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schematic

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?

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    \$\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
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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}\$.

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  • \$\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
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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.
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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:

schematic

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:

schematic

simulate this circuit

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  • \$\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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