Thevenin impedance with mutual inductance

Joined Dec 18, 2008
11
Hello. I attached my question to this thread.

My attempt was [ (-10j+5j) || (6j) ] + 12j - 16j == -34j, but it seems wrong compared to the book's answers. The book gave -162j as a result.
Can you help me figure out what is wrong?

Thank you TurboDsm4g63

Joined Dec 10, 2008
4
If the attachment was better it would help

Joined Dec 18, 2008
11
I'm afraid I lost the original picture. Below are the values:

W = 10^6 rad/s
3 inductors 12microH, 5microH, 6microH, Mutual is 8microH
Capacitor is 0.1microF

I thought the circuit was clear enough The Electrician

Joined Oct 9, 2007
2,724
You can't solve this problem by just calculating parallel and series combinations of impedances; you'll have to use the loop method.

Assume that a voltage source of 1 volt and a frequency of w=10^6 is applied to the top of the 12 uH inductor. Then your circuit will have two loops, one on the left consisting of the source, the 12 uH inductor and the 6 uH inductor. Let the current in that loop be designated I1, and let the current be oriented clockwise. Let the current in the right hand loop be designated I2, and also oriented clockwise.

Then the voltage across the 12 uH inductor will be j*12*I1 - J*8*I1 + J*8*I2.

The voltage across the 6 uH inductor will be J*6*I1 - J*8*I1 - J*6*I2.

Summing the drops across the 12 uH and 6 uH inductors and equating to the voltage of the source, we get as our first equation:

J*2*I1 + J*2*I2 = 1

For our second equation, the drop across the 6 uH inductor is J*8*I1 - J*6*I1 + j*6*I2. The drop across the 5 uH inductor is J*5*I2 and the drop across the capacitor is -j*10*I2.

This gives a second equation of:

j*2*I1 + j*1*I2 = 0

If you solve these two simultaneous equations, you will get I1 = j*1/2. Divide the value of the voltage source, which is 1, by this current and you get Zin = -j*2.

I verified this result with a spice simulation. I can't explain why your book has a different answer.

Thank you Electrician Thanks for the explanation I also figured out another way to do it - by using the T-equivalent of the 2 inductors with mutual inductance. It turns out to be: