Solved Problems

Thévenin’s Theorem – Circuit with Two Independent Sources

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Yaz

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Use Thévenin’s theorem to determine I_O.

Thevenin's Theorem - Circuit containing two independent sources
Fig. (1-27-1) – Circuit with two independent sources


Solution
Lets break the circuit at the 3\Omega load as shown in Fig. (1-27-2).
You may also watch the video of solving the problem below:

Also pdf version of the solution is available here.

Breaking circuit at the load
Fig. (1-27-2) – Breaking circuit at the load

Now, we should find an equivalent circuit that contains only an independent voltage source in series with a resistor, as shown in Fig. (1-27-3).

The Thevenin equivalent circuit
Fig. (1-27-3) – The Thevenin equivalent circuit


Unknowns are V_{Th} and R_{Th}. V_{Th} is the open circuit voltage V_{OC} shown in Fig. (1-27-2).
It is trivial that the current of 2\Omega resistor is equal to the current of the current source, i.e. I_{2\Omega}=-1A. Therefore, V_{OC}=V_{2\Omega}=2\Omega \times I_{2\Omega}=-2V. The Thévenin theorem says that V_{Th}=V_{OC}=-2V. Please note that it is not saying that V_{OC} is the voltage across the load in the original circuit (Fig. (1-27-1)).

To find the other unknown, R_{Th}, we turn off independent sources and find the equivalent resistance seen from the port, as this is an easy way to find R_{Th} for circuits without dependent sources. Recall that in turning independent sources off, voltage sources should be replace with short circuits and current sources with open circuits. By turning sources off, we reach at the circuit shown in Fig. (1-27-4).

Turning off the sources to find Rth
Fig. (1-27-4) – Turning off the sources to find Rth


The 6\Omega resistor is short circuited and the 5\Omega one is open. Therefore, their currents are zero and R_{Th}=2\Omega.
Now that we have found V_{Th} and R_{Th}, we can calculate I_O in the original circuit shown in Fig. (1-27-1) using the Thévenin equivalent circuit depicted in Fig. (1-27-3). It is trivial that
I_{O}=\frac{V_{th}}{R_{Th}+3 \Omega}=\frac{-2V}{2\Omega+3 \Omega}=-\frac{2}{5}A.

We used the Thévenin Theorem to solve this circuit. A much more easier way to find I_O here is to use the current devision rule. The current of the current source is divided between 2\Omega and 3\Omega resistors. Therefore,
I_{O}=\frac{2\Omega}{2\Omega+3 \Omega} \times (-1A)=-\frac{2}{5}A

Now, replace the current source with a -1V voltage source as shown below and solve the problem. The answers are V_{th}=\frac{4}{7} V, R_{th}=\frac{10}{7}\Omega and I_O=\frac{4}{31}A. Please let me know how it goes and leave me a comment if you need help 🙂

Homework
Fig (1-27-5) – Homework
Thévenins-Theorem-Circuit-with-Two-Independent-Sources-Solved-ProblemsDownload

Comments

134 responses to “Thévenin’s Theorem – Circuit with Two Independent Sources”

  1. mimohos Avatar
    mimohos

    i am sorry, but wouldn’t the Rth be 1.6 ohm ? not 10/7

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      6 ohm is short circuited by the voltage source, so Rth=2||5=(2*5)/(2+5)=10/7 ohm. Please let me know if you need more explanation.

      Reply
      1. vinaya Avatar
        vinaya

        sir plz can u explain me about super position theorem and thevenin’s thorem and problems…would u plz tell me the way to solve prblms

        Reply
        1. Dr. Yaz Z. Li Avatar
          Dr. Yaz Z. Li

          Please browse the site carefully. These topics are already covered.

          Reply
        2. sohag Avatar
          sohag

          thevenin’s theorem

          Reply
      2. Sruthy Prathap Avatar
        Sruthy Prathap

        Sir,
        I’m a 1st year student i need to get a page that explains Independent source
        along with examples for VDCS, CDCS , CDVS,VCVS

        Reply
      3. Utkarsh Verma Avatar
        Utkarsh Verma

        Since the 5 ohm resistance is open circuited why would the current would go in 5 ohm resistance, the current will only flow through 2 ohm resistance, hence the Rth will be 2ohm.

