simulate this circuit – Schematic created using CircuitLab

I have this circuit:

enter image description here

Sorry the current source got cut off when loading. It is 3 Amps. I was trying to apply thevenin's theorem to this circuit but I am confused on how to. Usually to apply Thevenin's you have a Vtest and two nodes that are open.

Can someone help me solve this?

  • 1
    \$\begingroup\$ Could you draw the circuit in the built in diagram editor? Just click edit below your question and hit Ctrl_M to add a circuit diagram. \$\endgroup\$ – jippie Sep 1 '14 at 19:16
  • \$\begingroup\$ Thanks, I didn't know I could do that. I just made it \$\endgroup\$ – JT Hiquet Sep 1 '14 at 19:24

To find the Thevenin equivalent resistance you need to turn off the independent current and voltage sources. A current source that has been turned off has 0A and is therefore an open circuit. A voltage source that has been turned off has 0V and is therefore a short circuit.

For this circuit, therefore, \$R_1 + R_2 = 20\Omega\$ is in parallel with \$R_3 = 5\Omega\$ (you can ignore \$I_1 = 0\$, and the lower ends of \$R_1\$ and \$R_3\$ are shorted since \$V_1 = 0\$).

\$(R_1 + R_2)||R_{3} = 20\Omega ||5\Omega = 4\Omega\$

This \$4\Omega\$ resistance is in series with \$R_5 = 1\Omega\$, giving \$5\Omega\$ resistance. This \$5\Omega\$ resistance is in parallel with \$R_4 = 10\Omega\$ so the Thevenin resistance across \$V_{o}\$ is \$5\Omega || 10\Omega\ = 3.33\Omega\$.

It's a similar procedure to find \$V_{TH}\$ except that the sources are left on. Combine the resistances where possible as I did for \$R_{TH}\$ to find the voltage at the node common to \$R_{2}\$, \$R_{3}\$, and \$R_{5}\$. Then \$R_{5}\$ and \$R_{4}\$ form a voltage divider which gives you the voltage \$V_{TH}\$ across \$R_{4}\$.

To calculate \$V_{TH}\$ use superposition: calculate \$V_{TH}\$ with the current source turned off (\$I_1 = 0\$), then calculate \$V_{TH}\$ with the voltage source turned off (\$V_1 = 0\$), and add the two results to find \$V_{TH}\$ as a result of both sources.

With \$I_1 = 0\$, \$R_1 + R_2\$ is in parallel with \$R_4+R_5\$. The voltage at the top node (call it \$V_{t1}\$) is calculated from a voltage divider formed by \$R_3\$ and \$(R_1 + R_2)||(R_4 + R_5)\$. Then looking back at the original circuit, you can calculate \$V_{TH1}\$ from the voltage divider of \$V_{t1}\$ formed by \$R_4\$ and \$R_5\$. \$V_{TH1}\$ is \$V_{TH}\$ due to \$V_1\$ only.

With \$V_1 = 0\$, \$R_4+R_5\$ is in parallel with \$R_3\$. \$(R_4+R_5)||R_3\$ is in series with \$R_2\$. \$((R_4+R_5)||R_3) + R_2\$ is in parallel with \$R_1\$ so use a current divider to find the current flowing into \$((R_4+R_5)||R_3) + R_2\$. This is the current flowing into \$R_2\$ in the original circuit, so use a current divider again to find the current flowing through \$R_3\$. This current multiplied by \$R_3\$ is \$V_{t2}\$, and you can calculate \$V_{TH2}\$ from the voltage divider of \$V_{t2}\$ formed by \$R_4\$ and \$R_5\$. \$V_{TH2}\$ is \$V_{TH}\$ due to \$I_1\$ only.

By superposition \$V_{TH} = V_{TH1} + V_{TH2}\$.

  • \$\begingroup\$ When you are finding the Vth after you have found Rth, how do you find the voltage at the node common to R2, R3, and R5? \$\endgroup\$ – JT Hiquet Sep 2 '14 at 0:11
  • \$\begingroup\$ @JTHiquet I've updated my answer with more information on how to calculate Vth. I've done it symbolically so you can calculate the numbers yourself. \$\endgroup\$ – Null Sep 2 '14 at 2:34
  • \$\begingroup\$ I got some crazy answer though so I don't know if I did the voltage dividers right. \$\endgroup\$ – JT Hiquet Sep 3 '14 at 0:31
  • \$\begingroup\$ @Ricardo No worries. I know you're not doing it on purpose, and the important thing is that the questions/answers are edited to be easier to read. Thanks. \$\endgroup\$ – Null Oct 15 '14 at 16:21

Well, if you want to find the Thevenin equivalent, I would suggest using source transformations.

  • In this case, start with the 3A current source and 16 ohm resistor. This pair forms a Norton equivalent source.
  • Transform it to a Thevenin equivalent by replacing them with a series combination of a voltage source and 16 ohm resistor. The voltage source will need to be I*R volts, in this case 48 volts.
  • Then combine the now series 16 ohm and 4 ohm resistors into a 20 ohm resistor.
  • Then transform both sources to current sources. The 48 volt source in series with the 20 ohm resistor becomes a 2.4 A source in parallel with a 20 ohm resistor. Likewise the 12 volt source and 5 ohm resistor become a 2.4 A source in parallel with a 5 ohm resistor.
  • Now combine the sources and resistors - two parallel 2.4 A sources are one 4.8 A source and 20 ohms in parallel with 5 ohms is 4 ohms.
  • Transform this back to a Thevenin equivalent - 19.2 volts in series with 4 ohms.
  • Combine the 4 ohm resistor with the 1 ohm resistor.
  • Then transform back again to a Norton equivalent - a 3.84 amp source in parallel with 5 ohms.
  • Now combine the 5 ohm and 10 ohm resistors to get a 3.33 ohm resistor.
  • Transform the source back to get a 12.8 volt source in series with a 3.33 ohm resistor.

It looks like your Vtest in this case is the Vo across the 10 ohm resistor. It might make more sense to you if those leads are extended further out:


simulate this circuit – Schematic created using CircuitLab

From there, you can use your regular Thevenin methods to find the equivalent circuit. CircuitLab tells me that Vtest is 6.40V; if I add a short across those leads, it shows a current of 1.92A. Since \$R = V/I\$, that gives me an equivalent resistance of 3.33 ohms, so the Thevenin equivalent is:


simulate this circuit

  • \$\begingroup\$ Thank you for this, this did help. My issue is now how to combine the the resistor's with 2 sources. \$\endgroup\$ – JT Hiquet Sep 1 '14 at 19:39
  • \$\begingroup\$ Usually, I find the Thevenin equivalent by solving for the open circuit voltage (ie. Vtest exactly the way I drew it) and the short circuit current (add a short across Vtest to make it 0 volts and find the current through it). The Thevenin equivalent can easily be found from there. Which resistors are you trying to combine? \$\endgroup\$ – Greg d'Eon Sep 1 '14 at 19:41
  • \$\begingroup\$ Well see thats my question. I don't really understand Thevenin. Do I need to combine the resistors to find the thevenin resistance? \$\endgroup\$ – JT Hiquet Sep 1 '14 at 19:43
  • \$\begingroup\$ I added an explanation of how I would find the equivalent circuit. Does that make sense now? \$\endgroup\$ – Greg d'Eon Sep 1 '14 at 19:55
  • \$\begingroup\$ How did you find 3.33 and 6.4 though. I want to understand the concept of thevenin so I know for future reference? Like what are those Thevenin methods you mentioned? \$\endgroup\$ – JT Hiquet Sep 1 '14 at 19:57

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