Thevenin’s Theorem Equation and Examples Electric Circuits

Thevenin’s theorem may help us greatly when dealing with electrical analysis in real life experience. It is a normal thing that sometimes a component in a circuit is variable (its load may change from time to time) while other components are fixed.

The simplest example of this thing is our household electrical load. We will plug our electrical devices to our electrical and it can be a lamp, charger, TV, refrigerator, computer, and many more. Our point here is, the load will be a variable component for our household’s electrical circuit.

It will be a tiresome thing if we want to analyze our electrical system every time we plug in different devices. This is where Thevenin’s theorem kicks in! We will analyze the fixed component in the electrical circuit into a simplified equivalent circuit and we don’t have to analyze it all over again when the variable component changes. This will save a lot of our precious time.

What is Thevenin’s Theorem

Observe the illustration below where the linear two terminal circuit is our fixed components while the load may change frequently.

thevenin's theorem

We can replace the linear two terminal circuit into its Thevenin equivalent circuit as shown below:

thevenin's theorem

The load may be just a resistor or even another circuit. This Thevenin equivalent circuit is our main focus to analyze electric circuits. If the load is replaced by any component, the Thevenin equivalent circuit will remain as it is.

Thevenin’s theorem states that:

A linear two terminal circuit can be simplified into a circuit that consists only of a voltage source VTh, connected series with an equivalent resistance RTh between the two observed terminals.

The true essence of this theorem is to simplify a circuit analysis, that is to make a substitute circuit consisting of voltage source connected in series with its equivalent resistance.

How to do Thevenin Equivalent Circuits

Our main objectives are to find:

  • Thevenin equivalent voltage, VTh
  • Thevenin equivalent resistance, RTh

Keep in mind that we can call them equivalent if the voltage and current are still the same at their terminals.

thevenin's theorem

From the illustration above, with the substitution theorem we can see the circuit B can be replaced with a voltage source with the same value when the current flowing through circuit B at the two terminals we are observing (terminal a-b)

After we get the substitution circuit, using Superposition theorem we get that:

1. If a voltage source V is active then the linear circuit A is not active (all its independent sources are replaced with their inner resistance), thus we can calculate the equivalent resistance. This step is about finding Thevenin resistance or finding R Thevenin.

thevenin's theorem

2. If the linear circuit A is active then the independent voltage source is replaced with its inner resistance that is zero or short circuit.

thevenin's theorem

After unify these two conditions (superposition theorem) we get:

(1)   \begin{align*}i&=i_{1}+i_{s}\\i&=-\frac{V}{R_{Th}}+i_{sc}\end{align*}

When terminal a-b is open circuit (OC), then i flowing in the circuit is zero (i=0), then

thevenin's theorem

(2)   \begin{align*}i&=-\frac{V}{R_{Th}}+i_{sc}\\0&=-\frac{V_{oc}}{R_{Th}}+i_{sc}\\V_{oc}&=i_{sc}.R_{Th}\end{align*}

This is about finding Thevenin voltage. From Equations.(1) and (2), we get

    \begin{align*}i&=-\frac{V}{R_{Th}}+i_{sc}\\&=-\frac{V}{R_{Th}}+i_{sc}\frac{R_{Th}}{R_{Th}}\\&=\frac{1}{R_{Th}}(-V+i_{sc}.R_{Th})\\i.R_{Th}&=-V+V_{oc}\\V&=V_{oc}-i.R_{Th}\end{align*}

Note:

It is not a weird thing when the Thevenin resistance RTh has negative value. The negative resistance will cause the voltage (v=-iR) negative. It means the circuit is supplying power. Thevenin negative resistance is possible if the circuit has dependent sources.

How to Find Thevenin Equivalent Resistance

We need to get the value of Thevenin resistance as the equivalent resistance for the circuit. First thing is we turn off all the independent sources in linear circuit A

  • Replace the voltage source with its inner resistance R=0 or short circuit
  • Replace the current source with its inner resistance R=∞ or open circuit.

If the circuit has dependent sources then we need to find the “short circuit current” (isc) first. From this current we will get the Thevenin equivalent resistance (RTh) from the voltage across the desired terminal divided by the current flowing through the short-circuited terminal (isc).

Thevenin’s Theorem Procedures

1. Find and determine the terminal a-b where the parameter is asked or observed.

2. Remove the component at the terminal a-b, make it open circuit at that terminal and calculate the voltage across that terminal a-b (Vab=Voc=VTh).