        Reply
        1. Manohar gangikunta Avatar
          Manohar gangikunta

          Absolutely correct

          Reply
    2. sadiq hussain Avatar
      sadiq hussain

      hi, Rth is 2ohm because there is a short circuit in network.

      Reply
      1. Dr. Yaz Z. Li Avatar
        Dr. Yaz Z. Li

        That is not a reason

        Reply
        1. rashmi jain Avatar
          rashmi jain

          whts the reason dat we got rth =2

          Reply
          1. Pavan Avatar
            Pavan

            As current passing through 5ohm and 6 ohm is zero as they are open circuited

      2. Niloy Avatar
        Niloy

        I also think that the Rth is 2 ohm because all the other resistances goes into short circuit. . . .! isn’t is?

        Reply
    3. Nikhil Avatar
      Nikhil

      You have replaced -1A current source to -1V voltage source that is why values you got are mismatched.

      Reply
  2. Solomon Ikem Avatar
    Solomon Ikem

    its very good

    Reply
  3. mrehan Avatar
    mrehan

    i need help to find Vth

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      6 \Omega resistor is in parallel with 3V voltage source. the series combination of 2 \Omega and 5 \Omega resistors is connected to the series combination of 3V and -1 volt voltage sources. Therefore, V_{5\Omega}+V_{2\Omega}=3V+(-1V)=2V. V_{Th} is in fact the voltage across the 2\Omega that can be found by voltage division rule: V_{Th}=V_{2\Omega}=\frac{2\Omega}{2\Omega + 5\Omega} \times 2 V= \frac{4}{7} V

      Reply
      1. faiz Avatar
        faiz

        i am not able to understand

        Reply
        1. awais Avatar
          awais

          i also …. at the end what you do when 2v.vth after this line i cant under stand what you have done.. please help me…

          Reply
          1. awais Avatar
            awais

            also explain the voltage division rule…

      2. Younas khan Avatar
        Younas khan

        Sir please solve problem clearly and to the point please we are not phd doctor to understand

        Reply
  4. mrehan Avatar
    mrehan

    that`s very helpful can you give us more examples 😀

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      Have you checked out this example?

      Reply
      1. parmvir singh Avatar
        parmvir singh

        please explain me thevenin theorem with two volt sources and only with three resistors.

        Reply
      2. junaid Avatar
        junaid

        how we will know that it is short or open circuit plz

        Reply
  5. mrehan Avatar
    mrehan

    yes but i need examples where there are two independent sources

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      I am going to add some soon 🙂
      I am also in process of writing a new ebook which will include Thévenin’s Theorem too.

      Reply
      1. Saad Salfi Avatar
        Saad Salfi

        wow that would be great sir , me waiting for that ebook

        Reply
  6. mrehan Avatar
    mrehan

    that would be great but please harry i had an exam soon and i`m completly lost and when you finish the book please tell us 🙂

    Reply
    1. Yaz Z Li Avatar
      Yaz Z Li

      I’ll try my best 🙂

      Reply
  7. Gajendra Rajput Avatar
    Gajendra Rajput

    How do we solve Thevenin equivalent circuit problem with combination of dependent and independent sources?
    (voltage and current)

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      The solution follows a similar procedure, you need to find V_{th} and R_{th}. To find V_{th} any method can be used. However, to find R_{th}, the best way is to connect a 1 V voltage(or 1 A current) source to the output and find the current passing through it (or voltage drop on the terminal, if you are using a current source). R_{th} is \frac{V_{voltage source}}{I_{voltage source}}=\frac{1 V}{I_{voltage source}} (or \frac{V_{current source}}{I_{current source}}=\frac{V_{current source}}{1 A}, if you are using a current source). I am planning to add an example soon.

      Reply
  8. moin Avatar
    moin

    this problem is really good

    Reply
  9. Shrawan raut Avatar
    Shrawan raut

    In fig(1-27-4) 6 ohm resistance is in series wid 5 ohm resistance w.r to 2 ohm….if 6 ohm is open thn how 2 ohm is parallel wid 5 ohm….?

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      It is not precise to say that 6 ohm is in series with 5 ohm with respect to 2 ohm. Two resistors cannot be in series with respect to some other element. Two resistors (or any element) are in series if one of their terminals are connected to each other and the have same current.
      2 ohm is not parallel with 5 ohm either.

      Reply
  10. Sophie Kovalevsky Avatar
    Sophie Kovalevsky

    I thought that this problem it’s good to begin, after to do more problems, the thevenin’s theorem will be eaasy.

    However, this it’s a good begin for the newbies.

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      Thank you Sophie.

      Reply
      1. elamaruthu Avatar
        elamaruthu

        sir i want solve the norton’s theorems in three loop cricuit../

        Reply
  11. Ryan Avatar
    Ryan

    Why did u replace the voltage source with the current source ???

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      That is a new problem. Has nothing to do with the original problem. Just to practice more.

      Reply
  12. thirulok Avatar
    thirulok

    sir , i couldnt find Io=4/31 value….
    i tried by the steps given below….
    1. find total current I = Vth/(Rth || RL) = 30/31
    2.find current flowing thru load IL= I * 2/(2+3) = 12/31( my answer)

    sir pls let me know, where i did mistake

    Thanks & regards,
    THIRU.

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      Vth, Rth and RL all are in series (always) as shown in Fig. (1-27-3). Therefore, IL=Vth/(Rth+RL)= 4/31A

      Reply
  13. mohamed Avatar
    mohamed

    excellent !!!!!!
    i didnt understand when ma teacher taught on board…..nw i got!!!!
    thnk u vry much sir…

    Reply
  14. nisha Avatar
    nisha

    Thank you sir. now i can understand better

    Reply
  15. shrenik Avatar
    shrenik

    isnt the 5ohm resistor also short circuited in the 2nd example? how do u find rth for it?

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      No, 5 \Omega is not short circuited. Only 6 \Omega is short circuited. Hint: R_{th}=2 \Omega || 5 \Omega. Let me know if you still need help.

      Reply
  16. sujatha Avatar
    sujatha

    i dont knw wat to do first when i see this problem in question paper sir….
    pls tell me the procedures to do like hint sir….. thank you

    Reply
  17. Mohammad Zohaib Avatar
    Mohammad Zohaib

    The Rth is 2ohm not the 10/7ohm

    Reply
  18. Mohammad Zohaib Avatar
    Mohammad Zohaib

    You have replied on the first comment that it wouldn’t be 2 ohm but how it is possible
    When we short all voltage source than it would be only 2ohm so The total Rth will be 2ohm……If i am wrong plz explain……!!!

    Reply
  19. Mohammad Zohaib Avatar
    Mohammad Zohaib

    Now i understand sir sorry sorry

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      It is ok 🙂

      Reply
  20. vinal Avatar
    vinal

    sir its very gud example to know about the thevnins example & get idea to how to calculate this example wid simple steps..so,thnx a lot

    Reply
  21. Saad Salfi Avatar
    Saad Salfi

    Sir , why is 6 ohm source is shot with the voltage source and we just take 2 and 5 ohm resistances for equivalent resistance

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      In calculating, replace voltage sources with short circuits. Therefore, element which are in parallel with voltage sources are short circuited.

      Reply
  22. camille bustamante Avatar
    camille bustamante

    im confused. in the problem you said that RTH=2 ohms but in your replies you say that RTH=10/7. So what is it really?

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      Fo the homework (Fig (1-27-5) ), it is R_{th}=\frac{10}{7}\Omega but for the solved problem (Fig. (1-27-1)) it is R_{th}=2\Omega.

      Reply
  23. Idk Rebollos Avatar
    Idk Rebollos

    can u give us an example on how to get Rth with dependent sources?

    Reply
  24. mahesh Avatar
    mahesh

    helpful and thank you professor:)

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      You are very welcome!

      Reply
  25. anna poorani Avatar
    anna poorani

    can u help me with some more problems to workout… kindly give a reply

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      Sure. I can help as much as my free time allows me.

      Reply
      1. Atit Patel Avatar
        Atit Patel

        sir, Can you please help me solving the circuit with two loops, two sources and Load Resistance is independent of both of them.

        Reply
  26. brian Avatar
    brian

    dear Dr. Yaz am having problems in coming up with equations, especially when using nodal analysis. is there a solution to this problem sir?

    Reply
  27. sOMEONE Avatar
    sOMEONE

    can you solve any thevenin problem with a dependent source in it?

    Reply
  28. Tathagata Avatar
    Tathagata

    Sir i couldn’t find Rth. Although i could calculate Vth and Io correctly. Plz explain how Rth is 10/7 ohm. i’m gettin 22/13ohm.

    Reply
    1. non Avatar
      non

      Hint: It short circuit at the 6ohm resistor.

      Reply
  29. florence Avatar
    florence

    i wiuld like to be help with analysing nortons two loop network

    Reply
  30. florence Avatar
    florence

    Nortons two loop network analysis

    Reply
  31. anu Avatar
    anu

    can you please help me out wit more examples…….am clear with this problem……thank you sir

    Reply
  32. jaylo lag-asan Avatar
    jaylo lag-asan

    sir can you give me some of your reference books about electrical engineering

    Reply
  33. jaylo lag-asan Avatar
    jaylo lag-asan

    [img]doc1[/img]
    sir how can i solve the current at 25 ohm by using thevenin’s theorem???

    Reply
  34. abhinav chaturvedi Avatar
    abhinav chaturvedi

    awesome sir plz it will be my pleasure if u can upload some more examples thank you..

    Reply
  35. Banajyotshna Devi Avatar
    Banajyotshna Devi

    sir,what are the steps to solve a network having only dependent sources using Thevenin’s theorem.

    Reply
  36. soni.sonali92@gmail.com Avatar
    soni.sonali92@gmail.com

    thanku sir…bt i m very weak in these…can u give some more examples so that i can undrstnd it more..

    Reply
  37. Gurpreet Singh Avatar
    Gurpreet Singh

    Sir, may i know what to do when we have 5 indepntent sources
    3 voltage source and 2 current source????
    please reply soon!!!!

    Reply
  38. desmond Avatar
    desmond

    circuit theorem is a hand tool for electrical/electronics engineering,

    Reply
  39. shahid Ali Khan Avatar
    shahid Ali Khan

    its good but u solve the problem by using formulas line by line

    Reply
  40. poorwa gupta Avatar
    poorwa gupta

    here i did not get that how the Vth is found.i thought that Vth =-1v.is this correct?

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      No, sorry!
      It is -2 for the example and \frac{4}{7} for the homework.

      Reply
  41. DTB76 Avatar
    DTB76

    What would Rth be if the current and voltage were switched around or would Rth be the same?

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      In that case, R_{th}=2\Omega || (5\Omega+6\Omega)=\frac{2\times 11}{2+11}=\frac{22}{13}=1.69 \Omega. No, it won’t be the same.

      Reply
  42. UMAIR SHAHZAD Avatar
    UMAIR SHAHZAD

    YOU HAVE DONE A GOOD JOB SIR……… JUST ADD MORE SOLVED EXAMPLES … … THOUGH I HAVE MY PAPER ON 31ST OCT ,i am asking it coz for the benefits of other students,,, btw thank you..

    Reply
  43. viknesh Avatar
    viknesh

    In finding Vth, can i presume that the 3V source has no effect. why the -1A current source is only taken into account for calclation of Vth.

    Thanks in advance

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      This is true only for this example of course. Here, V_{th} is the voltage of a resistor whose current is known. So, V_{th} can be easily found and the 3V source has no effect in V_{th}.

      Reply
  44. M.Ali Avatar
    M.Ali

    how 5ohm resister is open in Rth….???

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      To see this clearly, remove the 6\Omega resistor because it is short circuited and no current would pass though it.

      Reply
  45. fawad humayun Avatar
    fawad humayun

    sir please tell me why we short the voltage source and open the current source while finding the R thevenin?

    Reply
  46. fawad humayun Avatar
    fawad humayun

    sir please tell me
    when we find the resistance equivalent why we short the voltage source and open the current source

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      This method to find the R_{th} is only valid for circuits without dependent sources. In this method, we are in fact replacing the sources with their equivalent internal resistance. The internal resistance of a voltage source is zero. The reason is that there can be a situation (short circuit) where the voltage across the source is a constant number and the current passing through it is infinite, Therefore, the internal resistance is \frac{V}{\infty}=0. For a current source, the internal resistance is infinite. This is because there can be a situation (short circuit) where the voltage across a current source is zero while the current is a constant number \frac{0}{I}=0.

