Maximum Power Transfer Theorem

In applications such as communications it is desirable to maximize the power delivered to a load.
We will address the problem of delivering the maximum power to a load when given a system with known internal losses.
It should be noted that this will result in significant internal losses greater than or equal to the power delivered to the load.
The Thevenin equivalent is useful in finding the maximum power a linear circuit can deliver to a load assuming that $R_L$ is adjustable.

$p=i^{2} R_{L}=\left(\frac{V_{\mathrm{Th}}}{R_{\mathrm{Th}}+R_{L}}\right)^{2} R_{L}$

To prove differentiate $P$ w.r.t $R_L$ and equate to zero $\begin{aligned} p=i^{2} R_{L}&=\left(\frac{V_{\mathrm{Th}}}{R_{\mathrm{Th}}+R_{L}}\right)^{2} R_{L}\\ \frac{d p}{d R_{L}} &=V_{\mathrm{Th}}^{2}\left[\frac{\left(R_{\mathrm{Th}}+R_{L}\right)^{2}-2 R_{L}\left(R_{\mathrm{Th}}+R_{L}\right)}{\left(R_{\mathrm{Th}}+R_{L}\right)^{4}}\right] \\ &=V_{\mathrm{Th}}^{2}\left[\frac{\left(R_{\mathrm{Th}}+R_{L}-2 R_{L}\right)}{\left(R_{\mathrm{Th}}+R_{L}\right)^{3}}\right]=0 \\ &= \left(R_{TH}-2R_{L}\right)=0 \\ & \Rightarrow \boxed{R_{L} = R_{Th}} \end{aligned}$
The maximum power transferred $\boxed{P_{max} = \dfrac{V_{Th}^2}{4R_{Th}}}$