- #1
GeoMike
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Homework Statement
The problem, as given in the textbook is:
http://www.mcschell.com/circuit.jpg
Homework Equations
KCL, KVL
The Attempt at a Solution
The issue I'm having with this problem is that I get two different values for Ix depending on how I set up the equations for this circuit.
If I do 2 KVL loops and KCL using:
Supermesh: Ix(1) + I1(1) + I1(j1) - I2(j1) + 12 = 0
Loop I2: I2(j1 - j1 + 1) - I0(1) - I1(j1) = 0
KCL at left node: Ix = I1 + 2
Given I0: I0 = 4+j0 = 4
For this system I get Ix = -9/5 + j(13/5)
However, if I replace the supermesh with a KVL loop around the outer edges of the circuit (leaving the other loops unchanged):
Outer (edges) loop: Ix(1) + I1(1) + I2(-j1) + I0(1) = 0
For this system I get Ix = j(4/3)
Worse still, the solutions manual has:
KCL at left node: Ix - I1 = 2 + j0 = 2
I0 loop: 12 = I0(2) - I2
Outer (edges) loop: Ix(1) + I1(1) + I2(-j1) + I0(1) = 0
Given I0: I0 = 4+j0 = 4
With Ix = -1 - j2
I double checked my mesh equations and solved using Mathematica. I don't understand what is "wrong" with my approaches. Why am I getting different values for Ix? Shouldn't they be the same regardless of the analysis method used?
Thanks,
GeoMike
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