Mutual Inductance (Parallel) Analysis Process

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SUMMARY

The discussion focuses on the analysis of mutual inductance in a parallel circuit involving three mesh currents (I1, I2, I3) and their interactions. The user expresses confusion regarding the correct formulation of mesh equations, particularly how to incorporate mutual inductance effects and voltage/current losses. Key insights include the understanding that pseudo voltage sources from mutual inductance do not behave as impedances and that the dot notation indicates the polarity of these sources. The user successfully reformulates the mesh equations after further analysis, demonstrating a clearer understanding of the circuit dynamics.

PREREQUISITES
  • Understanding of mesh analysis in electrical circuits
  • Familiarity with mutual inductance concepts
  • Knowledge of complex impedance and phasor representation
  • Ability to interpret circuit diagrams and dot notation
NEXT STEPS
  • Study the principles of mutual inductance in coupled inductors
  • Learn how to apply mesh analysis with mutual inductance in circuits
  • Explore the use of phasors in AC circuit analysis
  • Review examples of circuit analysis involving pseudo voltage sources
USEFUL FOR

Electrical engineering students, circuit designers, and professionals working on AC circuit analysis and mutual inductance applications will benefit from this discussion.

dwn
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Homework Statement



Please see attached images.

2. Relevant Questions

I'm fairly certain the mesh equations are incorrect. I'm having a hard time accounting for the mutual inductance and the voltage/current loss between each mesh. When we account for the lost voltage (using j300 in equations) do I include I3 in the first mesh equation? As I analyze the circuit, I must account for the loss from the 3rd mesh since there is a mutual inductance between L1 and L2. Is that correct?

If you could please walk me through the analysis process for the given equation, I think that would at least help me to understand the physical properties a little better.



The Attempt at a Solution



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dwn said:

Homework Statement



Please see attached images.

2. Relevant Questions

I'm fairly certain the mesh equations are incorrect. I'm having a hard time accounting for the mutual inductance and the voltage/current loss between each mesh. When we account for the lost voltage (using j300 in equations) do I include I3 in the first mesh equation? As I analyze the circuit, I must account for the loss from the 3rd mesh since there is a mutual inductance between L1 and L2. Is that correct?

If you could please walk me through the analysis process for the given equation, I think that would at least help me to understand the physical properties a little better.

The current that is associated with the mutual inductance pseudo voltage sources is the net current flowing through the "other" inductor. In this case, for example, mesh currents I2 and I3 flow through the second inductor, so their sum is that net current. These pseudo voltage sources do not behave as impedances for the mesh they are in, they act as voltage sources and are unaffected by the local mesh current.

Note that the dot notation tells you what polarity the pseudo source should have. Your text or notes should have a diagram similar to the following which shows the relationship:

attachment.php?attachmentid=70885&stc=1&d=1403731730.gif


If a net current flows "into" the dot on one inductor, then it should cause a current (via mutual inductance) to flow "out" of the dot on the other inductor. The voltage sources are oriented to reflect this, namely the polarity is such as to cause the current to flow in the desired direction with respect to the dots and net currents that flow into/out of them on the inductors.
 

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Thank you for the information gneill...I do remember seeing a lot of this in the text, but applying the principles to unfamiliar circuits can get very confusing!

After analyzing the circuit a second time:
Code:
-20+I[SUB]1[/SUB](3+j600) - I[SUB]2[/SUB](j600) + j300(I[SUB]2[/SUB]-I[SUB]1[/SUB]) = 0
-I[SUB]1[/SUB](-j600) + I[SUB]2[/SUB](5+j600+j600)+j300(I[SUB]3[/SUB]-I[SUB]2[/SUB])=0
-10∠25° + 0I[SUB]1[/SUB] - I[SUB]2[/SUB](j600) + I[SUB]3[/SUB](-j/200-j300) = 0

I found the net current through each inductor. The second and third meshes are the most difficult to translate and understand. Hopefully, I'm a little closer this time.
 

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