced Three-phase circuit.

1. May 7, 2010

### jfierro

[SOLVED] Unbalanced Three-phase circuit.

1. The problem statement, all variables and given/known data

This comes from "practice problem" 12.10 from Alexander's and Sadiku's Fundamentals of Electric circuits 3rd edition, chapter 12.

The problem statement is:

"Find the line currents in the unbalanced three-phase circuit of Fig. 12.26 and the real power absorbed by the load."

http://img413.imageshack.us/img413/8428/circuits.png [Broken]

2. Relevant equations

- LCK

3. The attempt at a solution

The problem is I can't get to the shown answer. I have followed two methods:

Getting I_AB and I_CA from the known phase voltages:

I_AB = (220|0º) / (5|-90º)
I_CA = (220|-120º)/(10|0º)

And then doing I_a = I_AB - I_CA.

The second method is to construct a 3x3 matrix with complex coefficients based on the three meshes shown in the circuit. For those of you familiar with mathematica notation my matrix is:

LinearSolve[{{-5 I, 5 I, 0}, {5 I, 10 + 5 I, -10 I}, {0, -10 I, 10 I}}, {220, 0, -110 + Sqrt[3]*0.5*I}]

where I is the imaginary number and the first solution would be I_a.

Both methods yield 11 + 44.866i, far different from the listed result.

Any idea of how the authors got to (64, 80.1º)?

Last edited by a moderator: May 4, 2017
2. May 7, 2010

### xcvxcvvc

I just did exactly what you said using my beautiful vector-capable TI-89 and got the book's answer. I assume you did something wrong with your vector math.
I_AB = (220|0º) / (5|-90º) =(44|90º)
I_CA = (220|-120º)/(10|0º) = (22|-120º)
I_a = I_AB - I_CA = (44|90º) - (22|-120º) = (64|80.104º)

3. May 7, 2010

### jfierro

Hmm it was indeed a problem of my vector algebra ^_^u. I was doing them by hand and when getting the imaginary part of I_CA I did only sin(-120º) and not 22sin(-120º)...

I made the same mistake when constructing the matrix, the matrix should've been:

$$\begin{bmatrix} -5j & 5j & 0 \\ 5j & 10+5j & -10j \\ 0 & -10j & 10j \end{bmatrix} \begin{bmatrix} I_a\\ I_b\\ I_c \end{bmatrix} = \begin{bmatrix} 220 \\ 0\\ -110 + 220\frac{\sqrt{3}}{2}j \end{bmatrix}$$

Since sin(120º) = 0.5sqrt{3}.

Thanks! :D