Thevenin and Norton

What and why?

Thevenin and Norton methods are used to simplify any linear circuit to make an equivalent circuit with a single voltage source and resistor connected to a load. Thevenin's Theorem is useful in analyzing circuits where one particular load resistor is subject to change.

When we combine resistors in parallel or in series, we don't care about any of the nodes or branches inside the black box containing the resistive network. We only care about the two terminals that we keep fixed. The resistive network across these terminals is replaced with a single, equivalent resistor.

Now, using Thevenin and Norton, we are trying to find an equivalent circuit to an initial circuit that not only has resistors, but a variety of other sources as well.

Consider attaching an arbitrary voltage to a Thevenin and Norton circuit:

We can find an expression relating i and v for both circuits. For Thevenin, Ohm's law gives us:

$i = \frac{v-V_{Th}}{R_{Th}} = \frac{v}{R_{Th}}-I_N$

From Norton, we apply KCL and Ohm's law across $R_N$:

$i = \frac{v}{R_N}-I_N = \frac{v}{R_{Th}}-I_N$

As expected, both circuits give us the same relationship!

Example: Thevenin Equivalence

Let's try an example, lifted from All About Circuits:

Returning to our example circuit from the superposition chapter, let's assume that $R_2$ is a variable resistor. While we could still use superposition or nodal analysis to solve for the voltage difference across $R_2$, we would have to repeat this process every time the value of $R_2$ changed, which is time-consuming and tedious:

It would be much simpler to instead replace the rest of the circuit with a Thevenin equivalence, reducing the entire circuit down to a single voltage supply and resistor in series:

Let's begin! First, we will replace $R_2$ with an open circuit:

Now we can determine the voltage differences throughout the circuit using nodal analysis. We should get the following:

Understanding checkpoint: Do you understand how to use the other circuit analysis methods at our disposal to find these values?

Note the $11.2 V$ voltage difference across what used to be the load resistor terminals. This value is our $V_{th}$, or $E_{Thevenin}$ in the diagram:

Now we just need to find $R_{th}$! We do so by removing the voltage supplies and load resistor from our original circuit:

We can see that $R_1$ and $R_3$ are in parallel - thus, the equivalent resistance between what used to be the load resistor's terminals is $R_{eq} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_3}} = 0.8 \Omega$.

Understanding checkpoint: How do we know $R_1$ and $R_3$ are in parallel?

Substituting $R_{eq}$ for $R_{th}$, we have our completed Thevenin equivalent circuit:

Example: Norton Equivalence

We will now find the Norton equivalence for the same circuit as above:

Finding the Norton equivalence requires us to reduce the entire circuit down to a single current source and resistor in parallel:

Again, we begin by removing $R_2$, our load resistor. This time, we will replace it with a short circuit, and repeat our circuit analysis. We should get the following:

The current across the the short circuit becomes our $I_{No}$ in the Norton equivalent circuit:

Note that in this diagram, the arrow notation of the current source points opposite the current flow.

We now find $R_{no}$ in much the same style as we did $R_{th}$:

Giving us our complete Norton equivalent circuit: