The electric motor consumes a current of 20 A at a voltage of 220 V, determine the full work of the motor current

| January 5, 2021 | education

The electric motor consumes a current of 20 A at a voltage of 220 V, determine the full work of the motor current and the amount of heat released in 30 minutes if the resistance of the motor winding is 0.75 Ohm

Electric current work:
A = I * U * t, where I is the current strength (I = 20 A), U is the voltage in the network (U = 220 V), t is the operating time of the electric motor (t = 30 min = 30 * 60 s = 1800 s ).
A = I * U * t = 20 * 220 * 1800 = 7920000 J = 7.92 MJ.
According to the Joule-Lenz law, the amount of heat that will be released on the winding:
Q = I ^ 2 * R * t, where R is the winding resistance (R = 0.75 Ohm).
Q = I ^ 2 * R * t = (20 ^ 2) * 0.75 * 1800 = 540,000 J = 540 kJ.
Answer: The work of the current strength is 7.92 MJ, heat of 540 kJ will be released on the winding.



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