3. If there are only independent sources, then the resistance is measured at the terminal a-b when all of the sources are turned off and replaced by their inner resistance. We will get Thevenin equivalent resistance (Rab = RTh)

  • Replace voltage source with short circuit.
  • Replace current source with open circuit.

4. If there is dependent source, we use equation below to find the Thevenin equivalent resistance,

    \begin{align*}R_{Th}=\frac{V_{Th}}{I_{sc}}\end{align*}

5. In order to find the value of Isc (short circuit current), we make terminal a-b short-circuited and calculate the current flowing through that terminal (Iab=Isc).

6. Redraw the Thevenin equivalent circuit as a series circuit consists of:

  • The removed component we did on Step.(2).
  • Thevenin voltage/ open circuit voltage/ a-b voltage (VTh=Voc=Vab)
  • Thevenin equivalent resistance (RTh).

7. Solve the simplified circuit.

Thevenin with Independent Source

1. Find the value of i with Thevenin’s theorem!

thevenin's theorem

Solution:

Determine terminal a-b on R where i is observed. Remove the component and make it open circuit. Calculate the voltage across terminal a-b when open circuit:

thevenin's theorem

We get the Vab or Voc:

    \begin{align*}V_{ab}=V_{oc}&=-5+(4\times6)\\&=-5+24=19V\end{align*}

Find the Thevenin resistance RTh when all the independent sources are turned off (replace them with their inner resistance). From the terminal a-b perspective:

thevenin's theorem

Thus we get the Thevenin resistance

    \begin{align*}R_{Th}=4\Omega\end{align*}

We redraw the circuit into Thevenin equivalent circuit:

thevenin's theorem

Hence,

    \begin{align*}i=\frac{19}{8}A\end{align*}

2. Find the value of i with Thevenin’s theorem!

thevenin's theorem

Solution:

Determine the terminal a-b on R where i is observed. Remove the component and calculate the voltage across terminal a-b when open circuit:

thevenin's theorem

With nodal analysis:

thevenin's theorem

Observe node voltage v1:

    \begin{align*}\frac{v_{1}}{6}+\frac{v_{1}-12}{12}-3&=0\\2v_{1}+v_{1}-12-36&=0\\3v_{1}=48\rightarrow v_{1}=\frac{48}{3}&=16V\end{align*}

Thus

    \begin{align*}V_{ab}=V_{oc}&=(4\times3)+v_{1}\\&=12+16\\&=28V\end{align*}

Find the Thevenin resistance RTh when all the independent sources are turned off and replace them with their inner resistance. From the terminal a-b perspective:

thevenin's theorem

Then

    \begin{align*}R_{Th}&=\frac{6\times12}{6+12}+4\\&=4+4\\&=8\Omega\end{align*}

Redraw the Thevenin equivalent circuit:

thevenin's theorem

Thus,

    \begin{align*}i=\frac{28}{8+6}=2A\end{align*}

3. Find the voltage across terminal a-b with Thevenin’s theorem!

thevenin's theorem

Solution:

Find Vab when terminal a-b is open circuit:

thevenin's theorem

Then the Vab:

    \begin{align*}V_{ab}&=V_{oc}=V_{ax}+V_{xb}\\V_{ax}&=\frac{24}{24+24}\times24=12V\\V_{xb}&=\frac{48}{48+24}\times24=16V\end{align*}

Thus

    \begin{align*}V_{ab}=V_{oc}=-12+16=4V\end{align*}

Find the Thevenin resistance when all the independent sources are turned off and replaced with their inner resistance. From the a-b perspective:

thevenin's theorem

Then

    \begin{align*}R_{Th}=\frac{24\times24}{24+24}=28\Omega\end{align*}

Then Thevenin equivalent circuit:

thevenin's theorem

Hence,

    \begin{align*}V_{ab}=-4+(28\times2)=-4+56=52V\end{align*}

Thevenin with Dependent Source

1. Find the value of V with Thevenin’s theorem!

thevenin's theorem

Solution:

Find Vab where voltage across the R=3Ω then we make that terminal open circuit:

thevenin's theorem

Then

    \begin{align*}&V_{ab}=V_{oc}=-2i_{1}-1i_{1}+12=-3i_{1}+12\\&\mbox{where:}\quad i=-6A\\&V_{oc}=(-3\times-6)+12=18+12=30V\end{align*}

