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Homework Help: Long winded Thevenin question

  1. Nov 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Essentially I've been attempting to solve the first question.

    2. Relevant equations
    I'm mostly concerned on the methodology to solve such a question; So far I've been attempting the following:

    As always with these questions I'm seeking the Open circuit voltage, and the short circuit current and saying V/I=Rthevenin.

    The trick is in finding Voc, and Isc

    For Voc I first found the voltage at the node connected to the positive of the capacitor; this was a long process involving finding the complementary and particular solutions and adding them. I wasn't sure how to solve the constants that arose from this, so I made the assumption that at time = 0, capacitor voltage will be 0. (Note the question doesn't mention initial conditions)

    Then I simply multiplied Vc by 3, because Va=vc+(2*Vc),Va=3vc.

    To find Isc, I realized that Va =0 volts, and because Va=(vc*3) then Vc=0. (i'm fairly sure I'm far offtrack with this one)

    To find the current through the 6ohm resistor we can simply say Vs/6. (with Vs being a sinusoid). subtract from this the current through the capacitor (KCL at a node).. and I assumed that would be ISC.

    Where am I wrong, if I was ever on the right track, and what's the preferred method to solve something like this.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Nov 20, 2013
  2. jcsd
  3. Nov 20, 2013 #2


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    Staff: Mentor

    Okay, a bit of a tricky one here. Since you've got the angular frequency of the source you can determine the impedance of the capacitor. No need to go solving differential equations if you're looking for the steady-state conditions. The controlled source, being ideal, isn't going to care about the load it sees, it'll want to faithfully produce twice the voltages it sees at the top of the capacitor. So intuitively it feels like the Thevenin impedance should be zero... something to confirm by analysis.

    One approach to finding Thevenin equivalents when you've got controlled sources involved is to place a load resistor RL across the output and solve for an expression for the output voltage. Now, a Thevenin equivalent with a load attached is just a source and potential divider:


    If you can force your expression for the output voltage into the form of the voltage divider expression, then you can pick out the Thevenin voltage and impedance by inspection.

    If the Thevenin impedance turns out to be zero then your expression for the output voltage will not contain the variable RL at all, and you'll be left with just a Thevenin voltage :wink:

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  4. Nov 20, 2013 #3
    I would solve for the output voltage by frequency domain analysis right?
  5. Nov 20, 2013 #4


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    Staff: Mentor

    Sure, but the only frequency involved is 5 rad/sec as determined by the argument to the cosine function of the voltage source. In other words, s = ωj, and ω = 5 rad/sec.

    This lets you determine the impedance (complex value) of the capacitor. Then proceed as you would for any circuit where you know the impedances.
  6. Nov 21, 2013 #5
    I see..

    Say that I wanted to solve this by finding Voc/Isc.. When I short the two ports. It makes one side of the dependent voltage source = 0 (directly connected to ground)...

    So how does that work? If the equation governing the top of the capacitor and the other side of the dependent source is

    V2 = 3*V1,
    (with V1 being the top of the capacitor, and V2 being port A)

    If V2 is 0 (because it's shorted directly to the ground)... what does that make V1? 0 aswell? Or does the dependent circuit take priority, and make port a equal to 2x whatever v1 is despite being connected to ground.
  7. Nov 21, 2013 #6

    Here's my working, where have I gone wrong.

    A bit of an explanation of what I've done is basically considering a supernode across the dependent source, and then performing KCL on that node.

    Relating V1 and V2, then solving for V1. solving for V2, and declaring V2 and my Voc.

    Shorting the ports, calculating Isc (this is the part I'm least sure about). (I've assumed current going through the capacitor in the event of a short circuit across the ports is 0, as both ports of the capacitor are at 0volts.)
    Then saying Zth = Voc/Isc
  8. Nov 21, 2013 #7


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    Staff: Mentor

    It looks as though you've reached the correct values, if I'm reading your work correctly.

    That's the idea. If you write KVL around the second loop (output shorted) you have v1 + 2v1 = 0, which means v1 is zero. In order for v1 to be zero, the mesh current in loop 2 must be equal to the mesh current in loop 1 (that is, i2 = i1, where i1 and i2 are the mesh currents for the two loops).
  9. Nov 21, 2013 #8

    The Electrician

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    Gold Member

    It looks to me like the angle of Zth is -15.945°, not -74°.
  10. Nov 21, 2013 #9


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    Staff: Mentor

    You're right. I missed that in the fine print Good catch.
  11. Nov 21, 2013 #10
    Hmm? Does this mean I've done something wrong with my phase angle on the thevenin resistance?
    How are you reaching -15.945?
  12. Nov 21, 2013 #11

    The Electrician

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    Gold Member

    Calculate (1.166666 +.33333 j) in polar coordinates; what angle do you get?
  13. Nov 21, 2013 #12
    Oh right,

    No idea how I ended up with -74.

    Thanks guys, this helped out a lot.
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