Undoing a thevenin?

Discussion in 'Homework Help' started by Kevin2341, Apr 23, 2013.

  1. Kevin2341

    Thread Starter Member

    Nov 1, 2009
    19
    0
    reversethevenin.png

    I am given this circuit, I'll try to explain what's going on here as I couldn't fully label everything as it is on my homework.

    This is a phasor problem, my Voltage source is unknown, however, I know the voltage at Vo to be 8∠ 45º

    Both resistors are 2Ω, the inductor is j2Ω and the capacitor is -j2Ω

    The full problem is asking for the current through each branch, which that will be easy. However, I am stuck with where to go to find my Vs value.

    I understand in a thevenin equivalent, the voltage at the terminals in series with the equivalent resistance is the same as having the original voltage source with the network of resistors and all that other good stuff, however, I've never had to reverse one of these things.

    I have calculated my equivalent impedance to (4+j2)Ω, now, where I am stuck, I cannot figure out how to calculate the correct current. I thought to try plugging in the equivalent impedance with the terminal voltage, but that gives me a funky number that doesn't feel right.

    V=IZ, I = V/Z
    V = 8∠ 45º, or 4√2+j4√2
    I = (4√2+j4√2)/(4+j2)Ω = (6√2+j2√2)/5

    My reasoning behind why this is not correct is because this is the current from the terminal voltage and not from the voltage source. Undoing the calculation and solving for V from that set of numbers brings us back to 4√2+j4√2

    My next thought was to try a voltage division at the Vo terminal. Here is what I tried:

    We know voltage division is:
    Va = (R1)/(R1+...Rn)*Vs

    So what I tried was similar to that concept

    (bare with the unholy amount of ugly computer-typed math, basically, I'm calling the top impedance everything but the last 2 ohm resistor divided by equivalent impedance).

    4√2+j4√2 = [2-(1/j2-1/j2)^-1]/(4+j2)*Vs, solving for Vs I got:

    Vs = 4√2+j12√2
    Vs = 17.89∠ 71.56º

    This number seems like it could be correct. It is higher than the terminal voltage, so it satisfies that requirement.

    Anyone have any comments? Does it look correct? I'm going to try calculating the thevenin equivalent from this to check my answer and I suppose that will be the ultimate test to see if it's correct. If not... back to the scratch paper.
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,795
    By Vs, do you mean the V1 in the diagram?

    If you know the current in each branch and you know the component impedance in each branch, isn't it just a simple matter of walking along a path from node to node until you reach the node you want the voltage far?

    What you are being asked to do doesn't really have anything to do with a Thevenin equivalent circuit. If you try to force it into that mold, you will probably just confuse yourself.

    See what I mean?

    My next thought was to try a voltage division at the Vo terminal. Here is what I tried:

    We know voltage division is:
    Va = (R1)/(R1+...Rn)*Vs

    So what I tried was similar to that concept

    (bare with the unholy amount of ugly computer-typed math, basically, I'm calling the top impedance everything but the last 2 ohm resistor divided by equivalent impedance).

    4√2+j4√2 = [2-(1/j2-1/j2)^-1]/(4+j2)*Vs, solving for Vs I got:

    Vs = 4√2+j12√2
    Vs = 17.89∠ 71.56º

    This number seems like it could be correct. It is higher than the terminal voltage, so it satisfies that requirement.

    Anyone have any comments? Does it look correct? I'm going to try calculating the thevenin equivalent from this to check my answer and I suppose that will be the ultimate test to see if it's correct. If not... back to the scratch paper.[/QUOTE]

    If I tell you that I have a battery connected to two resistors in series that are 8Ω and 12Ω and that the voltage across the 12Ω resistor is 16V, can you tell me what the battery voltage is? Without doing anything Thevenin-like?

    Same goes for this problem.
     
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