A direct-current generator has an emf of 120 V ; that is, its terminal voltage is 120 V when no current is flowing from it. At an output of 20 A , the terminal potential is 115 V . (a) What is the internal resistance $r$ of the generator? (b) What will be the terminal voltage at an output of 40 A ?
The situation is much like that shown in Fig. 26-3. Now, however, $\ell=120 \mathrm{~V}$ and $I$ is no longer 25 A .
(a) In this case, $I=20 \mathrm{~A}$ and the p.d. from $A$ to $B$ is 115 V . Therefore,
$$
115 \mathrm{~V}=+120 \mathrm{~V}-(20 \mathrm{~A}) \mathrm{r}
$$
from which $r=0.25 \Omega$.
(b) Now $I=40 \mathrm{~A}$. So
Terminal p.d. $=E-I r=120 \mathrm{~V}-(40 \mathrm{~A})(0.25 \Omega)=110 \mathrm{~V}$
Fig. 26-3