Why is power dissipated in a circuit maximum when external resistance is equal to the internal resistance in the circuit? [![question 24][1]][1]Suppose in a circuit the battery has emf 6V and internal resistance 3ohm. It is connected to an external resistance = R ohm. According to my book the maximum power dissipated is when R = 3 ( I.e. internal resistance). They have derived the result and it seems fine. Now power = I^2 ( R+3) and I = 6/(R + 3) so power should be 36/(R+3). But if we put R=3 power is 6 w and if we put R = 0 power is 12 w. So shouldn't power be maximum when external resistance is 0?
 A: Your confusion is between two related concepts.


*

*Power dissipated in total = internal power + external power. If that is the power you are talking about, then an external short circuit will maximize the current and therefore maximize total power, $V\cdot I$.

*Power delivered to the load. That is the thing addressed by the maximum power transfer theorem, and it requires internal resistance = external resistance.


The proof follows simply. If we have internal resistance $R_i$ and external resistance $R_o$, then the total resistance is $R_i+R_o$. The current is $\frac{V}{R_i+R_o}$ and the voltage across the external resistor is current times resistance. It follows that power in the external resistor is $\frac{V}{R_i+R_o}\frac{V\cdot R_o}{R_i+R_o}$
To find the maximum of that power, we take the derivative w.r.t $R_o$ and set it to zero:
$$\begin{align}\frac{dP}{d(R_o)}\propto \frac{-2R_o}{\left(R_i+R_o\right)^3}+\frac{1}{\left(R_i+R_o\right)^2}&=0\implies\\
-2R_o+R_i+R_o&=0\implies\\
R_o&=R_i\end{align}$$
