understanding maximum power transfer theorem

Discussion in 'Homework Help' started by PG1995, May 28, 2011.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Hi :)

    I'm trying to understand 'background' working of the maximum power transfer which says that power transfer takes place across R_L when it is equal to R_th (R_L is load resistance and R_th is Thevenin resistance). It is easily proved by doing some math.

    P=VI, P=I^2.R, P=V^2/R, V=IR

    Let's focus on P=V^2/R_L. To have maximum power transfer to R_L, V should be as large as power (V is volts dropped across R_L). V across R_L could be found using voltage divider rule: (R_L x E)/(R_L + R_th). Further R_L should be as little as possible because it is denominator. But also note that making R_L smaller would reduce the volts dropped across R_L.

    But now focus on P=I^2.R_L. Compare it with the previous analysis of P=V^2/R_L. In (I^2.R_L), R_L should be as large as possible which is in contrast with the previous analysis which required R_L to be minimum. So some compromise is needed. But there is another point to note. R_L and R_th are in series so if R_L is made too big then the value of I would drop.

    I hope you could see where I'm coming from (or, rather trying to come from! :) ). Could you please help me to really understand the working of maximum power transfer theorem? Many thanks.
     
  2. jegues

    Well-Known Member

    Sep 13, 2010
    735
    43
    I = \frac{V_{th}}{R_{th} + R_{L}}

    P_{L} = I^{2}R_{L} = \left( \frac{V_{th}}{R_{th} + R_{L}} \right) ^{2} \times R_{L}

     = \frac{V_{th}^{2}}{\frac{R_{th}^{2}}{R_{L}} + 2R_{th} + R_{L}}

    To simplify the math, determine RL such that the denominator is minimized. (Thus giving maximum power)

    \frac{d}{dR_{L}} \left( \frac{R_{th}^{2}}{R_{L}} + 2R_{th} + R_{L} \right)

    = 1 - \left( \frac{R_{th}}{R_{L}} \right)^{2} = 0

     \Rightarrow R_{L} = R_{th}
     
Loading...