      Reply
      1. fawad humayun Avatar
        fawad humayun

        thanxs sir

        Reply
  47. fawad humayun Avatar
    fawad humayun

    Sir please give me the examples of dependent source on the superposition topic.

    Reply
    1. Dr. Yaz Z. Li Avatar
      Dr. Yaz Z. Li

      Superposition Method – Circuit With Dependent Sources

      Reply
      1. crystalguardian Avatar
        crystalguardian

        sir,
        in rth 5ohm resistor is left bcoz current will enter it flow through short circuit but will then stop flowing due to open circuit am i right plz tell me?

        Reply
        1. Dr. Yaz Z. Li Avatar
          Dr. Yaz Z. Li

          Yes, you are right. The current cannot flow due to the open circuit. 6 ohm parallel with short circuit (0 ohm) is short circuit. So, we can remove that 6 ohm and see that the 5 ohm is not connected anywhere, so current cannot pass through it.

          Reply
  48. fawad humayun Avatar
    fawad humayun

    sir plaese give some more tips to solve the problems i don’t understand it

    Reply
  49. Rajeev Avatar
    Rajeev

    can u discuss the problem related to find power to a particular resistanceor any element,and a lot of voltage source and current source are given.

    Reply
  50. rashid Avatar
    rashid

    thanks. from that example i will be able to do my exam briliantly.

    Reply
  51. pankaj kumar Avatar
    pankaj kumar

    sir,

    i am facing problem with dependent source while solving Rth . sir i have also prob. while solving maximum power transfer theorem question. i am preparing for GATE 2012.
    please mail me with some prob. and solution.

    Reply
  52. rafiqahmad Avatar
    rafiqahmad

    thanks a lot. i have solved the example and assignment also.
    my answers are correct as you have given.

    Reply
  53. rafiqahmad Avatar
    rafiqahmad

    kindly tell me how can we select the refrence node in node-voltage method????

    Reply
  54. sridhar Avatar
    sridhar

    what is mean by independent?

    Reply
  55. damans52 Avatar
    damans52

    hello, sir
    do you have any doc file related to using Laplace Transform in electric circuits like a Transistors or an Op-Amp. if yes kindly mail me aur send me a link.
    i’ll be very greatful to you.

    Reply
  56. Oladele Temitope Michael Avatar
    Oladele Temitope Michael

    Sir pls what about kirchoff‘s law? I don‘t understand it all…pls help me out sir

    Reply
  57. Aqsaa Khattak Avatar
    Aqsaa Khattak

    sir another method we hav read to find Rth is to first find short circuit (isc) and then by relation Rth=Vth/isc.
    now how in the above given example we can actually CALCULATE the isc first and then Rth?
    please its confusing me.do reply sir
    thank u.

    Reply
  58. shibananda Avatar
    shibananda

    sir iwant some problems on thbnins theorem

    Reply
    1. sakib Avatar
      sakib

      dnt knw da spelling of”thevenin”>?????????
      gadha 😛

      Reply
  59. Mukesh kumar Avatar
    Mukesh kumar

    first thanks u very-very much,sir. Please tell me some more example;which is based on thevenin’s theorem.

    Reply
  60. Anifowose Taiwo Avatar
    Anifowose Taiwo

    it really great,openin this page,pls hw can we calculate a resistance equivalent where only different resistances are given between a and b and the circuit is short circuited

    Reply
  61. Joyanta Avatar
    Joyanta

    Hello sir, To find the Vth for the problem in Fig 1-27-5, I used superposition theorem. First I shorted the 3V source and found E1= 13/7 V. Then I shorted the -1V source and found E2= -2/7 V. Adding them I found the Vth = 11/7 and Io= 11/31. That doesn’t match with your answer Vth=4/7. Where did I make mistake? Is it okay use superposition theorem for this problem. Is there any easier way? Please give me an answer.

    Reply
  62. Tehreem Raja Avatar
    Tehreem Raja

    Sir plz suggest me some of your reference books as your books are very helpful and easy to understand.I really need them.My exams are imminent…….. yours way of explaining a topic is awesome.

    Reply
  63. sakib Avatar
    sakib

    too easy problem..evn my baby promy can solve thhat 😉

    Reply
  64. gordon Avatar
    gordon

    i want solutions of different nodal circuit diagarms

    Reply
  65. Avik Sinha Roy Avatar
    Avik Sinha Roy

    I don’t understand how we get the value of Rth i.e. 2ohm?

