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Thevenin’s Theorem. Step by Step Procedure with Solved Example

Thevenin’s Theorem in DC Circuit Analysis

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 VTH and a Series Resistance RTH.

Steps to Analyze Electric Circuit through Thevenin’s Theorem

  1. Open the load resistor.
  2. Calculate / measure the open circuit voltage. This is the Thevenin Voltage (VTH).
  3. Open current sources and short voltage sources.
  4. Calculate /measure the Open Circuit Resistance. This is the Thevenin Resistance (RTH).
  5. Now, redraw the circuit with measured open circuit Voltage (VTH) in Step (2) as voltage source and measured open circuit resistance (RTH) 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.
  6. Now find the Total current flowing through load resistor by using the Ohm’s Law: IT = VTH/ (RTH + RL).

Solved Example by Thevenin’s Theorem:

Example:

Find VTH, RTH and the load current flowing through and load voltage across the load resistor in fig (1) by using Thevenin’s Theorem.

Click image to enlargeThevenin's Theorem. Easy Step by Step Procedure with Example (Pictorial Views)

Solution:-
 
Step 1.

Open the 5kΩ load resistor (Fig 2).

Click image to enlargeThevenin’s Theorem 
 
Step 2.

Calculate / measure the open circuit voltage. This is the Thevenin Voltage (VTH). 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 well as 4kΩ resistor. Therefore 12V will appear across the AB terminals. So,

VTH = 12V 

Click image to enlargeThevenin's Theorem may be stated below:

Step 3.

Open current sources and short voltage sources as shown below. Fig (4)

Click image to enlargeThevenin's Theorem: Simple explanation

Step 4.

Calculate / measure the open circuit resistance. This is the Thevenin Resistance (RTH)

We have removed the 48V DC source to zero as equivalent i.e. 48V DC source has been replaced 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)

 
RTH = 8kΩ +  [(4kΩ x 12kΩ) / (4kΩ + 12kΩ)]
RTH = 8kΩ + 3kΩ

RTH = 11kΩ

Click image to enlargeThevenin's Theorem solved examples

Step 5.

Connect the RTHin series with Voltage Source VTH 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

Thevenin's equivalent circuit
Thevenin’s equivalent circuit

Step 6.

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

IL = VTH / (RTH + RL)

= 12V / (11kΩ + 5kΩ) → = 12/16kΩ

IL= 0.75mA

And

VL = ILx RL

VL = 0.75mA x 5kΩ

VL= 3.75V

Click image to enlargethevenin's theorem examples thevenin's theorem solved problems

Now 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.

You may also read: 

 

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95 comments

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

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

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

  4. simple and good steps for easy understanding

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

  6. How did u get 3mA?

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

  8. 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!!

  9. 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!!

  10. tq admin..mechanical stdent frm malaysia 🙂

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

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

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

  14. thank u very much sir……

  15. plesae provide telligan’s theorem

  16. 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.

  17. outstanding sir

  18. 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

  19. It is too much easy

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

  21. sir this is awesome sir
    i had clarify my doubts succesfully

  22. it is soo easy to understand

  23. Thank you very much!!!

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

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

  26. 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.

  27. will you please add step by step procedure for dual loops that will help alot
    Thanx in advance

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

  29. Raghavendra Varun

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

  30. Thanks…

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

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

  33. please inform more and more about electrical and electronics

  34. 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

  35. simply superb

  36. Thank you very much,
    this was the best explanation available on the internet

  37. Was very helpful…thanks a billion times ?

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

  39. 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

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

  41. 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

  42. wow. thank you. now i understand. 🙂

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

  44. Please how was the 3mA current gotten ?

  45. can you explain to me about how did you get 12V as it is 48V?

  46. IT IS GOOD BUT I WILL LIKE IT IF IT SOLVED USING SMALL NUMBERS LIKE 3,4,2 TO MAKE IT LOOK MORE SIMPLER

  47. Thevenin made simply. thanks dear.

  48. A/C to my point of view, Its very simple and easy method for understanding..

  49. Simplified example…great work

  50. if I find current across 5 ohm resister there is a greater value then the total load current that is 0.75 can anyone answer

  51. thanks for the example.

  52. thank you dear

  53. A really Understanding work!

  54. Thanks for keeping the simplicity in the explination I thoroughly understood the conpect.

  55. Thanks soooooo much, it really help me out.

  56. Thank you so much to help me in easy way

  57. Akshay Chaudhari

    And if there is an ammeter in parallel with the resistance, then what to do

  58. If there are a voltage or current source between A & B terminals or in the place of load resistor, Then how can i solve this????????

  59. VERY SIMPLY AND THOROUGHLY EXPLAINED THE CONCEPT OF THEVENINS THEOREM. PLEASE INCLUDE THE NOTES OF TWO POLE NETWORKS ALSO

  60. Am happy for the steps of thevenins i cam now calculate but following the steps

  61. sir please tell me….is this the right and only method to solve problems by thevenin’s theorem?

  62. Very good explanation. I tend to solve complex circuits using Kirchoff’s laws…Are there particular cases when one method is favourably used instead of the other? ( I have not read all posts…maybe you have already gone into this)

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