# Geogebra App on Maximum Power Theorem

This simulation demonstrates the power dissipated in a variable resistor given that the battery has an internal resistance (made variable in this app as well).

Since the power dissipated by the resistor is given by

$P=I^2R$

and the current is given by

$I=E(R+r)$,

$P=E^2\times\dfrac{R}{(R+r)^2}=\dfrac{E^2}{\dfrac{r^2}{R}+R+2r}$

This power will be a maximum if the expression for the denominator $\dfrac{r^2}{R}+R+2r$ is a minimum.

Differentiating the expression with respect to R, we get
$\dfrac{d(\dfrac{r^2}{R}+R+2r)}{dR}=-\dfrac{r^2}{R^2}+1$

When the denominator is a minimum,
$-\dfrac{r^2}{R^2}+1=0$

Therefore,
$r=R$ when the power dissipated by the resistor is highest.