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In: Electrical Engineering

A 5-kW, 250-V shunt motor has an armature resistance of 0.1 ? and a field-circuit resistance...

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 operating at rated voltage, the speed is observed to be 1200 r/min when the machine is loaded such that the armature current is 30 A. In order to protect both the motor and the dc supply under starting conditions, an external resistance will be connected in series with the armature winding (with the field winding remaining directly across the 250-V supply). The resistance will then be automatically adjusted in steps
so that the armature current does not exceed 300 percent of rated current. The step size will be determined such that, until all the external resistance is switched out, the armature current will not be permitted to drop below 150 percent of rated current. In other words, the machine is to start with 300 percent of rated armature current and as soon as the current falls to 150 percent of rated current, sufficient series resistance is to be cut out to restore the current to 300 percent. This process will be repeated until all of 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 the starting operation and
iii. At what speed should each step change occur?
iv. Plot the following graphs; time versus speed, time versus the generated Voltage (Ea) and time versus the armature current.

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