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

Thevenin’s Theorem in DC Circuit Analysis

A French engineer, M.L Thevenin, made one of these quantum leaps in 1893. Thevenin’s Theorem (also known as Helmholtz–Thévenin Theorem) is not by itself an analysis tool, but the basis for a very useful method of simplifying active circuits and complex networks. This theorem is useful to quickly and easily solve complex linear circuits and networks, especially electric circuits and electronic networks.

Thevenin’s Theorem may be stated below:

Any linear electric network or a complex circuit with current and voltage sources can be replaced by an equivalent circuit containing a single independent voltage source VTH and a Series Resistance RTH.

• VTH = Thevenin’s Voltage
• RTH = Thevenin’s Resistance

Steps to Analyze an Electric Circuit using Thevenin’s Theorem

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 it.
6. Now find the Total current flowing through the 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 IL flowing through and load voltage across the load resistor in fig (1) by using Thevenin’s Theorem.

Solution:-

STEP 1.

Open the 5kΩ load resistor (Fig 2).

STEP 2.

Calculate / measure the open circuit voltage. This is the Thevenin Voltage (VTH). Fig (3).

We have already removed the load resistor in 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 and current will not flow in the 8kΩ resistor as it is open.

This way, 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 an 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. i.e,

VTH = 12V

STEP 3.

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

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

STEP 5.

Connect the RTH in 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.

STEP 6.

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

IL = VTH / (RTH + RL)

IL = 12V / (11kΩ + 5kΩ) → = 12/16kΩ

IL = 0.75mA

And

VL = IL x RL

VL = 0.75mA x 5kΩ

VL= 3.75V

Now compare this simple circuit with the original circuit shown in figure 1. Do you see how much easier it will be to measure and calculate the load current in complex circuit and network for different load resistors by Thevenin’s Theorem? Yes and only yes.

Good to know: Both Thevenin’s and Norton’s theorems can be applied to both AC and DC circuits containing difference components such as resistors, inductors and capacitors etc. Keep in mind that the Thevenin’s voltage “VTH” in AC circuit is expressed in complex number (polar form) whereas, the Thevenin’s resistance “RTH” is stated in rectangular form.

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Electrical Technology

1. allforu4 says:

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

1. Electrical Technology says:

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

1. Tom Burrows says:

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

2. Tom Burrows says:

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

1. Electrical Technology says:

2. Ted Juma says:

If the voltage source is more than one you calculate two components of Nortons current that is I’ and I”, add them to get the total Nortons current and repeat the procedure explained above.

2. John says:

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

1. Electrical Technology says:

Thanks for appreciation…

3. Calvin says:

Thevenin&#39;s Theorem is not very intuitive for a newbie, good to have this nice easy to follow explanation.

4. yathiraj BK says:

simple and good steps for easy understanding

5. Anonymous says:

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

6. Anonymous says:

How did u get 3mA?

1. Electrical Technology says:

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 />

1. Hi! says:

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.

2. deeksha says:

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

3. hassam says:

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

7. Bilal Ali Khan says:

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

1. Kasen says:

Use Kirchhoff law

8. Krishna Chetty says:

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!!

1. Electrical Technology says:

9. Krishna Chetty says:

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. Tapas says:

Thank you

1. Raju says:

Dhuurrrrrr!!

11. faiz says:

tq admin..mechanical stdent frm malaysia :)

1. Electrical Technology says:

Most welcome dear….

1. Zamik says:

How can I find you in Facebook?

12. Qaisra says:

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

13. Huat says:

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

14. kuna chaitanya says:

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

15. UTTEJ says:

thank u very much sir……

16. king says:

plesae provide telligan’s theorem

17. Ejaz says:

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.

18. fakhr uddin says:

outstanding sir

19. Maheshwari says:

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

20. Aambar Jutt says:

It is too much easy

21. femi says:

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

1. prashant says:

best tutors site

1. Electrical Technology says:

Thank you…

22. sekharr says:

sir this is awesome sir
i had clarify my doubts succesfully

23. nivetha says:

it is soo easy to understand

24. Smik says:

Thank you very much!!!

25. vic12 says:

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

26. Sydney says:

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

1. Electrical Technology says:

Thanks for appreciation….

27. abubakar says:

why you solve 12 and 4 as parallel?

1. Electrical Technology says:

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.

28. Kaisar Rather says:

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

29. Pronoy says:

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

30. Raghavendra Varun says:

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

1. Electrical Technology says:

Thanks…

32. sowjanya says:

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

1. Electrical Technology says:

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

33. gopal misal says:

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

34. Nurul says:

35. ANKIT SAROHA says:

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

1. Saqib Jaweed says:

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)

1. Electrical Technology says:

Saqib Javeed is Right…. Thanks

let me know if you want to know more… Thanks

36. sivaapparao says:

simply superb

37. Saqib Jaweed says:

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

1. Electrical Technology says:

Thanks for appreciation …

38. Joyal George says:

Was very helpful…thanks a billion times ?

1. Electrical Technology says:

Most Welcome….

39. Divine says:

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

1. Electrical Technology says:

Most welcome… And thanks for appreciation….

40. Iron Longwe says:

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

1. Electrical Technology says:

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

41. faith fenna says:

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

1. Electrical Technology says:

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

42. jelbin says:

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

1. Electrical Technology says:

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

43. anonymous says:

wow. thank you. now i understand. :)

44. ghoax says:

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

45. Gideon says:

Please how was the 3mA current gotten ?

46. Rachael says:

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

47. DICKSON HENRY says:

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

48. Benard says:

49. Zakir Ullah says:

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

50. Arinze says:

Simplified example…great work

51. syed umer ali shah says:

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

52. naomi says:

thanks for the example.

53. Beezus says:

thank you dear

54. Prosperity says:

A really Understanding work!

55. Yash mhatre says:

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

56. Tim says:

Thanks soooooo much, it really help me out.

57. Suraj Kumar says:

Thank you so much to help me in easy way

58. Akshay Chaudhari says:

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

59. Teja says:

Really good

60. yamu says:

thnks sir

61. RaJu says:

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????????

62. ANJANA says:

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

63. TUGUME PAUL says:

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

64. mukund says:

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

1. Electrical Technology says:

Yes! But you can analyze the same problem by different methods as well.

1. mukund says:

can you please tell me the different methods sir?

2. mukund says:

65. Vincent Aquilina says:

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)

66. irant says:

how did u get 3mA

1. Electrical Technology says:

3mA current will flows in both 12kΩ and 4kΩ resistors as it is a series circuit because current will not flow in the 8kΩ resistor as it is open (Fig 2)

67. zeroy ulnga says:

thankyou its help full.
request.. how can I calculate if there are voltage source in the circuit?

68. lock says:

thank u so much

69. Daveed says:

Where does the load resistor come from? Can any load resistor be used?

70. Chris says:

If point A and point B are at 4kohm, how to get V(TH) and R(TH)?
Thanks

71. Deependra says:

Sir RL=5kom kese find ?

1. It is the given data in the first circuit (fig 1).

72. Bala says:

Sr i need more explanation on how you get 12v

73. Clearo Clearion says:

Inspiring lessons,
Thanks so much

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