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What is the back emf of a 120 V motor that draws 8.00 A at its normal speed and 20.0 A when first starting?

a) 50 V
b) 60 V
c) 70 V
d) 80 V

User Mjv
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1 Answer

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Final answer:

The back electromotive force (back EMF) of a 120 V motor that draws 8.00 A at its normal speed and 20.0 A when first starting is 60 V.

Step-by-step explanation:

Back electromotive force (back EMF) is the voltage generated in the opposite direction to the applied voltage in a motor or generator. It is caused by the change in magnetic flux through the motor's windings. When a motor draws 8.00 A at its normal speed, it means that the current I flowing through the motor is 8.00 A. The back EMF can be calculated using the formula:

Back EMF = Applied Voltage - (Current x Resistance)

Given that the applied voltage is 120 V, and the current is 8.00 A, we can substitute these values into the formula to find the back EMF:

Back EMF = 120 V - (8.00 A x Resistance)

To find the value of the resistance, we need another equation. When the motor draws 20.0 A when first starting, we can write:

20.0 A = 120 V / (Resistance)

Solving for resistance gives us:

Resistance = 120 V / 20.0 A

Substituting this value back into the original equation for the back EMF:

Back EMF = 120 V - (8.00 A x (120 V / 20.0 A))

Simplifying the equation gives the back EMF:

Back EMF = 60 V

User Gogognome
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