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QUESTION:

I'm having a hard time figuring up what the Thevenin Voltage is along with Thevenin Resistance for this circuit below. I'm also confused on what I'm suppose to do when there is a current source thrown in or multiple sources of any kind thrown in. Do I convert them all to voltage sources before I even start figuring Thevenin Equivalent?

I arrived at the following answers. If they are wrong could you provide a brief explanation on how you figured each answer and a solution to my last question above.

R (TH) = 1.57K ohms V (TH) = 44.8 Volts

NOTE: I didn't know how to enter the Load Resistor in my Simulator. Therefore the break in the circuit on the far right is where it's placing the Load Resistor at.

Circuit

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You need to consider the sources separately, one at a time, and then combine (add together) their effects to determine the final result.

For example, when considering the 16V source (reference designators would be helpful here), replace the current source with an open circuit. What is the Thevenin equivalent now?

Next add the current source back in. What kind of offset does this introduce into the Thevenin equivalent resistance?

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  • \$\begingroup\$ if I take out the current source and figure R(TH) I get 1.57949k ohms. It seems that no matter what this values is the same either way cause voltage sources are shorted and current sources are open when determining R(TH). If I'm going to figure E(TH) and I take out the current source then I would assume the V(TH) across the load would be the same across the 5.6k resistor. Would I be correct in saying this? \$\endgroup\$
    – Shane Yost
    Apr 19 '13 at 18:34
  • \$\begingroup\$ No, Vth is not the voltage across the load, it's the source voltage on the other side of Rth. \$\endgroup\$
    – Dave Tweed
    Apr 19 '13 at 18:37
  • \$\begingroup\$ Taking the current source out I used the voltage divider rule and got 11.4872 Volts for the V(TH). Then I went back and took out the voltage source. Doing this I assume that I should convert the current source to a voltage by using it's parallel resistor which would be the 5.6K. Next I rewrite the circuit with a new source which is -44.8Volts with the 5.6K in series with it. Since V(TH) is saying what's across this certain branch I would say -44.8 Volts is. Take the difference between the two since they are of opposite polarity and I get a value of 33.3128 Volts. This is my V(TH)? \$\endgroup\$
    – Shane Yost
    Apr 19 '13 at 18:53
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    \$\begingroup\$ Change the voltage source to a short instead of 'taking it out'. \$\endgroup\$
    – jippie
    Apr 19 '13 at 18:55
  • \$\begingroup\$ The second part where I configure it with the current source is confusing. Not for sure if I should convert it over to a voltage source or not. \$\endgroup\$
    – Shane Yost
    Apr 19 '13 at 18:56
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By inspection, using superposition:

\$V_{TH} = 16V \dfrac{5.6k \Omega}{2.2k \Omega + 5.6k \Omega} - 8mA (2.2k \Omega ||5.6k \Omega) \$

\$R_{TH}=2.2k \Omega ||5.6k \Omega\$

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  • \$\begingroup\$ I follow now. I was not configuring it correctly before. V(TH) is -1.14872...It seems almost easier to not convert and just to figure up V(TH) for each source separately. Then use those two answers to find the difference. \$\endgroup\$
    – Shane Yost
    Apr 19 '13 at 19:25
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It seems you are making this too complicated and not seeing the forest for all the trees in the way. Solving this is 3 simple steps:

  1. Start at the left with just the 16 V source and R1 (you really need to put component designators on your schematics). By inspection it should be obvious that is a Thevenin source of 16 V and 2.2 kΩ.

  2. Now add R2. Add resistor accross the Thevenin source from #1 and transform that into a new single Thevenin source. You should know how to combine the resistance and how to determine the new open circuit voltage of the new combined Thevenin source.

  3. Take the result from #2, which is just a single Thevenin source and add the 8 mA load. The result can again be expressed as a single Thevenin source, which is also the answer to the overall problem.

    You may have to think a little about what happens when you put a current source on a Thevenin source. If you are having problems, consider the effect on impedance and voltage separately. What is the impedance of a current source?

Ask more specific questions if you get stuck, but I think I've given you enough hints for now.

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  • \$\begingroup\$ Determining the effect that the voltage source plays on the load will come out to 12.6359 Volts when I make the current source open. Then I determine the effect with the current source by shorting out the voltage source. The result will be -11.4872 Volts. Since the current source is supplying current towards ground this is a negative polarity. Therefore by finding the difference I get -1.14872 Volts as my V(TH). I hope I'm getting the terminology right here. Could you explain what you mean by impedance? \$\endgroup\$
    – Shane Yost
    Apr 19 '13 at 19:32

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