Thévenin's Theorem - Circuit with Two Independent Sources

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

Lets break the circuit at the  3\Omega load as shown in Fig. (1-27-2).

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 🙂


Fig (1-27-5) - Homework

121 thoughts on “Thévenin's Theorem - Circuit with Two Independent Sources

      1. 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

      2. 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

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

        1. 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...

  1. 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 🙂

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

    1. The solution follows a similar procedure, you need to find $latex V_{th}$ and $latex R_{th}$. To find $latex V_{th}$ any method can be used. However, to find $latex R_{th}$, 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). $latex R_{th}$ is $latex \frac{V_{voltage source}}{I_{voltage source}}=\frac{1 V}{I_{voltage source}}$ (or $latex \frac{V_{current source}}{I_{current source}}=\frac{V_{current source}}{1 A}$, if you are using a current source). I am planning to add an example soon.

  3. 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....?

    1. 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.

  4. 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.

  5. 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,

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

  6. 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

  7. 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......!!!

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

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

  10. 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?

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

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

  12. 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.

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

  14. 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..

  15. 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

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

    1. This method to find the $latex R_{th}$ 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 $latex \frac{V}{\infty}=0$. 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 $latex \frac{0}{I}=0$.

      1. 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?

        1. 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.

  16. 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.

  17. 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.

  18. 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.

  19. 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 reply sir
    thank u.

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

  21. 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

  22. 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.

  23. 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.

  24. 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

  25. 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 ??


  27. 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

  28. 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.

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