A 5-kW, 250-V shunt motor has an armature resistance of 0.1 ?and a field-circuit resistance of 80 ?. When the motor is operatingat rated voltage, the speed is observed to be 1200 r/min when themachine is loaded such that the armature current is 30 A. In orderto protect both the motor and the dc supply under startingconditions, an external resistance will be connected in series withthe armature winding (with the field winding remaining directlyacross the 250-V supply). The resistance will then be automaticallyadjusted in steps
so that the armature current does not exceed 300 percent of ratedcurrent. The step size will be determined such that, until all theexternal resistance is switched out, the armature current will notbe permitted to drop below 150 percent of rated current. In otherwords, the machine is to start with 300 percent of rated armaturecurrent and as soon as the current falls to 150 percent of ratedcurrent, sufficient series resistance is to be cut out to restorethe current to 300 percent. This process will be repeated until allof the series resistance has been eliminated.
i. Find the maximum value of the series resistance.
ii. How much resistance should be cut out at each step in thestarting operation and
iii. At what speed should each step change occur?
iv. Plot the following graphs; time versus speed, time versus thegenerated Voltage (Ea) and time versus the armature current.