Mesh Analysis AC, solving simultaneous equations....

[kibara]

Homework Statement

Hi guys I'm doing the mesh analysis of AC circuit and looking for some guidance.
Here is the mesh picture and some components data:

Homework Equations

Kirchoff's voltage law and current law.

The Attempt at a Solution

I decided that mesh currents are going in clockwise direction.

mesh1: V1-I1(Z1+Z4)+I2Z4=0

120-I1(2-J5)+I2(-J5)=0 eq(1)

mesh2: -V3-I2(Z4+Z5)+I1Z4+I3Z5=0

-14.14+J14.14-I2(J4-J5)+I1(-J5)+I3(J4)=0 eq(2)

mesh3: -V2-I3(Z3+Z5)+I2Z5=0

-J120-I3(4+J4)+I2(J4)=0 eq(3)

Are these equations correct ?

 

[gneill]

Looks good (after a cursory scan).

See also the thread: https://www.physicsforums.com/threads/ac-circuit-analysis-mesh-and-nodal.791744/

 

[kibara]

Thanks for your response gneill.

I've seen the above link previously that;s why I doubted my above equations as they are different so is my whole calculation which is as follows:

Multiplying eq(2) by 4+j4 and eq(3) by j4

-113.12-I2(4-j4)+I1(20-j20)+I3(-16+j16)=0 new eq(2)

480-I3(-16+j16)+I2(-16) new eq(3)

Adding (3) to (2)

366.88+I1(20+j20)-I2(20-j4)=0 eq(4)

Multiplying eq(1) by (20-j4) and (4) by (-j5)

120(20-j4)-I1(2-j5)(20-j4)+I2(-j5)(20-j4)=0

2400-j480-I1(20-j108)+I2(-20-j100)=0 new eq(1)

366.88(-j5)+I1(20+j20)(-j5)-I2(20-j4)(-j5)=0

-j1834.4+I1(100-j100)-I2(-20-j100)=0 new eq(4)

Subtracting new eq(4) from new eq(1)

j1834.4-j480+2400-I1(20-j108)-I1(100-j100)=0

2400+j1354.4-I1(120-j208)=0

I₁ = 0.1090 + j11.4756 (to 4 d.p.)

Substituted value of I₁ into initial eq(1) and obtained

120-(0.1090+j11.4756)(2-j5)+I2(-j5)=0

-57.596-j22.4062+I2(-j5)= -120

I2(-j5) = -120+57.596+j22.4062

I2= -4.4812-j12.4808

Substituted value of I₂ into initial eq(3) and solved for I3:

-j120+(-4.4812-j12.4808)(j4)-I3(4+j4)=0

49.9232-j17.9248-I3(4+j4) = j120

I3=-11.0002-j23.481 Can someone tell me what did I do wrong. Also I'm not sure what to think about negative value of current when expressed in complex notation. I mean if the circuit would be pure resistance I would swap negative value for positive, right ? let's say if I= -12 A i would swap for I=12 A.

Another thing. In what format can I input complex notation in Octave ? Can it be:

octave:1> A = [4+j4, -j5+j2]

Regards.

 

[gneill]

Taking another look at your mesh equations, I noticed that you've made $V_3 = 14.14 - j14.14$ for your mesh2 equation. Why the negative imaginary component? Should be $10\sqrt{2}(1 + j)$.

 

[scottdave]

As far as negative values in complex numbers, it could be thought of as a phase shift (or opposite direction). A phase shift of 180° is in the opposite direction.
As @gneill pointed out, it looks like you have a mistake, which needs fixing. But the value of -11 - j23.48 converted to polar form is 25.93 at an angle of -115.1°, if that helps you to visualize it.

 

[kibara]

Hello again.

gneill V3 = 20 <45 = 14.14 + j14.14 V3 is going anticlockwise and I2 clockwise hence -V3 = -(14.14+j14.14) = -14.14-j14.14 right ?

scottdave I completely forgot about converting back to polar form.

Thanks for your input guys.

 

[gneill]

gneill V3 = 20 <45 = 14.14 + j14.14 V3 is going anticlockwise and I2 clockwise hence -V3 = -(14.14+j14.14) = -14.14-j14.14 right ?

Right. But in your mesh equations you've written:

mesh2: -V3-I2(Z4+Z5)+I1Z4+I3Z5=0

-14.14+J14.14-I2(J4-J5)+I1(-J5)+I3(J4)=0 eq(2)

(My emphasis in red)

Note that only the first term of that complex number has been negated.

 

[kibara]

Hi. Thanks for all your help. I have now solved for I as follows:

mesh 1: 120-I1(2-j5)+I2(-j5)=0
mesh 2: -14.14-j14.14-I2(-j)+I1(-j5)+I3(j4)=0
mesh 3: -j120-I3(4+j4)+I2(j4)=0

Multiplying (2) by (4+j4) and (3) by (j4)

-j113.12-I2(4-j4)+I1(20-j20)+I3(-16+j16)=0 equation (2A)

480-I3(-16+j16)+I2(-16)=0 equation (3A)

Adding (3A) to (2A)

480-j113.12+I1(20-j20)+I2(-20+j4)=0 equation (4)

Multiplying (1) by (-20+j4) and (4) by (-j5)

-2400+j480-I1(-20+j108)+I2(20+j100)=0 equation(1A)

-565.6-j2400+I1(-100-j100)+I2(20+j100)=0 equation (4A)

Subtracting (4A) from (1A) and solving for I1

-1834.4+j2880+I1(120-j8)=0

I1=16.8119-j22.8792

Substituting value of I1 into equation (1) and solving for I2

I2=25.9636-j40.1544

I=I1-I2=-9.1517+j17.2752 A

or I=19.5496 <117.9128^o A

Now moving to node analysis :)