# 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)

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

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

$\Rightarrow R_{L} = R_{th}$