 # Analyzing AC Circuits using Thevenin and Norton Equivalent Circuits (Example 1)

### For the following circuit, find the Thevenin equivalent circuit at terminals a-b: Recall the process used to determine a Thevenin equivalent circuit:

1. Determine the open-circuit voltage across terminals a-b.
2. OR:
3. Determine the short-circuit current through terminals a-b.
4. (Choose whichever is easiest for steps 1,2)
5. Determine the equivalent impedance at terminals a-b when independent sources are "turned off".
6. (Do not "turn off" dependant sources.)
7. Solve for the unknown using Ohm's Law $$\mathbb{Z}_{th} = \mathbb{Z}_n = \frac{\mathbb{V}_{th}}{\mathbb{I}_n}$$

When considering steps one and two we will choose option one (open-circuit voltage at terminals a-b).

### Step 1: Get the open-circuit voltage at terminals a-b:

We start by applying Kirchhoff's Current Law (KCL) at terminal/node "a": $$\frac{\mathbb{V}_{th}}{10} + \frac{j\mathbb{V}_{th}}{20} + \frac{\mathbb{V}_{th}-20}{10+j10} = 2$$ Multiply the third term in the above equation by its complex conjugate: $$\frac{\mathbb{V}_{th}}{10} + \frac{j\mathbb{V}_{th}}{20} + \frac{\mathbb{V}_{th}-20}{10+j10} \Big( \frac{10-j10}{10-j10} \Big) = 2$$ $$\frac{\mathbb{V}_{th}}{10} + \frac{j\mathbb{V}_{th}}{20} + \frac{10\mathbb{V}_{th}-j10\mathbb{V}_{th}-200+j200}{200} = 2$$ $$\frac{\mathbb{V}_{th}}{10} + \frac{j\mathbb{V}_{th}}{20} + \frac{\mathbb{V}_{th}}{20} - \frac{j\mathbb{V}_{th}}{20} - 1 + j = 2$$ Multiply the above expression by 20 $$2\mathbb{V}_{th} + j\mathbb{V}_{th} + \mathbb{V}_{th} - j\mathbb{V}_{th} -20 + j20 = 40$$ $$3\mathbb{V}_{th} + j20 = 60$$ $$3\mathbb{V}_{th} = 60 - j20 = 63.25 \angle (-18.43^{\circ})$$

$$\mathbb{V}_{th} = 21.08 \angle (-18.43^{\circ})$$

### Step 2: Get the equivalent impedance at terminals a-b (with independent sources "turned off"):

With the independent sources "turned off" we have the following circuit with labeled impedances: The equivalent impedance can be summarized by the following parallel/series impedance combination: $$\mathbb{Z}_{th} = (10) \;||\; (-j20) \;||\; (10+j10)$$ $$= \frac{1}{\frac{1}{10}+\frac{j}{20}} \;\Big|\Big|\; (10+j10)$$ We now multiply the complex term in the denominator by its complex conjugate: $$= \frac{1(\frac{1}{10}-\frac{j}{20})}{\frac{1}{10}+\frac{j}{20}(\frac{1}{10}-\frac{j}{20})} \;\Big|\Big|\; (10+j10)$$ $$\mathbb{Z}_{th} = \frac{0.1-j0.05}{0.0125} \Big|\Big| 10+j10$$ $$= 8-j4 || 10+j10$$ Applying the rule of impedances in parallel and multiplying by the complex conjugates at the same time gives us: $$\mathbb{Z}_{th} = \frac{1}{\frac{1}{8-j4}\frac{8+j4}{8+j4} + \frac{1}{10+j10}\frac{10-j10}{10-j10}}$$ $$= \frac{1}{\frac{8+j4}{80} + \frac{10-j10}{200}}$$ $$= \frac{1}{0.1+j0.05+0.05-j0.05}$$

$$\mathbb{Z}_{th} = 6.667 \; \Omega$$

Now that we have determined the Thevenin voltage and impedance, our Thevenin equivalent circuit will look like the following: Let's take a look at another example problem.

Continue on to Thevenin and Norton circuits example 2