Max. Possible Avg. Power to Load

Discussion in 'Homework Help' started by tquiva, Dec 23, 2010.

  1. tquiva

    Thread Starter Member

    Oct 19, 2010
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    I know the answer for this is ii. I really don't remember seeing this problem in my readings though. Can someone please help me how to get √2 ?
     
  2. tyblu

    Member

    Nov 29, 2010
    199
    16
    Power transfer is maximized when load impedance is conjugately matched to source impedance: when power factor is 1. I would start this by transforming the source into a Thevenin source.
    V_{S,Th.} = cos(t + 45^o) \cdot (1 - j/\omega) = cos(t + 45^o) \cdot |1 - j/\omega| \angle arctan(-1/1) = cos(t + 45^o) \cdot \sqrt{2} \angle -45^o = \sqrt{2} \angle 0^o V
    Z_{S,Th.} = (1 - j/\omega) \Omega
    Set load impedance for conjugate matching:
    Z_L = Z_{S,Th.}*
    1\Omega + Z_X = (1 - j/\omega)* = (1 + j/\omega) \Omega
    Since \omega = 1, Z_X = j/\omega = j.
    An inductor has impedance Z_{inductor} = j\omega L, so L = 1 H.

    If the answer truly is ii, I'd like to know where I went wrong.
     
    tquiva likes this.
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    First, isn't the 45° a red herring? 45° relative to what?
    Second, I cheated and ran a simulation. Current through RL was maximum when L=1.
     
  4. tyblu

    Member

    Nov 29, 2010
    199
    16
    Normally I would ignore this number as well, but you'll see that it cancels out with the -45° impedance branch, making subsequent steps (step?) slightly simpler. Good to know the math holds up...
     
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