Max Power Proof

Discussion in 'Homework Help' started by ihaveaquestion, May 5, 2009.

  1. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    http://img140.imageshack.us/img140/3572/scan0028.jpg

    a) the only proof I can come up with is that for RS = 0, RL would have max power put into it because there would be no voltage drop across RS since it doesn't exist, so since p=vi, the max power (source) would be going into RL.

    b) I'm not sure

    c) I'm not sure

    Doing some analysis on the circuit I know that:

    VL = Vi*(RL/RS+RL)
    VS = Vi*(RS/RS+RL)
    I = Vi/(RS+RL)

    We did this problem before.. you have to get a function of something and take the derivative to maximize the function...
     
  2. steveb

    Senior Member

    Jul 3, 2008
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    469
    Part A is easy since Rs=0 results in the greatest current and greatest voltage across RL. Since P=VI, power is clearly maximized.

    Part B is the tricky one. The usual approach is to write the power in RL as a function of RL and other known quantities. Then take the first derivative of power with respect to RL. Set this to zero and solve for RL. It will be clear that RL=Rs maximizes power.

    Part C is easy once you prove B because the resistors are equal in value and the have equal current flowing in them. P=I^2*R, hence power is the same in both resistors.
     
  3. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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    Ok I'll try that right now:

    In the meantime I have a quick question about current divider

    http://img134.imageshack.us/img134/4711/73592586.jpg

    1) Are those two circuits equivalent?

    I think they are... if I moved the R2 resistor over to the left side, they wouldn't be equivalent because I'm not allowed to do that right? I can only rearrage R1 like that...

    2) Are these expressions correct for both circuits?

    i1 = I*(R2/R1+R2)
    i2 = I*(R1/R1+R2)
     
  4. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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  5. steveb

    Senior Member

    Jul 3, 2008
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  6. steveb

    Senior Member

    Jul 3, 2008
    2,433
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    For #2 above, the formula look correct.

    For #1 above, I think there is a mistake. You can move R1 to the other side (as you showed in the figure), but you can also move R2 to the other side. Why do you say it's not allowed to move R2 and but R1 is OK? This would seem to violate symmetry since both resistors are connected the same in terms of node connections. In general, the physical location of components will not matter for circuits, only the nodal connections are important. (Note: Every rule has exceptions, but you are nowhere near a situation with transmission line effects, electromagnetic radiation or Faraday Law inductive coupling effects)
     
    Last edited: May 5, 2009
  7. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
    314
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    Ahh I see... I understand, thanks
     
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