Question

A dc motor with its rotor and field coils connected in series has an internal resistance of 3.2 ω. when running at full load on a 120-v line, the emf in the rotor is 105 v.

Answer 1

Missing question: what is the current drawn by the motor from the line?

Solution:
The potential difference between the two ends of the motor is
$\Delta V= \epsilon + Ir$
where $\epsilon$ is the emf, I the current and r the internal resistance of the motor.

By re-arranging the equation and using the data of the problem, we can solve to find the current I:
$I= \frac{\Delta V-\epsilon}{r}= \frac{120 V-105 V}{3.2 \Omega}=4.7 A$