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. 




  • 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.

  • afsal 2017-06-28

    How can i simplify a T network with a current source given in a branch and load resistor at the other branch in between which a resistor and a dependent source is being connected in series to it?

  • ABINANDHU M 2018-01-27

    If the dependent source is (voltage control voltage source)or (CCVS) or (VCCS) ,what we have to do?

  • zorro_phu 2018-06-20

    How come V0=0.888? When i slove it, it comes V0=-1.846…V.

  • zorro_phu 2018-06-20

    How come V0=0.888? When i slove it, it comes V0=-1.846…V.

    • zorro_phu 2018-06-20

      I used equation 2,3 and 4 to find V0.

      • RK37 2018-06-21

        I am getting an answer similar to yours (-1.859 V). I’ll ask another AAC technical editor to confirm, then we’ll have to see if we can get this article sorted out.

        • Mark Hughes 2018-06-21

          If I ignore the article & diagrams, and focus strictly on the math, I can get to the 0.88888 answer.  It’d be easier to demonstrate in Latex, but I don’t think we can use that in comments.

          Substitute equation 4 into equation 2 yields V1=-5 V0.  Substitute that result into equation 3 produces 0.88888, as the author stated.

          If, during the substitution, you lose track of your signs in the numerator of the first fraction of equation 3 (subtracting a positive number instead of subtracting a negative number), that produces the -1.846 value.  I was unable to find the source of the -1.859 value, which leads me to believe it is the source of a rounding error.

          In my opinion, a larger issue exists with this article—I do not believe this article clearly illustrates how to apply Thevenin’s theorem.  I’ll discuss next steps with Robert via email.

        • RK37 2018-06-22

          Thanks Mark. I looked at it again and I did indeed lose track of a minus sign.