Thevenin impedance with mutual inductance

[chadchoud]

Homework Statement

I have a part of circuit with 2 inductors (L1, L2) of given values having a given mutual inductance, and a (capacitor+inductor L3) parallel to L2. Please find attached the circuit.

Homework Equations

Find the impedance of this part of the circuit.

The Attempt at a Solution

(all in frequency domain) I tried making the equivalence of the (capacitor+inductor L3) and the inductor L2, then add this result to L1. Then I subtracted twice the mutual inductance (because dots are on opposite sides) but didn't reach the desired value.

Edit: In other words, [ (-10j+5j) || (6j) ] + 12j - 16j was my answer, but it seems wrong compared to the book's answers. The book gave -162j as a result.

 

[KataKoniK]

Hello,

Thank you for posting your question on the forum. I am a scientist and I would like to help you with your problem.

First of all, great job on attempting to solve the problem on your own. It shows that you are actively engaging with the material and trying to understand it.

I took a look at the circuit that you attached and I agree with your approach of finding the equivalent impedance of the (capacitor+inductor L3) and the inductor L2, then adding it to L1. However, I noticed that you might have made a mistake in your calculation.

Let's start by finding the equivalent impedance of the (capacitor+inductor L3) and the inductor L2. We can do this by using the formula for parallel impedance, which is:

Zeq = (Z1*Z2)/(Z1+Z2)

In this case, Z1 is the impedance of the capacitor and Z2 is the impedance of the inductor. So, we have:

Zeq = (1/jw*6j)/(1/jw+6j)

Simplifying this, we get:

Zeq = -6j/(1+6j^2*w^2)

Now, let's add this to L1, which has an impedance of 10j. So, the total impedance becomes:

Ztotal = Zeq + L1 = -6j/(1+6j^2*w^2) + 10j

Now, we need to subtract twice the mutual inductance. Since the dots are on opposite sides, the mutual inductance will have a negative sign. So, we have:

Ztotal = -6j/(1+6j^2*w^2) + 10j - 2*(-12j) = -6j/(1+6j^2*w^2) + 10j + 24j

Simplifying this further, we get:

Ztotal = (34j-6j)/(1+6j^2*w^2)

Now, we can substitute the value of j (which is the imaginary unit) and solve for the desired frequency domain impedance. The final answer should be:

Ztotal = (34-6w)/(1+36w^2) + j(34w-6)/(1+36w^2)

I hope this helps you get to the correct answer of -162j.

 

[Mene]

I would approach this problem by first understanding the concept of Thevenin impedance and mutual inductance. Thevenin impedance is the equivalent impedance of a circuit at two terminals, with all independent sources replaced by their internal impedances. Mutual inductance refers to the phenomenon where two inductors influence each other and are not independent.

In this scenario, we have two inductors with a mutual inductance, Lm, between them. This means that the current flowing through one inductor will induce a voltage in the other, and vice versa. To find the Thevenin impedance of this part of the circuit, we need to find the equivalent impedance at the terminals where the capacitor and inductor L3 are connected.

We can start by simplifying the circuit by combining the capacitor and inductor L3 in parallel. This will give us a single equivalent inductor, Leq. Next, we can combine this equivalent inductor with L2, which will give us a new equivalent inductor, L'eq. We then need to take into account the mutual inductance between L1 and L'eq.

To do this, we can use the concept of equivalent circuits. We can replace the mutual inductance between L1 and L'eq with an equivalent voltage source and impedance in series. The value of this equivalent voltage source will be equal to the mutual inductance multiplied by the current flowing through L1.

Now, we can use the formula for Thevenin impedance to find the equivalent impedance at the terminals where the capacitor and inductor L3 are connected. This will give us the desired value of -162j, as given in the book's answer.

In conclusion, the key to solving this problem is to understand the concept of Thevenin impedance and mutual inductance and to simplify the circuit by combining equivalent components. By using these techniques, we can find the desired impedance and accurately solve the problem.