Question

A battery with an EMF of 100 V and a resistance of 1  is closed to a resistance of 200 . Find
the total and useful current power.

Answer 1

The total power is

$P=I^2(R+r),$

where I - current in the circuit;

R - resistance of external circuit;

r - internal resistance of battery.

$I=\frac{E} {R+r}$,

where E - EMF of battery.

So,

$P=\frac{E^2}{(R+r)^2} \cdot (R+r) =\frac{E^2}{(R+r)}=\frac{100^2}{200+1}=49.75\space W$

The useful power is

$P_u=I ^2R=\frac{E^2}{(R+r) ^2}\cdot R=\frac{100 ^2}{(200 +1)^2}\cdot 200=49.5\space W$

Answer: total power 49.75 W, useful power 49.5 W.