Question

A battery of emf $E$ and identical resistance $r$ is connected across a pure resistive device (such as an electric heater) of resistance $R$. Prove that the power output of the device will be maximum if $R=r$.

Answer 1

Total resistance of the circuit $=r+R$

Let the current flowing through the circuit be $i$

$\therefore$ $i=\frac{E}{(r+R)}$

Output power of the device, $P=i^{2}R=\frac{E^{2}R}{(R+r{)}^2}$

For power to be maximum, $\frac{dP}{dR}=0$

$\Rightarrow$$E^2(\frac{(r+R{)}^2\times 1-R\times 2(R+r)}{(R+r{)}^4})=0$

$\Rightarrow (\frac{r^2+R^2+2rR-2R^2-2rR}{(R+r{)}^4})=0$

$\Rightarrow \frac{r^2-R^2}{(R+r{)}^4}=0$

$\Rightarrow$$R^2=r^2$

$\Rightarrow$$R=r$