A 227 VRMS AC motor in a domestic vacuum cleaner is rated to produce 502 W of me

Question

A 227 VRMS AC motor in a domestic vacuum cleaner is rated to produce 502 W of mechanical power while operating at 4885 rpm. Calculate the torque produced by this motor at this speed. Give your answer in Nm rounded to 2 decimal places.

A 295 VRMS AC motor in a domestic vacuum cleaner is rated to produce 506 W of mechanical power while operating at 4350 rpm. If the motor is 100% efficient, what would be the RMS value of the current into the motor? Give your answer in A rounded to 2 decimal places.

Explanation / Answer

a)

Given Voltage input to AC motor is Vrms = 227 V

Mechanical Power output is P = 502 W

Speed of the motor is N = 4885 rpm

We know P = T*W

where P is power

T is torque

W is speed in rad/sec = 2*pi*N/60

T = P/W = (502*60)/(2*pi*4885)

T = 0.98 N-m

b)

Given Voltage input to AC motor is Vrms = 295 V

Mechanical Power output is P = 506 W

Speed of the motor is N = 4350 rpm

Given motor is 100% efficient

We know P = V*I

where P is power

V is Voltage

I is current

I = 506/295 = 1.71 A

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