# Homework Help: Mutual Inductance (Parallel)

1. Jun 25, 2014

### dwn

1. The problem statement, all variables and given/known data

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.

3. The attempt at a solution

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2. Jun 25, 2014

### Staff: Mentor

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:

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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3. Jun 25, 2014

### dwn

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 (Text):
-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.