- #1
jfierro
- 20
- 1
[SOLVED] Unbalanced Three-phase circuit.
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
- LCK
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)?
Homework Statement
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
Homework Equations
- LCK
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: