**Thevenin’s Theorem**

Step by Step Procedure with Examples.

A French Engineer,

**M.L Thevenin**, made one of these quantum leaps in 1893.**Thevenin’s Theorem**is not by itself an analysis tool, but the basis for a very useful method of simplifying active circuits and complex networks because we can solve complex linear circuits and networks especially electronic networks easily and quickly.**Thevenin’s Theorem**may be stated below:

*Any Linear Electric Network or complex circuit with Current and Voltage sources can be replaced by an equivalent circuit containing of a single independent Voltage Source V*

_{TH }and a Series Resistance R_{TH}.**Simple Steps to Analyze Electric Circuit through Thevenin’s Theorem**

- Open the load resistor.
- Calculate / measure the Open Circuit Voltage. This is the Thevenin Voltage (V
_{TH}). - Open Current Sources and Short Voltage Sources.
- Calculate /measure the Open Circuit Resistance. This is the Thevenin Resistance (R
_{TH}). - Now, Redraw the circuit with measured open circuit Voltage (V
_{TH}) in Step (2) as voltage Source and measured open circuit resistance (R_{TH}) in step (4) as a series resistance and connect the load resistor which we had removed in Step (1). This is the Equivalent Thevenin Circuit of that Linear Electric Network or Complex circuit which had to be simplified and analyzed by Thevenin’s Theorem. You have done. - Now find the Total current flowing through Load resistor by using the Ohm’s Law I
_{T}= V_{TH}/ (R_{TH}+ R_{L}).

**Example:**

Find V

_{TH}, R_{TH}and the load current flowing through and load voltage across the load resistor in fig (1) by using Thevenin’s Theorem.*Click image to enlarge*

* *

**Solution:-**

**Step 1.**

Open the 5kΩ load resistor (Fig 2).

**Step 2.**

Calculate / measure the Open Circuit Voltage. This is the Thevenin Voltage (V

_{TH}). Fig (3).We have already removed the load resistor from figure 1, so the circuit became an open circuit as shown in fig 2. Now we have to calculate the Thevenin’s Voltage. Since 3mA Current flows in both 12kΩ and 4kΩ resistors as this is a series circuit because current will not flow in the 8kΩ resistor as it is open.

So 12V (3mA x 4kΩ) will appear across the 4kΩ resistor. We also know that current is not flowing through the 8kΩ resistor as it is open circuit, but the 8kΩ resistor is in parallel with 4k resistor. So the same voltage (i.e. 12V) will appear across the 8kΩ resistor as 4kΩ resistor. Therefore 12V will appear across the AB terminals. So,

**V _{TH} = 12V **

**Step 3.**

Open Current Sources and Short Voltage Sources. Fig (4)

**Step 4.**

Calculate /measure the Open Circuit Resistance. This is the Thevenin Resistance (R

_{TH})We have Reduced the 48V DC source to zero is equivalent to replace it with a short in step (3), as shown in figure (3) We can see that 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor. i.e.:

8kΩ + (4k Ω || 12kΩ) ….. (|| = in parallel with)

R

_{TH}= 8kΩ + [(4kΩ x 12kΩ) / (4kΩ + 12kΩ)]R

_{TH}= 8kΩ + 3kΩ**R _{TH} = 11kΩ**

**Step 5.**

Connect the R

_{TH}in series with Voltage Source V_{TH}and re-connect the load resistor. This is shown in fig (6) i.e. Thevenin circuit with load resistor. This the Thevenin’s equivalent circuit*Click image to enlarge*

* *

**Step 6.**

Now apply the last step i.e Ohm’s law . calculate the total load current & load voltage as shown in fig 6.

I

_{L}= V_{TH}/ (R_{TH}+ R_{L})= 12V / (11kΩ + 5kΩ) → = 12/16kΩ

**I**

_{L}= 0.75mAAnd

V

_{L}= I_{L}x R_{L}V

_{L}= 0.75mA x 5kΩ**V**

_{L}= 3.75VNow compare this simple circuit with the original circuit of figure 1. Can you see how much easier it will be to measure/calculate the load current for different load resistors by Thevenin’s Theorem? Yes and only yes.

thanks, good easy to understand.but the voltage source have more than one. how to do it.

It is simple example for ref & explanation purpose only…Wait for upcoming posts….Thanks

Surely when the load resistor is replaced the circuit changes and the 8k and 5k are in parallel with the 4k causing the cu

Surely when the load resistor is replaced the circuit changes and the 8k and 5k are in parallel with the 4k causing the current and voltage to change oops just figured it out the long way and the 5k gets 3.75 out of the 9.75 volts across that part of the circuit your way of showing thevenin is better thank you

Thanks for your kind words…

Admin. You are doing great work. Your posts related to electricals are easy to learn. . . Keep updating. . .

Thanks for appreciation…

Thevenin's Theorem is not very intuitive for a newbie, good to have this nice easy to follow explanation.

simple and good steps for easy understanding

Good one for basic understanding of theorems :)<br />

How did u get 3mA?

I = V / R …<br />= 48V / (12kΩ + 4kΩ )<br />= 3mA <br /><br />So 3mA Current will flow in both 12kΩ and 4kΩ resistors as this is a series circuit because current will not flow in the 8kΩ resistor as it is open<br /><br />

If the rounded off value of 3mA is used to find Vth you get 12V.

If the less rounded off value of 3.273mA is used we’d get 13V.

when the load terminals are open then all the current passes through only first circuit and the current is from v=i/r can be calculated

1st of all find req than by v=ir , i=v/r we can get the value of current

Thanku. <br />What happens when there is 2 voltage sources?

Use Kirchhoff law

i am proud that u can make this subject such intresting..<br /><br />hope many teachers of ur kind should be there..<br /><br />i am learning this during my job period ..if it was during my college it would have been a big deal for me..<br /><br />anyway thanks!!

Thanks for your kind words…

hope many teachers of ur kind should be there..<br /><br />i am learning this during my job period ..if it was during my college it would have been a big deal for me..<br /><br />anyway thanks!!

Thank you

tq admin..mechanical stdent frm malaysia 🙂

Most welcome dear….

step by step process is easy way for students to understand the problem

Great job there, sometimes is easy to understand through other source.

thank you sir…. this is really a helpful information..

thank u very much sir……

plesae provide telligan’s theorem

Thanks for your very informative post. I had a very bad going with this topic, but now I can assure keep my hands dirty on complex circuits related to this phenomenon by using your idea.

outstanding sir

it is very easy to understand the concept,but if more than one voltage sources are present then how will be the solution and procedure

It is too much easy

boss i love this your tutorials its the best i have ever seen you simplified the sturf to the last five star 4 u *****

best tutors site

Thank you…

sir this is awesome sir

i had clarify my doubts succesfully

it is soo easy to understand

Thank you very much!!!

could you give another exercise with 2 suppliers? and show steps

Thumps up to yo tutorials,now I understand the theorem way better than I previously did.Thanx so Much

Thanks for appreciation….

why you solve 12 and 4 as parallel?

If you see in step 4, 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor.

i.e. If you solve for the parallel connection of 4kΩ resistor and 12k Ω, It becomes in Series with 8kΩ.

In step 2, The Circuit is Open, Hence, Current will not flow in 8kΩ, Therefore, That’s Why it is in parallel with 4kΩ.

will you please add step by step procedure for dual loops that will help alot

Thanx in advance

Great job. you make the determination of thevenin’s voltage is very very easy. Thank you…

I could not solve even a single problem .But this helped me a lot .Thank u

Glad it helps You….

Thanks…

can i get more relevant problems using thevinin’s theorem & may i know that thevinin’s theorem can use for AC supplied circuits.

Yes…. It can be used on both AC and DC Circuits,,, We will post about Thevinin’s theorem about AC Circuits in coming days… Thanks

Told how to solve thevinins them in ac by using kvl.

please inform more and more about electrical and electronics

sir I have a confusion when you calculate Rth because I dont think that Rth=8+4||12. But It should be Rth=12+4||8. Am I wrong Sir ?, if so then please guide me.

Thanks

Ankit

Hi Ankit,

As mentioned above that We can see that 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor. i.e.:

8kΩ + (4k Ω || 12kΩ)

Saqib Javeed is Right…. Thanks

let me know if you want to know more… Thanks

simply superb

Thank you very much,

this was the best explanation available on the internet

Thanks for appreciation …

Was very helpful…thanks a billion times 😁

Most Welcome….

Awesome explanation. It’s very useful to us, Thank you!!

Most welcome… And thanks for appreciation….

have we already been provided with 3Am or you have solved it? if you have solved it how have you done it?? a ain’t sure if I have followed you on that one

It is not provided.

In Step 2, You may see that this is an Open Circuit, I.e. Current will not flow in the 8kΩ.

So the circuit of 48V, 4kΩ and 12kΩ becomes a series circuit.

By using Ohms Law, (I = V/R), We found the value of current as 3mA.

Thanks

what will be total R in the circuit if 4k shud the load resistance in revolving towards the source

You may just change the value of the resistance and solve again same as above…. No Difficulties…. 🙂

you said 8k resistor is parallel to 4k resistor in step 2.in step 4 you said 8k resistor is sereis to 4k.why?

Yes Dear…. If you see in step 4, 8kΩ resistor is in series with a parallel connection of 4kΩ resistor and 12k Ω resistor.

i.e. If you solve for the parallel connection of 4kΩ resistor and 12k Ω, It becomes in Series with 8kΩ.

In step 2, The Circuit is Open, Hence, Current will not flow in 8kΩ, Therefore, That’s Why it is in parallel with 4kΩ.

Thanks.

let me know if you want to know more… Thanks

wow. thank you. now i understand. 🙂

thank you for this. but how about 3 sources using this theorem?

Please how was the 3mA current gotten ?