AC Sources - Thevenin & Max Power

Discussion in 'Homework Help' started by jegues, Nov 22, 2010.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    Could someone please check my work on this question? There are no solutions given and I'd like to check my answer.

    Please see figures attached for my work and the problem statement.

    I started by computing the impedances of each element.

    After doing so, I opened the circuit up and used mesh analysis as well as a few KVL's to solve for the voltages across the elements/source in the open Voc loop.

    Once I had obtained all the voltages I simply applied KVL in the open loop and solved for Voc.

    After obtaining Voc I found the equivalent impeadance at the terminals of the load by shorting out the voltage sources and opening up the current sources.

    With Voc and Zeq I was able to create the Thevenin equivalent and connect it to the load.

    The condition for maximum power transfer to the load is that the load, Zl, be the complex conjugate of Zt, which I labeled accordingly.

    Is my work correct? Can anyone spot any errors?

    Also, how do I determine the power transfered to Zl?
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    Your calculation titled "KVL in bottom loop" is incorrect. The current in the 6mH inductor is just what comes from the current source, 3 amps.

    The value of Voc I get is (148 -j12)/13.

    To calculate the power delivered, calculate the magnitude of the current in the impedance of the load, then the power is I^2*Realpart(Zl).
  3. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    I'll reattempt the problem from that KVL equation and see if I can get the same Voc.

    Thank you for taking the time to check. I'll see if I can get the power as well.

    I'll return and post my results.
  4. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    Here's my second attempt.

    I'm still not getting the same Voc as you.

    Our real parts are the same, but the imaginary parts are not.

    Did I make another mistake somewhere?

    EDIT: I found one mistake in my KVL in open loop.

    I should have a 18j not a -18j.

    I'm finding small mistakes all over the place, I'll try it again!

    EDIT: Posted attempt number 2, I don't think I made any mistakes this time. Are you sure your Voc is correct? I think I can spot your mistake, you must have wrote down a positive twelve over 13 while I have a negative twelve over 13.

    136 + 12 = 148,

    136 - 12 = 124.

    I hope the Voc I found is in fact correct.
    Last edited: Nov 26, 2010
  5. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    I think you used the wrong sign for the voltage across the inductor.

    Here's another way to find Voc.

    You know the current i2; you have -0.692+j0.462

    Use that current to calculate the voltage across the 2 ohm resistor and then add that voltage (observing proper polarity) to the voltage Vs(t). That will be Voc.
  6. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    BAH! Thats so much easier! I wish I would have seen that loop from the start... I'll give it a shot right now and see what I get.

    I'll edit this post with my results.

    EDIT: I finally got the same Voc. That was MUCH simplier. I made an attempt to calculate the power delivered to Zl as well.

    Is my work correct?

    Also, where did that formula for power delievered come from? Is that always what it is?

    The equation in my textbook for average power delivered to an element is,

    P = (\frac{I_{m}^{2}}{2})Re(Z)

    Is this the formula you were reffering to? If it was then I have to divide my answer posted by 2.

    Thanks again!
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    Last edited: Nov 27, 2010