Answer
13 A
Work Step by Step
It is apparent from equation 21–5 that the back emf is proportional to the angular speed of rotation. If the angular speed is halved, so is the back emf, The new back emf is 54 volts.
The back emf opposes the applied emf of 120 volts. The combined effect gives the net voltage across the motor, which is IR.
$$\epsilon_{applied}-\epsilon_{back}=IR$$
$$I=\frac{\epsilon_{applied}-\epsilon_{back}}{R} $$
$$=\frac{120V-54V}{5.0\Omega}=13A$$