Mutual inductors can be a lot of fun, and sometimes a bit of an headache if you mess something up or represent them in a complicated way. Take for instance the following circuit designed with Circuit Lab a great online circuit design and simulator tool

This is one of the first excercises we were taught in our circuit class. As you can probably see, the circuit is composed by two current sources with different frequency, some resistors and a mutual inductor. The aim of the analysis is to find out the total power consumed by the resistors. Let's analyse each problem carefully before proceeding.

## The current sources

The circuit uses current sources only, good start, we do not have to worry about voltage sources and in some part of the circuit the currents are already known, however, as you might have noticed, the two current sources do not have the same frequency. I1 has a frequency of about 160 Hz while I2 is a constant current source. If you were to solve the circuit by hand, you would have to take this into account and solve the circuit two times: the first time with only I1 on and the second time with only I2 on, i.e. you would be using the superposition principle and assume that the circuit is linear.## The resistors

Resistors are assumed to be linear so they are not a problem in the slightest## The mutual inductor

This is where the fun begins!In cases such as this one, the current entering the inductor is known, in fact, the current entering the inductor is I1 when I2 is turned off, and I2 when I1 is turned off. Furthermore, as you could probably guess, when the current entering in L1 is constant, no voltage is induced in L2, therefore no current will be going through R3 in that case.

As to how to deal with the mutual inductor, since the current is known, you can use the following representation in the time domain

$$\begin{cases}

\psi_1 = L_{11}i_1 + L_{m}i_2\\

\psi_2 = L_{m}i_1 + L_{22}i_2

\end{cases}$$

which translates in the following equations in the phasors domain

$$\begin{cases}

V_1 = j\omega L_{11}I_1 +j\omega L_{m}I_2\\

V_2 = j\omega L_{m}I_1 +j\omega L_{22}I_2

\end{cases}$$

where

$$L_m = k \sqrt{L_{11}L_{22}}$$

However, we are not going to use these since the aim of this post is to use LTspice to solve the circuit.

## The LTspice model and the simulation

This is the circuit modelled in LTspice IVAs you can see, the spice directive "K L1 L2 0.9" is coupling the two inductors as specified by the data given in the problem while the ".tran 20ms" directive tells you that the analysis will last 20 milliseconds.

By clicking "run simulation" you can start the simulation.

First of all you can see that the current through the resistor R1 is the same as the one from the source I1 (since they are connected in series). Furhtermore, the current through R2 can be obtained as the sum of I1 and I2 using Kirchhoff's current law, again, as expected.

As far as the power is concerned, by holding alt and clicking on the component you can actually see the simulated power consumption in the time domain.

R1 and R3 are easy to figure out. The average power consumption of R1 is 100 watt while R3 consumes around 81 watt.

R2 is a litte bit tougher to figure out. LTspice lets you calculate the average value by just holding ctrl and clicking on the label on the graph.

The total power consumption is then 100+81+271 = 452 watt. By making the calculation manually, one finds out that actually the true result should be 450 watt but that is fine since the approxximation I did is quite accurate I would say and took a very short time to be made.

Note also that LTspice calculates the integral of the power and by remembering that an ideal resistor is characterized by the following fundamental equation

$$\delta W_E = \delta Q$$

you can actually approximate how much heat is dissipated by the resistors.

I hope this was interesting. You can download the LTspice model here.