# Maximum power transfer derivation

In my textbook, the author showed the calculus derivation of what value of load resistance is needed such that it will absorb maximum power from the source, while keeping the Thevenin voltage and Thevenin resistance fixed. After that, the author asked us to repeat the derivation for the case where the load resistance is fixed while the source resistance is variable. It seems logical that the the source resistance has to be zero for maximum power transfer to the load. However, I can't seem to get the answer via differentiation. I tried differentiating with respect to the source resistance but the numerator has no Rt term.

• What is your question? Show your work. – WhatRoughBeast Jul 13 '18 at 13:49
• Look at equation 3.34 you don't need to differentiate to see when this is a maximum if you change the source impedance. – Warren Hill Jul 13 '18 at 14:07

You don't even need to find the first derivative to find the max of this function (even a plot of the power as a function of $R_T$ would show this) but you're still after the mathematical proof since the problem asks for it...

You take expression 3.35 and find the derivative using the quotient rule, that's all:

$$\dfrac{dP}{dR_T}=-\dfrac{2v_T^2R_L}{(R_L+R_T)^3}$$

Just like you showed in your post. From here on, this becomes one of those math problems...There is clearly no $R_T$ in the numerator but as $R_T\to \infty$, the derivative goes to zero—that is a critical point.

The list of possible values for $R_T$ are [0,$\infty$)—can't consider negative resistance values— and these are the endpoints of the interval. In order to find the relative/absoulte max and min of a function, you need to evaluate the orginal function, $P(R_T)$, at the critical points (the ones you get by setting the derivative to zero) and the endpoints, see this.

That is, you now need to find $P(0)$ (an endpoint) and $P(\infty)$ (this is both an end point and a critical point). That clearly will give you a max value at $P(0)$.

There is a theoretical answer, which is not realisable:

$$\small P_L=\frac{V_T^2\:R_L}{(R_T+R_L)^2} =\frac{V_T^2\:R_L}{R_T^2+2R_TR_L+R_L^2}$$ Divide by $\small R_L^2$: $$\small P_L= \frac{V_T^2/R_L}{R_T^2/R_L^2+2R_T/R_L+1}$$ Let $\small U=R_T/R_L$ $$\small P_L =\frac{V_T^2}{R_L}\:.\:(U^2+2U+1)^{-1}$$ $$\small \frac{dP_L}{dU}= \frac{V_T^2}{R_L}\left(-\frac{2U+2}{(U^2+2U+1)^2}\right)$$

Setting $\large \frac{dP_L}{dU}\small=0$ gives $\small2U+2=0$

Hence: $$\small R_T=-R_L$$ This means the loop resistance is zero, current is infinite and power dissipated by $\small R_L$ is infinite.