An electric motor operating on 50 V DC supply draw a current of 12 A is the efficiency of motor is 30% then resistance of the winding
of the motor
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Answer:
Solution :
given V=50V
I=12A
n=30%
Input power = VI
=50×12=600walt
Since n=30% , the energy dissipated as heat
P=70100
=70100×600
=420w
P=I2R
R=PI2
=42012×12
=2.9Ω
or 3Ω (approx)
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