    Reply
  66. chanaka Avatar
    chanaka

    if Vth is 2 v that would result in a 7v on the branch where I source is place.But the voltage that it can have is 3V ,,from the left branch!!!!!is their any mistakes

    Reply
  67. Arijit Bhadra Avatar
    Arijit Bhadra

    Why aren’t you considering the 3V source while determining the Vth or Voc?

    Reply
  68. Raiha Zahra Avatar
    Raiha Zahra

    plz give a detailed lecture on source transformation in linear circuits.

    Reply
  69. Rahul Avatar
    Rahul

    Sir, if the 5 ohm resistor is not taken, even the 2 ohm resistor should not be taken too. right ? because one side of the 2 ohm resistor is in the open circuit ??

    Reply
    1. Yaz Avatar
      Yaz

      actually it can be said that both sides are open circuits. However, we need R_{th} seen from two terminals of 2 \Omega.

      Reply
  70. SILAYO Avatar
    SILAYO

    sorry sir how can i solve a question which have two independent (voltage source & current source) and dependent voltage source?

    Reply
  71. yakubu auwal Avatar
    yakubu auwal

    THE SOLVED PROBLEMS THAT ARE UPLOADED ON THE WEBSITE WILL NOT BE ENOUGH FOR US TO GET THE CONCEPT PROPERLY.I SUGGEST THAT YOU ADD MORE RELEVANT QUESTIONS SO THAT WE W’LL KEEP OURSELVE BUSY ALWAYS

    Reply
  72. yeswanth Avatar
    yeswanth

    sir ,,,, is voltage source is considered if it is present near the terminals A-B across the the thevenins is calculated

    Reply
  73. noah Avatar
    noah

    sir just explain how
    you know that 6 ohm resistor is parallel to the voltage source,and explain why 5 ohm resistor is not short circuited

    Reply
  74. sharon Avatar
    sharon

    How to find Vth in problems when the AB terminals are open circuited in the middle of the circuit?

    Reply
    1. Yaz Avatar
      Yaz

      It is basically the same. Middle of circuit, left hand side, top, right, … do not matter!

      Reply
  75. gravemind Avatar
    gravemind

    Fairly certain there’s a mistake in the answer for the second problem:

    I calculate the same Req as you have, but using loop analysis, the current in the 2ohm resistor is 4/7A giving a Vth value of 8/7V not 4/7V. This gives a value of Io = 8/31A. I get the same value completing the whole problem with loop analysis.

    Reply
    1. Yaz Avatar
      Yaz

      the current in the 2ohm resistor is 4/7A

      No! it is not. The voltage is 3+(-1)=2 and the current becomes 2/(2+5)A.

      Reply
    2. Joey Mac Avatar
      Joey Mac

      Yes, you are correct. I also posted this. I verfied with SPICE.

      Reply
  76. H L D Sandakelum Avatar
    H L D Sandakelum

    Sir,what is the meaning of short circuit.I can’t understand it…

    Reply
  77. H L D Sandakelum Avatar
    H L D Sandakelum

    Sir what is the meaning of short circuit…

    Reply
  78. Joey Mac Avatar
    Joey Mac

    Hello Yaz, the last circuit with the -1 volt substituted for the -1amp current source. You have Vth=4/7 but that is current that goes through the 5 Ohm and 2 Ohm resistors, not Vth. I spiced up the circuit to verfiy. Please check it out when you have time. I know you did this a long time ago. Thanks for your examples.

    Reply
    1. Joey Mac Avatar
      Joey Mac

      My bad, the 3 volt and 1 volt source are opposing I drew the circuit incorrect.

      Reply
  79. Hemant Avatar
    Hemant

    How u change this -1A current to 1V in above ckt and except this I m getting ur method

    Reply
  80. ASWINKUMAR K Avatar
    ASWINKUMAR K

    Hi broo! In the 2nd one the -1A current source is changed into -1V voltage source sum the I2 value is 4/7 but we need to find Vth so Vth = I2*2=8/7 is the answer . I think so

    Reply
  81. hoho Avatar
    hoho

    even i got the same answer ……grave mind …8/31

    Reply

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