Thevenin and Norton Circuits for Independent Sources

 

Thevenin and Norton Equivalent circuits are most often used to simplify a circuit.  They are useful when you are interested about the behavior of an element connected to a rather complicated system through terminals a and b (this scenario is addressed in part c of the question).

First thing to look at is what is a Thevenin and Norton equivalent circuit.

Thevenin Circuits take the form of:

Where \(\mathrm{V}_{\mathrm{Th}}\) is the thevenin equivalent voltage, sometimes called \(\mathrm{V}_{\mathrm{OC}}\) the open circuit voltage, and \(Z_{\mathrm{Th}}\) is the total impedance (impedance being any opposition to current flow, whether through capacitance, resistance, or inductance) of the circuit, sometimes called \(\mathrm{Z}_{\mathrm{eq}}\).

Norton Equivalent circuits take the form of:

Where \(\mathrm{Z}_{\mathrm{Th}}\) is once again the total impedance, and \(I_{S C}\) is the short circuit current.

Basically, you are taking your original circuit (like the one given in the problem) and simplifying it down to one of these two forms.

Since \(\mathrm{Z}_{\mathrm{Th}}\) appears in both of them we will tackle how to solve that first (and in general starting at finding the total impedance is not a bad idea for reasons we will get to later).

To solve for \(\mathrm{Z}_{\mathrm{Th}}\) in a circuit with only independent sources, such as independent voltage sources and, in the case of the problem, an independent current source, all you have to do is deactivate the independent sources and simplify the remaining impedances into one. To deactivate the source, you set the voltage for voltage sources or current for current sources to 0, so voltage sources become short circuits/wires, and current sources become open circuits/breaks:

NOTE: dependent current sources are usually drawn in a diamond and depend on another voltage or current within the circuit while independent sources are drawn in a circle and are not modified by the circuit.

To find \(\mathrm{Z}_{\mathrm{Th}}\):

Remember that when dividing two angles you subtract the denominator from the numerator, when multiplying two angles you add them

NOTE: While this doesn’t play a huge part in calculating \(\mathrm{Z}_{\mathrm{Th}}\) in this problem, we are concerned with what the a/b terminal sees, so, if applicable, we want to start simplifying from the side farthest from the terminal:

NOTE: This panel is not part of the problem, it’s merely here to illustrate a point about the order in which you simplify the impedances.

Next we should calculate \(\mathrm{V}_{\mathrm{Th}}\) in order to find the Thevenin equivalent circuit. As stated before, \(\mathrm{V}_{\mathrm{Th}}\) is also called the open circuit voltage, and that’s because we have to find the voltage across terminals a and b:

This answers part a

Now to calculate the Norton equivalent. To do this, you need \(I_{S C}\), which you can get by calculating the current through a hypothetical short circuit drawn across terminals a and b:

This answers part b

You may have noticed that \(\mathrm{V}_{\mathrm{Th}}\) was just equal to the input current \(22<30\) times \(\mathrm{Z}_{\mathrm{Th}}\). The input current however, was equal to \(\mathrm{I}_{\mathrm{SC}}\), which means that \(\mathbf{V}_{\mathrm{Th}}=\mathbf{I}_{\mathbf{S C}} * \mathbf{Z}_{\mathrm{Th}}\), and this is true for all cases of finding the Thevenin and Norton equivalent. So theoretically, once you find two out of the three, you can use this relationship to solve for the last one.

Now for the last part, we need to insert a new impedance across a and b and calculate the current through it.

This really showcases why Thevenin and Norton equivalents can be so useful, as we now have a much simpler circuit we can use to solve the problem. For this particular case, I will use the Thevenin equivalent, as we will just have two impedances in series, and we can add them up and use \(\mathrm{V}=\mathrm{I}* \mathrm{z}\) to calculate the current (though you can technically use either):

This answers part c