Question

A D.C. series motor draws 45 A from a D.C. supply when developing the rated torque at 1500rpm. If the motor develops 1/4 rated torque, neglecting saturation and losses, evaluate the current drawn by the motor, and the motor speed

Answer 1

For a DC series motor, the armature current would reduce to half (22.5 A) if the motor develops one quarter of the rated torque, and the speed of the motor would increase due to reduced load. The exact new speed cannot be determined without additional specific motor characteristics.

Step-by-step explanation:

The question is about the performance of a DC series motor when the torque and hence load changes. Specifically, it inquires about the current the motor would draw when it's developing one quarter of the rated torque and the speed at which the motor would run under these conditions.

In a DC series motor, the generated torque is proportional to the square of the armature current (I). Therefore, if the torque is reduced to a quarter, the current would reduce to half of the original current, because torque is related to the current by a square relationship (T ≈ I^2). Original current was 45 A, so the motor would now draw 22.5 A if it develops one-quarter of the rated torque. As for the speed, series motors have a characteristic of increasing speed as the load (and thus torque) reduces. Neglecting saturation and losses, the speed of the motor would increase since it's now operating under a reduced load. An exact speed is not calculable without additional information about the motor's characteristics.