The objective of this post is to analyze circuits that can be transformed into simpler Thevenin equivalent circuits. We looked at several types of configurations and worked out the theoretical and simulated results of the Thevenin equivalent circuit, and examined specific points in the circuits including their resistance, current and voltage. As an additional experiment, I’m going to be playing around with the typography on this post to aid readability, especially since there are lots of inline math equations!
Thevenin equivalent circuits allow us to simplify a complex circuit into one that consists of an independent voltage source that is in series with a resistor. The independent voltage source and resistor are denoted by Vth and RTh respectively. To transform the circuit to a thevenin circuit it needs to be a linear circuit that has two terminals at its end. The Vth can be calculated with voltage division. This can be done because the voltage of the two terminals is equivalent to the Vth. To calculate the RTh we need to use the short circuit method at the terminals to find the isc which can be used in the equation . If the circuit only contains independent sources then the RTh equals the equivalent resistance. Another method that will be used this lab is to use the equation . To utilize this equation, all sources must be taken out for a test voltage source. After finding the test current, the equation can be used to find RTh.
A computer with the MATLAB software installed will be needed for this experiment. The first step in beginning this lab is to create the Simulink model of the circuit shown below. Remove the 60 V voltage source and then add a test voltage source across the terminals a and b in the above circuit to build the second circuit. To fill out the table, the test source had to be switched out with the given values and to measure the 𝐼test. Using the values from the table and the formula below, RTH can be solved.
For the third circuit, short the terminals a and b of the created circuit. Measure the current Isc and fill up the simulation in Table 3. Add an 8Ω resistor to the circuit as shown in the circuit below and find the current in the resistor. Add the 8Ω resistor to the Thevenin equivalent circuit you found in Steps 1, 2, and 3 and calculate the current 𝑖. Following directly is our filled out table with data and the built circuits.
|Calculation Results||Simulation Results|
|Vtest||1 V||5 V||10 V||100 V|
|Theoretical Solution||Simulation Solution|
For all of the circuits given in the experiment, we had to calculate values with the use of equations and then check with Simscape in MATLAB. For the first circuit we had to find the V thevenin. Voltage division was utilized to find the Vth is 48V with the equation . With the help of MATLAB we quickly verified the voltage to be 48V.
Next we found the RTh by putting a test voltage in between a and b and taking out the 60V to calculate the test current. The mesh current method proved to be useful in calculating the test current. After isolating the current, we get an equation for the test current in terms of the test voltage at . With this equation we found the test currents, tabulated in table 1, at the different test voltages given. Now we have both the test current and test voltage so we can find the RTh with the equation . The R thevenin was found to be 16Ω and will always be 16Ω no matter the change in voltage. With calculating the current of circuit 2 in MATLAB we were able to verify that our calculations were correct.
We now look back at Circuit 1 to find the isc that flows through the terminals a and b. The current isc can be found by dividing the Vthevenin by the Rthevenin in the equation . By using this equation we can find that isc is 3A. After setting up Circuit 3 in MATLAB we found that isc is 3A just like our initial calculations.
After looking at thevenin equivalent circuits we will now look at the actual circuit which can be seen in Circuit 4. Circuit 1 is quite similar to Circuit 4 except the fourth circuit has an 8Ω resistor in place of the terminals. To calculate the current going through the 8Ω where the terminals were, we configured the circuit in MATLAB. The result of running the programs was that the current displayed was 2A. Then we move on to thevenin equivalent circuit and add the 8 ohm resistor to it.
We use Simscape to display the current after the 8Ω resistor which turns out to be 2A. The current at 8 ohms for Circuits 4 and 5 are the same. The thevenin equivalent circuit is essentially the same as Circuit 4 excluding the 8Ω resistor at the end. These findings show that by finding the equivalent thevenin circuit we can simplify the circuit to make it easier to find the current at 8Ω.
In this lab we created thevenin equivalent circuits and assessed the Vth and Rth from the circuit. After we calculated the Vthevenin and Rthevenin by utilizing voltage division and the mesh current method to find itest, we configured the original circuit in MATLAB. We then notice that thevenin equivalent circuits are very useful in simplifying circuits to help find unknown values. Again, huge thanks to Nodebechukwu Paully-Umeh and Michael Rodriguez for helping write parts of this report back in 2020.