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Finding Thevenin equivalent circuit

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the Thevenin resistance.

    http://i.imgur.com/XoEfy.png

    2. Relevant equations

    Rth =Vex/Iex


    3. The attempt at a solution
    I found that the thevenin voltage is 869.304mV. I read in my textbook that in order to find the thevenin resistance in a circuit that contains a dependent source an external source is required, then the circuit can be solved using any method possible. So I added an external source to my circuit and then I tried using node-voltage to solve it, but I am stuck.

    kcl Node 1: [itex]\frac{V1}{70}[/itex] + [itex]\frac{V1-V3}{13}[/itex] + 5V2 = 0
    kcl node 2: -5V2 + [itex]\frac{V2}{50}[/itex] + [itex]\frac{V2 -V3}{18}[/itex] = 0
    kcl node 3: [itex]\frac{V3-V2}{18}[/itex] + [itex]\frac{V3-V1}{13}[/itex] + [itex]\frac{V3}{16}[/itex] - Iex = 0


    I ended up with Rth = -1.24118

    I know this isn't right. Even if it was right, I don't know how I can find Vex using that information.

    Can someone tell me what I'm doing wrong?


    Edit: I think I should use mesh current. I'll try that now.

    Using mesh current I ended up with 1.99 as Rth. None of my answers are right. I built the circuit and the resistance is supposed to be 12 ohms.

    I think it may just be my algebra that's messing my answers up. I had 4 equations for the mesh current analysis.

    70I1 + 5Vx + 50(I1-I2) = 0
    16(I2-I3) + 50(I2-I1) + 18(I2-14) = 0
    Vex + 16(I3-I2)=0
    18(I4-I2)-5Vx + 13I4=0
    Vx = 50(I1-I2)


    I think I overcomplicated this by using an external source. It would have been fine if I just tried to find the short circuit current parallel to the 16 ohm resistor.
     
    Last edited: Oct 2, 2012
  2. jcsd
  3. Oct 2, 2012 #2

    gneill

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    Staff: Mentor

    In your node 1 equation you've left out the 6V battery in the 1st term.

    If you're using a current source (Iex) to probe the circuit, then Vex will be identical to the node 3 voltage V3.
    This happens to be a type of circuit where the output resistance actually varies with the external conditions. That is, due to the sneaky operation of the active source, the apparent output resistance of the circuit will vary as you vary the external current or voltage supply you're using to determine the resistance. Because of this I don't know if there's a single value you can use as an answer! Perhaps the minimum resistance, which is about 12.03 Ohms, which is about the value you mentioned.

    Note that since the circuit contains an independent fixed source it can "stimulate" itself without an external probing source. To find the "base" values for the Thevenin voltage and resistance you can just slap resistive load RL onto the output and then solve for the output voltage in terms of that load resistance. The idea is to wrestle the resulting expression into the form of a voltage divider where your RL plays the role of the "lower" resistor. It will yield an expression of the form:
    $$V3 = Vth \frac{RL}{RL + Rth} $$
    with Vth, Rth and RL playing the role of the voltage divider. You can pick both Vth and Rth out of the expression all in one go.
     
  4. Oct 2, 2012 #3
    That equation you wrote looks similar to the one in my textbook, but it deals with maximum power transfer. The way my professor wanted me to solve for the thevenin resistance was to find the thevenin voltage and divide by the open-circuit current.

    http://i.imgur.com/OkJNF.png
     
  5. Oct 2, 2012 #4

    gneill

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    Staff: Mentor

    Surely that would be the short-circuit current. You'd find the open-circuit voltage (Thevenin voltage) and short-circuit current (Norton current) and then divide.

    The resulting Thevenin model won't account for the variation in output resistance that I mentioned above. But it could be that this issue is not relevant at this point in your coursework.

    To find the short-circuit current, try placing a load resistor RL on the output and then solving for the node 3 voltage (it will be an expression containing RL) The current will be V3/RL. Take the limit as RL goes to zero.
     
  6. Oct 2, 2012 #5
    Yes, I meant short-circuit current. I mixed up the different steps in my textbook. For another method it says when deactivating a current source, it should be replaced with an open circuit.

    I'll give that a try.
     
  7. Oct 3, 2012 #6

    uart

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    Science Advisor

    Yes this is correct. Let me summarize.

    - If you use the Voc Isc method (open circuit voltage and short circuit current) then you must leave the independent sources intact.

    - If you use the "test source" method (external source) then you should set all the other independent sources to zero (a zero voltage source is a s/c and a zero current source is an o/c). The preferred method is usually to use a 1A test source and perform nodal analysis. Your circuit gives a fairly easy 3 node system.

    As long as the sources are linear (that is, they depend only on constant multiples of other voltages and currents) then the presence of dependent sources will NOT cause the Thevenin resistance to be non constant (operating point dependent).

    The Thevenin resistance in this circuit is a constant and equal to about 12.025 ohms (using nodal analysis the test source method).
     
  8. Oct 3, 2012 #7
    I got 12.025. I used mesh current analysis. I misread the section in my textbook about supermeshes. I thought that supermeshes were only used when an independent current source was shared by two meshes. I didn't know that dependent current sources were acceptable. Boy do I feel stupid; I spent about 5 hours on this problem :p
     
  9. Oct 3, 2012 #8

    uart

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    Science Advisor

    Ok, good that you got it. :)

    Nodal analysis is easier though, only three nodal equations and no supermesh.
     
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