Because the circuit has a dependent source, in order to get the RTh we can’t turn off all the sources. We need to find the short circuit current Isc first:

thevenin's theorem

Then

    \begin{align*}i_{sc}=i_{2}+6=4+6=10A\end{align*}

Thus

    \begin{align*}R_{Th}=\frac{V_{oc}}{i_{sc}}=\frac{30}{10}=3\Omega\end{align*}

We redraw the Thevenin equivalent circuit:

thevenin's theorem

And we get,

    \begin{align*}V=\frac{3}{3+3}\times30=15V\end{align*}

2. Find the value i with Thevenin’s theorem!

thevenin's theorem

Solution:

Find Vab when terminal a-b is open circuit:

thevenin's theorem

We get

    \begin{align*}V_{ab}=V_{oc}&=12-(3\times6)\\&=12-18=-6V\end{align*}

Because there is a dependent source, in order to find RTh we can’t turn off all the sources directly. We first find the value of isc:

thevenin's theorem

We get:

    \begin{align*}&\Sigma v=0\\&2i_{sc}+3(i_{sc}+6)-12=0\\&5i_{sc}+6=0\rightarrow i_{sc}=\frac{-6}{5}A\end{align*}

Then,

    \begin{align*}R_{Th}=\frac{V_{oc}}{i_{sc}}=\frac{-6}{-6/5}=5\Omega\end{align*}

We redraw Thevenin equivalent circuit:

thevenin's theorem

And we get

    \begin{align*}i=\frac{-6}{6}=-1A\end{align*}

Thevenin’s Theorem Examples

1. Redraw the circuit below into its Thevenin equivalent circuit to the left of the a-b. Find the current through the RL when RL=6,16,36 Ω.

thevenin's theorem

Solution:

First we remove the observed terminal and turn off all of its independent sources and replace them with their inner resistances:

  • Turn off the 32 V voltage source and replace it with a short circuit.
  • Turn off the 2 A current source and replace it with an open circuit.

The circuit becomes below:

thevenin's theorem

We get the Thevenin equivalent resistance, RTh as:

    \begin{align*}R_{Th}&=4||12+1\\&=\frac{4\times12}{16}+1=4\Omega\end{align*}

In order to find the VTh, we can use mesh analysis to the two loops on the left as shown below:

thevenin's theorem

We obtain

    \begin{align*}-32+4i_{1}+12(i_{1}-i_{2})=0,\\\mbox{where}\quad i_{2}=-2A\end{align*}

Use the i2 we can solve the i1 = 0.5 A. Hence,

    \begin{align*}V_{Th}&=12(i_{1}-i_{2})\\&=12(0.5+2)=30V\end{align*}

We can also use nodal analysis for simpler solutions. Ignore the 1 Ω resistor because there will be no current flowing through the open circuit. From the node VTh with KCL produces:

    \begin{align*}\frac{32-V_{Th}}{4}+2&=\frac{V_{Th}}{12}\\\\96-3V_{Th}+24=V_{Th}\quad&\rightarrow\quad V_{Th}=30V\end{align*}

The value is the same as before. Next we redraw the circuit into Thevenin equivalent circuit:

thevenin's theorem

The current flowing through RL is

    \begin{align*}I_{L}=\frac{V_{Th}}{R_{Th}+R_{L}}=\frac{30}{4+R_{L}}\end{align*}

RL = 6,

    \begin{align*}I_{L}=\frac{30}{10}=3A\end{align*}

RL = 16,

    \begin{align*}I_{L}=\frac{30}{20}=1.5A\end{align*}

RL = 36,

    \begin{align*}I_{L}=\frac{30}{40}=0.75A\end{align*}

2. Redraw the circuit below into its Thevenin equivalent circuit at terminals a-b.

thevenin's theorem

Solution:

Since the circuit has a dependent voltage source, we still replace all of the independent sources with their inner resistances but leave the dependent source alone.

