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A series motor having an armature and field resistance of 0.1Ω derives a fan for which torque varies as the square of the speed. On 220 V, it runs at 300rpm and takes 25 A. The speed is to be raised to 400rpm by increasing the voltage. Find the voltage and current assuming the flux is directly proportional to current and frictional losses are negligible.

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Final answer:

The question involves increasing the speed of a series motor from 300rpm to 400rpm by increasing voltage, considering the torque is proportional to the square of the speed, and the flux is proportional to the current. However, to calculate the exact new voltage and current, additional specific data or assumptions about the motor's constants are needed.

Step-by-step explanation:

The student's question is about finding the required voltage and current to raise the speed of a series motor from 300rpm to 400rpm, given that the motor's torque varies as the square of the speed while the motor's flux is assumed to be directly proportional to the current. Initially, the motor runs on 220 V at 300rpm, drawing 25 A. The problem states that the armature and field resistance of the series motor is 0.1Ω. To solve this, we can use the relationship of voltage, resistance, current, and speed in a series motor and the fact that torque is proportional to the square of the speed. However, the student's information doesn't provide a clear way to determine the precise voltage and current needed without additional data or assumptions regarding the machine's constants. However, we can establish that an increase in voltage is necessary to achieve a higher speed while keeping in mind that the relationship between speed and voltage in a DC motor is not linear due to the armature reaction and magnetic saturation.

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