Mastering Mesh and Nodal Analysis with KVL for Complex Circuits

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The discussion focuses on mastering mesh and nodal analysis using Kirchhoff's Voltage Law (KVL) for complex circuits. Participants share their attempts at formulating mesh equations, emphasizing the importance of consistent current direction and correctly identifying potential drops and rises. The conversation highlights the significance of supernodes when voltage sources are involved, which simplifies the analysis by allowing certain components to be excluded from the equations. Participants also discuss the necessity of clear notation to avoid confusion in their equations. Overall, the thread provides insights into solving circuit problems through careful application of KVL and nodal analysis techniques.
  • #31
M P said:
Lets stick to web. Let me just ask a quick question. my equation need to be equal to V30-V40/Z2?
Nope. In fact you'll find that, as was the case for mesh analysis, Z2 plays no role in the solution thanks to V3.

One thing, you should get into the habit of using parentheses to remove ambiguity from your equations. When you write: V30-V40/Z2 it could be interpreted as V30 - (V40/Z2) , or as (V30 - V40)/Z2. Now, the latter would be correct thanks to context in this case, but it is not always so simple to fathom the intended order of operations.
 
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  • #32
M P said:
V30-V1/Z1+V30/Z4+V40/Z5+V40-V2/Z3 = 0
V30-V40=V3
that is what I understood
That's looking good! Solve for V30.
 
  • #33
thank you for being patient..:s
 
  • #34
one more question is version of V1-V30/Z1-V30/Z4+V2-V40/Z3-V40/Z5=0 also work?
 
  • #35
M P said:
one more question is version of V1-V30/Z1-V30/Z4+V2-V40/Z3-V40/Z5=0 also work?
Yes, it is also a valid equation for the supernode. Of course, you could confirm that yourself by seeing if it yields the same results as the mesh analysis version. I'd still like to see more parentheses in the equation :rolleyes:
 
  • #36
Back on track after break with V40 = -60.1+5.31j and V30 = -45.9+19.51j in a need of your physics assistance
 
  • #37
M P said:
Back on track after break with V40 = -60.1+5.31j and V30 = -45.9+19.51j in a need of your physics assistance
Your value for V30 doesn't look right, and it's the important value here since it determines the current through Z4.

You can check your result by comparing the current through Z4 due to V30 to the result from the previous mesh analysis.
 
  • #38
Is it ok to show calcs? I did check it and I am not sure what is causing damage.
 
  • #39
M P said:
Is it ok to show calcs? I did check it and I am not sure what is causing damage.
Yes! Showing your work is highly encouraged. Trying to determine where things are going awry from the results alone requires a good deal more effort :) Show your calculations, preferably with notations as to what the steps are trying to accomplish.
 
  • #40
I have another one to check :confused: V40 = 60.1j - 214.5 V30 = 74.3j - 200.3 ?
 
  • #41
gneill said:
Yes! Showing your work is highly encouraged. Trying to determine where things are going awry from the results alone requires a good deal more effort :) Show your calculations, preferably with notations as to what the steps are trying to accomplish.

Now what calcs you prefer from first results or second ?
 
  • #42
M P said:
I have another one to check :confused: V40 = 60.1j - 214.5 V30 = 74.3j - 200.3 ?
You should be in a position to determine the node potentials from the results of your previous mesh calculations. V30, for example, is given by the potential across Z4. The mesh currents will yield that from Ohm's Law. Since your mesh calculations were successful, you have a "standard result" against which you can compare new results.

If there's a problem with the method by which you're finding new results, then while it's easy to declare the result wrong or right, it's more difficult to tell you why it's wrong and what to do to fix it from the result alone. Can you explain what you're trying?
 
  • #43
M P said:
Now what calcs you prefer from first results or second ?
If by "first results" you mean the mesh analysis portion of the problem, then that's not required since I believe that you found the correct results there. If you are having difficulty achieving the same results with nodal analysis, then those are the calculations that need to be examined.
 
  • #44
M P said:
one more question is version of V1-V30/Z1-V30/Z4+V2-V40/Z3-V40/Z5=0 also work?
I was trying to do complex numbers on this ...
 
  • #45
M P said:
one more question is version of V1-V30/Z1-V30/Z4+V2-V40/Z3-V40/Z5=0 also work?
and on this ? that is what I understood to obtain V3 and V4 of this with complex numbers..
 
  • #46
none of them seems ok as you explained?
 
  • #47
that is what coursework shows using substitution
 
  • #48
I can't comment on what I don't see. Can you show more of your work?
 
  • #49
gneill said:
I can't comment on what I don't see. Can you show more of your work?
 

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  • #50
gneill said:
I can't comment on what I don't see. Can you show more of your work?

Page 2
 

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  • #51
I bet you smiling now ;)
 
  • #52
x1.gif


Check the sign of the term.
 
  • #53
gneill said:
View attachment 76995

Check the sign of the term.
before I go further -55.74 - 20.06j + 0.25 V40 - 0.05 j V40 ?
 
  • #54
M P said:
before I go further -55.74 - 20.06j + 0.25 V40 - 0.05 j V40 ?
I'm not seeing where that's coming from. Sorry.

Did you make a V30 substitution? Really you should eliminate V40 instead, since you need V30 to find the current through Z4.
 
  • #55
thank you I will do as suggested and see how it goes..
 
  • #56
thank you for all your help
 
  • #57
can you please confirm if I have conducted my kvl loops correctly as per diagram on post #9

V1 - I1Z1 - Z4(I1-I2) = 0...Eq 1

-V3 - Z5(I2-I3) - Z4(I2-I1) = 0...Eq 2

I3Z3 - V2 - Z5(I3-I2) = 0...Eq 3
 
  • #58
It all looks fine except for the sign of the Z3 term in the third equation. I3 causes a potential drop in Z3.
 
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  • #59
thnx gneill
 
  • #60
I3(-Z3) - V2 - Z5(I3-I2) = 0...Eq 3 new and revised eq 3
 

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