Because there is a dependent source, we energize the circuit with a voltage source vo connected to the terminal as shown below. We set the voltage source vo=1 V to make it simpler for calculation because the circuit is linear so the voltage-current relationship will not change.

thevenin's theorem

Next what we need to do is to find the value of io through the terminals a-b to obtain the value of Thevenin equivalent resistance:

    \begin{align*}R_{Th}=1/i_{o}\end{align*}

We can also connect current source io on the terminals a-b and find the VTh to get the RTh:

    \begin{align*}R_{Th}=v_{o}/1\end{align*}

Using the mesh analysis to the loop 1 produces

    \begin{align*}-2v_{x}+2(i_{1}-i_{2})=0\\v_{x}=i_{1}-i_{2}\end{align*}

But

    \begin{align*}-4i_{2}=v_{x}=i_{1}-i_{2}\end{align*}

Thus,

(1)   \begin{align*}i_{1}=-3i_{2}\end{align*}

For loop 2, KVL results in

(2)   \begin{align*}4i_{2}+2(i_{2}-i_{1})+6(i_{2}-i_{3})=0\end{align*}

For loop 3, KVL results in

(3)   \begin{align*}6(i_{3}-i_{2})+2i_{3}+1=0\end{align*}

Solving these three equations gives

    \begin{align*}i_{3}=-\frac{1}{6}A\end{align*}

But

    \begin{align*}i_{o}=-i_{3}=\frac{1}{6}A\end{align*}

Thus,

    \begin{align*}R_{Th}=\frac{1V}{i_{o}}=6\Omega\end{align*}

In order to get the Thevenin equivalent voltage (VTh) we will find the open circuit voltage, voc in the circuit below:

thevenin's theorem

Using the mesh analysis for the three loops produces:

Loop 1:

    \begin{align*}i_{1}=5\end{align*}

Loop 2:

    \begin{align*}-2v_{x}+2(i_{3}-i_{2})=0 \quad\rightarrow\quad v_{x}=i_{3}-i_{2}\\4(i_{2}-i_{1})+2(i_{2}-i_{3})+6i_{2}=0\end{align*}

Loop 3:

    \begin{align*}12i_{2}-4i_{1}-2i_{3}=0\end{align*}

But

    \begin{align*}4(i_{1}-i_{2})=v_{x}\end{align*}

Solve these equations will produce

    \begin{align*}i_{2}=\frac{10}{3}A\end{align*}

Thus,

    \begin{align*}V_{Th}=v_{oc}=6i_{2}=20V\end{align*}

We can redraw the circuit into its Thevenin equivalent circuit as shown below

thevenin's theorem

3. Convert the circuit below into its Thevenin equivalent circuit at terminals a-b.

thevenin's theorem

Just like the example before, we will energize the circuit with either 1 V voltage source or 1 A current source. For this case we will go with a current source with nodal analysis. The circuit becomes below

thevenin's theorem

Assume io = 1 A. Using nodal analysis produces

(3.1)   \begin{align*}2i_{x}+\frac{(v_{o}-0)}{4}+\frac{(v_{o}-0)}{2}+(-1)=0\end{align*}

Now we have two unknown variables but only one equation. We need the constraint equation

(3.2)   \begin{align*}i_{x}=\frac{0-v_{o}}{2}=-\frac{v_{o}}{2}\end{align*}

Substituting Equation.(3.2) into (3.1) produces

    \begin{align*}2(\frac{-v_{o}}{2})+\frac{v_{o}-0}{4}+\frac{v_{o}-0}{2}+(-1)&=0\\(-1+\frac{1}{4}+\frac{1}{2})v_{o}-1&=0\\v_{o}&=-4V\end{align*}

Since

    \begin{align*}v_{o}=1\times R_{Th}\end{align*}

Then

    \begin{align*}R_{Th}=\frac{v_{o}}{1}=-4\Omega\end{align*}

This is the example when the resistance has a negative value. It means our circuit is supplying power. Accurately the dependent source is the one supplying power. Thus the Thevenin equivalent circuit becomes

thevenin's theorem

Frequently Asked Questions

What is Thevenin’s theorem statement?

A linear two terminal circuit can be simplified into a circuit that consists only of a voltage source VTh, connected series with an equivalent resistance RTh between the two observed terminals.

What is VTH and RTH in Thevenin Theorem?

Thevenin equivalent circuit consists of Thevenin equivalent voltage (VTh) connected series with Thevenin equivalent resistance (RTh) and observed component.

What are the conditions to apply Thevenin’s theorem?

Thevenin’s theorem can simplify any linear circuit into a series circuit consists of single voltage source, a resistor, and a load.

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