Issue with mesh analysis of 4 loop circuit

In summary, to perform mesh analysis on a 4 loop circuit, label each loop with a variable and use Kirchhoff's Voltage Law to set up equations for each loop. The purpose of using mesh analysis is to calculate individual loop currents and identify any potential issues in a circuit. Dependent sources can be handled by incorporating their equations into the analysis. Mesh analysis cannot be used for circuits with non-linear elements and has limitations such as only being applicable to linear circuits and planar circuits.
  • #1
GeoMike
67
0

Homework Statement


The problem, as given in the textbook is:
http://www.mcschell.com/circuit.jpg [Broken]

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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  • #2


Dear GeoMike,

Thank you for bringing this issue to our attention. It is not uncommon to encounter discrepancies in circuit analysis problems, especially when using different methods. In this case, I believe the issue lies in the way the supermesh and outer loop have been defined.

When using the supermesh method, one must be careful to include all the necessary elements in the loop. In this case, it seems that you have missed including the 1Ω resistor in the supermesh loop. This results in a different value for Ix, as you have noticed.

Similarly, when using the outer loop method, one must ensure that the loop includes all the elements. In this case, it seems that the 1Ω resistor has been included twice in the loop, leading to a different value for Ix.

I would recommend double-checking your mesh equations and ensuring that all the elements are correctly included in the loops. This should result in the same value for Ix regardless of the method used.

I hope this helps. Good luck with your analysis!
 
  • #3


it is important to approach problems with a critical and analytical mindset. In this case, it seems like there may be an error in one of the equations or calculations, as the different approaches are yielding different results for Ix. It would be helpful to double check the equations and calculations to identify the source of the discrepancy.

Additionally, it is important to consider the assumptions and limitations of the mesh analysis method. Are all the necessary equations and variables being considered? Is the circuit too complex for mesh analysis to accurately solve? It may be helpful to try a different method of analysis, such as nodal analysis, to see if the results are more consistent.

Ultimately, as a scientist, it is important to thoroughly investigate and troubleshoot any discrepancies or issues with the analysis method being used to ensure accurate and reliable results.
 

1. How do I perform mesh analysis on a 4 loop circuit?

To perform mesh analysis on a 4 loop circuit, you will first need to label each loop with a variable. Then, use Kirchhoff's Voltage Law to set up equations for each loop, taking into account the direction of current flow. Solve the equations simultaneously to find the currents in each loop.

2. What is the purpose of using mesh analysis in circuit analysis?

The purpose of using mesh analysis is to calculate the individual currents in each loop of a circuit. This can help in determining the power dissipated in each loop and identifying any potential issues or malfunctions in the circuit.

3. How do I handle dependent sources in mesh analysis?

Dependent sources can complicate mesh analysis, but they can be handled by using the equations for the dependent source in each loop. For example, for a voltage-controlled voltage source, the equation would be V = kI, where k is the gain factor and I is the current in the loop.

4. Can mesh analysis be used for circuits with non-linear elements?

No, mesh analysis is only applicable to linear circuits. Non-linear elements such as diodes or transistors cannot be analyzed using mesh analysis. In these cases, other circuit analysis methods such as nodal analysis may be used.

5. Are there any limitations to using mesh analysis in circuit analysis?

Yes, there are a few limitations to using mesh analysis. It can only be used for circuits with independent or dependent sources, resistors, and linear elements. Additionally, the circuit must be planar, meaning all the elements and connections lie on a flat surface. If the circuit does not meet these criteria, other methods of circuit analysis may need to be used.

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