### 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 (V_{Th}) in series with a resistor (R_{Th}) where V_{Th} is the open-circuit voltage at terminals A-B and R_{Th} 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 R_{Th}, 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 R_{Th} 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 I_{o} 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 terminal. The equivalent resistance then becomes a simple Ohm's law calculation, seen in Equation 1. If R_{Th} takes a negative value, it means the circuit is supplying power to the terminals. Then to find V_{Th}, 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 R_{Th }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.

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

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

where,

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

Solving for V_{o }yields V_{o }= .888V or 888.8mV. Remember that R_{Th} = V_{o }/ I_{o}, so R_{Th} is equal to 888.8 mOhms.

Next, to find V_{Th}, 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 V_{1 }in Equation 6 yields V_{1 }= 1.75*V_{Th}. Plugging this into Equation 5 and solving for V_{Th} yields V_{Th} = 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.

10 CommentsLoginsatish kurakula2017-03-07hie 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]

Daulet2017-03-08hi. 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.

afsal2017-06-28How 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 M2018-01-27If the dependent source is (voltage control voltage source)or (CCVS) or (VCCS) ,what we have to do?

zorro_phu2018-06-20How come V0=0.888? When i slove it, it comes V0=-1.846…V.

zorro_phu2018-06-20How come V0=0.888? When i slove it, it comes V0=-1.846…V.

zorro_phu2018-06-20I used equation 2,3 and 4 to find V0.

RK372018-06-21I 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 Hughes2018-06-21If 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.

RK372018-06-22Thanks Mark. I looked at it again and I did indeed lose track of a minus sign.