An electric motor operates on a 50 volt supply and a current of 12 A. If the efficiency of the motor is 30%, what is the resistance of the winding of the motor
V = 50 volt; I = 12 ampere
Total power consumed = 50x12 W = 600 W
Since efficiency of motor is 30%, therefore, 70% of power is lost in Joule heating.
If Q is the heat produced, then
Q = 70% of 600 W or Q = 70/100 x 600 W = 420 W
Also, Q = I2R
∴ 420 = (12)2R or R = 420/(12 x 12( Ω = 2.92 Ω
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