Imaginary Numbers/Mesh Analysis

[dwn]

Homework Statement

Image Attached

Homework Equations

Impedance/Reactance
Mesh Analysis

The Attempt at a Solution

KVL
-V + 2I1 + 4.7(I1 - I2) + j2(I1 - I3) = 0
4.7(I2 - I1) - j0.056(I2) + 2(I2 - I3) = 0
j2(I3 - I1) + 2(I3 - I2) + I3 = 0

(4.7 + j2)I1 - 4.7I2 - j2(I3) = 4V
4.7(I2-I1) - j0.056(I2) + 2(I2 - I3) = 0
-j2(I1) - 2I2 + (3 - j2)I3 = 0

Solve the second and third equation for I2 and I3, respectively.
I2 = (4.7I1 + 2I3)/(6.7 - j0.056)
I3 = (2(I2) + j2(I1)) / (3 - j2)

Substitute I3 into I2, in order to eliminate one of the variables.

(4.7I1 + 2(2I2 + j2(I1)/(3 - j2)) / (6.7 - j0.056)

This is where things get a little confusing for me, and I'm not sure how to proceed. I was going to multiply by the conjugate in order to simplify the equation, but am I able to do this with 2 variables (I1 and I2)?? I did it for the equation I3.

Also, is it possible to calculate this within Matlab? I know how to reduce linear equation (rref), but not sure how to account for the imaginary values (j) and the multiple currents (I1, I2, and I3). Any tips please? Hope it is easy to follow. Thanks so much!

 

[dwn]

Would I be better off converting the rectangular form to phasor form to make the calculation easier? I remember in class our professor mentioned addition and subtraction rectangular is convenient and polar when using multiplication and division.

 

[rude man]

First thing you do is label the componrnts R1, R2, R3, L and C. Don't work in numbers. You lose dimensional information and you're almost sure to make a math error.

Repost your KVL equations.

 

[dwn]

ω = 20
L = jwL = j2
C = -j/(wL) = -j0.056

Revised KVL
R1 = 2 R2 = 4.7 R3 = 2 R4 = 1

-V + (R1 * I1) + R2(I1 - I2) + L(I1 - I3) = 0
R2(I2 - I1) + CI2 + R3(I2 - I3) = 0
L(I3 - I1) + R3(I3 - I2) + I3*R4 = 0
==================================
I1((R1+R2) + L) - (R2*I2) - LI3 = V
(-R2*I1) + I2((R1+R2) + C) - 2I3 = 0
(-L*I1) - RI2 + I3((R3 +R4)+L) = 0
==================================
I1(6.7 + j2) - 4.7I2 - j2(I3) = 4∠0°
-4.7I1 + I2(6.7 - j0.056) - 2I3 = 0
-j2I1 - 2I2 + I3(3+j2) = 0

Is it best to convert these values to polar coordinates and then solve?

 

[dwn]

when converting j2 of the first equation into polar coordinates..specifically tan^-1(2/0). Do we define this as: 0∠90° ?
ThanksEDIT

I was able to solve the problem...but there must be a better way? Is there a shortcut to avoid "plugging and chugging"?

 

[rude man]

when converting j2 of the first equation into polar coordinates..specifically tan^-1(2/0). Do we define this as: 0∠90° ?
Thanks

To tell you the truth I use s=jw so there are no complex quantities involved in KCL or KVL or (what I use) sum of currents at each independent junction = 0 which is almost KVL. But that's because I'm familiar with the Laplace transform. Professional EE's all do that.

EDIT

I was able to solve the problem...but there must be a better way? Is there a shortcut to avoid "plugging and chugging"?

This is a pretty complex circuit so I'd say you did it the best way. Don't forget to use software for solving boring simultaneous equations. Again, with s=jw you don't have to worry about taking complex conjugates etc.

 

[rude man]

ω = 20
L = jwL = j2
C = -j/(wL) = -j0.056

Revised KVL
R1 = 2 R2 = 4.7 R3 = 2 R4 = 1

-V + (R1 * I1) + R2(I1 - I2) + L(I1 - I3) = 0
R2(I2 - I1) + CI2 + R3(I2 - I3) = 0
L(I3 - I1) + R3(I3 - I2) + I3*R4 = 0
==================================
I1((R1+R2) + L) - (R2*I2) - LI3 = V
(-R2*I1) + I2((R1+R2) + C) - 2I3 = 0
(-L*I1) - RI2 + I3((R3 +R4)+L) = 0
==================================
I1(6.7 + j2) - 4.7I2 - j2(I3) = 4∠0°
-4.7I1 + I2(6.7 - j0.056) - 2I3 = 0
-j2I1 - 2I2 + I3(3+j2) = 0

Is it best to convert these values to polar coordinates and then solve?

Definitely not. You're adding and subtacting terms.