Mesh Analysis AC, solving simultaneous equations....

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Discussion Overview

The discussion revolves around the application of mesh analysis in an AC circuit, focusing on the formulation and solution of simultaneous equations derived from Kirchhoff's laws. Participants share their equations, calculations, and seek clarification on specific aspects of their work.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents their mesh equations and seeks validation of their approach and calculations.
  • Another participant provides a cursory approval of the equations but references a related thread for further context.
  • A participant expresses doubt about their equations being different from those in the referenced thread and shares detailed calculations, questioning the correctness of their results.
  • Concerns are raised about the interpretation of negative current values in complex notation, with suggestions that they could represent phase shifts.
  • One participant points out a potential error in the sign of the imaginary component in one of the mesh equations, suggesting a need for correction.
  • Further clarification is sought regarding the negation of the voltage term in the equations, leading to a discussion about the correct representation of the voltage in the mesh analysis.
  • After receiving feedback, a participant revises their equations and shares updated calculations, indicating progress in solving for the mesh currents.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the initial equations and calculations, with some pointing out potential errors while others provide clarifications. The discussion remains unresolved regarding the accuracy of the calculations and the interpretation of negative values in complex notation.

Contextual Notes

Participants note the importance of correctly applying phase shifts and the implications of negative values in the context of AC circuit analysis. There are also references to the need for careful consideration of complex number representations in calculations.

kibara
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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:
bBF28L


mesh-Analysis.png


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 ?
 

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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.
 
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)##.
 
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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.
 
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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.
 
kibara said:
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:
kibara said:
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.
 
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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 <117.9128o A

Now moving to node analysis :)
 

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