# Answer to Question #84768 in Electric Circuits for yomna

Hi!

I think the *maximum power transfer theorem* is what will help you. It states that to get maximum external power, the resistance of the load must be equal to the resistance of the source. That is why despite the the fact that the efficiency @$\eta=0.5@$ in this case, the best parameters of your circuit are achieved with @$R=2000\text{ } \Omega@$. Look:

@$P_{ext}=\frac{E^2}{(R+r)^2}R,@$

find the maximum external power by calculating the derivative and equating it to 0:

@$\frac{dP_{ext}}{dR}=E^2\frac{r^2-R^2}{(R-r)^4}=0,@$

@$R=r.@$

So, you should determine the source resistance and compare it with the load resistance.

## Comments

## Leave a comment