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The armature windings of a dc motor have a resistance of 3.0 Ω. The motor is connected to a 120V line, and when the motor reaches full speed against its normal load, the back emf is 117V. What is the current into the motor when it reaches full speed?

Respuesta :

Answer:

0.6 A

Explanation:

As the motor gears towards full speed, the voltage of the circuit tends to become the difference that exists between the line voltage and the that of the back emf. Remembering Ohm's law, we then apply it to get the final, lower current that is based on the reduced voltage Ef. We use the provided resistance in the question, that is 3 ohms.

Ef = (120 V) - (117 V) = 3 V

If = Ef/R = (3 V) / (5.0 Ω) = 0.6 A

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