Thevenin's Theorem and its application in analyzing circuits with dependent sources.

Thevenin Theorem

When performing network analysis, Thevenin's theorem is a very helpful tool. It allows for a non-varying portion of a circuit to be replaced with a simplified design, thus simplifying the analysis of the entire network. This equivalent circuit performs the same way as the original circuit would. 

Thevenin's theorem states that any linear, two-terminal portion of a network can be replaced by a Thevenin equivalent circuit. A Thevenin equivalent circuit consists of a voltage source (VTh) in series with a resistor (RTh) where VTh is the open-circuit voltage at terminals A-B and RTh​ ​is the equivalent resistance at terminals A-B. This equivalent circuit can be seen in Figure 1 below. It is important to replace independent sources with their internal resistances when solving for RTh, i.e. current sources are replaced with open circuits and voltage sources are replaced with short circuits. 

Figure 1


Dependent Sources and Thevenin's Theorem

Thevenin's theorem can be applied when analyzing a circuit with dependent sources. In this case, all independent sources are turned off and the RTh is calculated by applying a current source or voltage source at the open terminal. When using a voltage source, it can be assumed to be 1V for simple calculations. Using mesh analysis, find the current Io at the output. When using a current source, it can be assumed to be 1A for easy calculations as well. Nodal analysis can be used to find the voltage at the a terminal. The equivalent resistance then becomes a simple Ohm's law calculation, seen in Equation 1. If RTh takes a negative value, it means the circuit is supplying power to the terminals. Then to find VTh, use mesh analysis with all independent/dependent sources included and solve for the open circuit voltage. 

$$R_{Th}=\frac{1\text{ } V}{I_{o}} \; \; \text{[Equation 1]}$$

Now to apply this theory to an example problem.

Figure 2

Solving for RTh first, redraw the circuit with the 12V source as a short circuit. Then excite the circuit using either a 1A current source or 1V voltage source at the a-b terminal. Using a current source at the a-b terminal produces the circuit in Figure 3 below. 

Figure 3

Now to write the nodal analysis equations. 

$$\frac{V_{1}}{10\text{ }\Omega}-1.5*I_{x}+I_{x}=0\; \; \text{[Equation 2]}$$

$$\frac{V_{o}-V_{1}}{6\text{ }\Omega}+\frac{V_{o}}{8\text{ }\Omega}=1\text{ }A\; \;\; \;\;\text{ [Equation 3]}$$


$$I_{x}=\frac{V_{1}-V_{o}}{6\text{ }\Omega}\; \; \; \;\; \;\; \;\; \;\; \;\;\; \;\; \;\;\text{[Equation 4]}$$

Solving for Vyields V= .888V or 888.8mV. Remember that RTh = V/ Io, so RTh is equal to 888.8 mOhms. 

Next, to find VTh, replace the 12V source and remove the current source from the a-b terminal. Again, utilizing nodal analysis, the equations are as follows. 

$$\frac{V_{1}}{14\text{ }\Omega}+\frac{V_{1}-12}{10\text{ }\Omega}-1.5*\frac{V_{1}}{14\text{ }\Omega}=0\; \; \text{ [Equation 5]}$$

$$\frac{V_{Th}}{8\text{ }\Omega}-\frac{V_{1}-V_{Th}}{6\text{ }\Omega}=0\; \; \; \;\; \;\; \;\; \;\; \;\;\; \;\; \;\;\; \;\; \;\;\text{[Equation 6]}$$

Solving for Vin Equation 6 yields V= 1.75*VTh. Plugging this into Equation 5 and solving for VTh yields VTh = 10.66V. 

This theorem allows for the simplification of resistor and source configurations into to a one source and one resistor equivalent circuit. It is commonly used with varying loads, that way the load current and power dissipation can be calculated easily. 



1 Comment

  • satish kurakula 2017-03-07

    hie why how did you took vi/14 2 times in the equation V1/14 Ω+(V1−12)/10 Ω−1.5∗V1/14 Ω=0 [Equation 5]

    • Daulet 2017-03-08

      hi. Remember that we have depentent source here Ix. The term −1.5∗V1/14 Ω is -1.5 * Ix. And from above we know that Ix = V1/14Ω. Hope it helps.