Frequency domain and mesh currents

[Cocoleia]

Homework Statement

Homework Equations

The Attempt at a Solution

So far I was able to transform each element into the frequency domain:

I am just unsure how to work with the dots to find the mesh currents. Can someone explain? Thanks.

 

[cnh1995]

Have you studied dependent sources?
In circuits with mutual inductance, you represent mutually induced emf using current controlled voltage source(s).

 

[Cocoleia]

Have you studied dependent sources?
In circuits with mutual inductance, you represent mutually induced emf using current controlled voltage source(s).

I believe we have studied this but I am definitely not clear on the subject.

 

[cnh1995]

I believe we have studied this but I am definitely not clear on the subject.

Ok. Think about what the dot convention means here.
When the current "enters" a coil at the dot, the dotted terminal of the other coil becomes positive i.e.both the dots become positive (and negative) simultaneously.

You have assumed i2 "leaving" the dot of the 6H coil. What will be the polarity of the dot?

 

[Cocoleia]

Ok. Think about what the dot convention means here.
When the current "enters" a coil at the dot, the dotted terminal of the other coil becomes positive i.e.both the dots become positive (and negative) simultaneously.

You have assumed i2 "leaving" the dot of the 6H coil. What will be the polarity of the dot?

Negative since it is leaving ?

 

[cnh1995]

Negative since it is leaving ?

Yes. So what is the polarity of the dot of the 8H coil? That would be the polarity of the mutual voltage induced in the 8H coil.i.e. the emf induced in 8H coil due to the current in the 6H coil. Add a current controlled voltage source in series with the 8H coil with the polarity you just found. Controlling current would be i2 for mutual emf in the 8H coil and i1-i2 for mutual emf in the 6H coil.

 

[Cocoleia]

Yes. So what is the polarity of the dot of the 8H coil? That would be the polarity of the mutual voltage induced in the 8H coil.i.e. the emf induced in 8H coil due to the current in the 6H coil. Add a current controlled voltage source in series with the 8H coil with the polarity you just found. Controlling current would be i2 for mutual emf in the 8H coil and i1-i2 for mutual emf in the 6H coil.

Would these be the two equations?

 

[gneill]

Your calculation of Zc is not correct. Zc should be of comparable magnitude to the other impedances in the circuit.

Your first loop equation does not include a mutual inductance term, and the signs for the currents flowing through the j200 inductor are not correct.

For your second loop you didn't account for I1's contribution to a mutual inductance term, and I2 should make two contributions (I2 flows through both inductors, hence though both dots).

I find that it can be helpful to re-draw the circuit with the mutual inductance inspired voltage sources inserted. Start out with them having their negative ends anchored to the dot of their inductor. Figure out their values by using labelled currents and the dots to assign appropriate signs (after which you can adjust the voltage source directions to eliminate negative terms if you wish), then write the loop equations. This way you can't drop terms if you do your KVL walks methodically around the loops.

 

[Cocoleia]

Your calculation of Zc is not correct. Zc should be of comparable magnitude to the other impedances in the circuit.

Your first loop equation does not include a mutual inductance term, and the signs for the currents flowing through the j200 inductor are not correct.

For your second loop you didn't account for I1's contribution to a mutual inductance term, and I2 should make two contributions (I2 flows through both inductors, hence though both dots).

I find that it can be helpful to re-draw the circuit with the mutual inductance inspired voltage sources inserted. Start out with them having their negative ends anchored to the dot of their inductor. Figure out their values by using labelled currents and the dots to assign appropriate signs (after which you can adjust the voltage source directions to eliminate negative terms if you wish), then write the loop equations. This way you can't drop terms if you do your KVL walks methodically around the loops.

View attachment 113697

Thanks for such a complete answer. One more question. Can you always attach the mutual inductance voltage source by the negative end to the dot? For example in this circuit:

Would I attach it to the dot, or do I need to take into account the sign of V1 and V2 ?

 

[gneill]

Attaching the negative end to the dot is just a convenient starting point. It follows the convention that a current flowing into one dot results in a current flowing out of the other dot, and really is just a matter of getting the sources in place without considering any details of the circuit currents. Once this is done it's up to you to define the mesh currents and then check whether they are flowing into or out of the dots. You assign signs to the voltage values on the sources accordingly.

You are of course free to reorient the polarities of the voltage sources as you like, provided that you change the signs of their values accordingly. In the end it's the currents flowing into or out of the dots that dictate the induced voltage polarities.

In your example i₁ is flowing into the dot on the 4H inductor, and so the voltage source value attached to the other inductor gets a positive sign. i₂, on the other hand is flowing out of the dot of the 6H inductor, so the voltage value assigned to the source on the 4H inductor gets a negative value.

The v₁ and v₂ shown in the diagram appear to be related to the potential drops caused by the mesh currents across the inductors (inductor impedances), or they're just labels for the net voltage across the inductor that you need to solve for with the indicated polarity being defined.