The three voltages of one equipment are 380V, the load current of the main switch is 160A, the rated capacity of the equipment is 65kva, and the power factor is about 0.95 The three voltage of a device is 380V, the main switch of load current is 160A, the rated capacity of the device is 65kva, and the power factor is about 0.95, Can this 160A be used as the total current?

The three voltages of one equipment are 380V, the load current of the main switch is 160A, the rated capacity of the equipment is 65kva, and the power factor is about 0.95 The three voltage of a device is 380V, the main switch of load current is 160A, the rated capacity of the device is 65kva, and the power factor is about 0.95, Can this 160A be used as the total current?

160A should be the maximum current
The useful power should be: 0.95 of 65000w = 61750w

The rated voltage of a motor is 220 v. the coil resistance R = 0.5. When the motor is in normal operation, the resistance of the motor coil is 4a. When the motor is in normal operation for 10 minutes, calculate the electric energy consumed by 1, the heat generated by 2, and the mechanical energy output by 3

1.
Power on time: T = 10min = 600s
Power consumption: w = uit = 220V × 0.5A × 600s = 66000j
two
Heat generation: q = i2rt = (4a) 2 × 0.5 ohm × 600s = 4800j
three
Output mechanical energy: W output = w-q = 66000j-4800j = 61200j

The rated voltage of a DC motor is 220 V and the rated power is 5.5 kW. How big is the current when it works normally? What is the power consumption for 2 hours continuous operation?

(1) When the motor works normally, the current I = Pu = 5.5 × 103w220 v = 25A; (2) the electric energy consumed by the motor for 2 hours is w = Pt = 5.5kw × 2H = 11kw · h. answer: the current of the motor is 25A when it works normally; the electric energy consumed by the motor for 2 hours is 11kw · H

The three-phase asynchronous motor is star connected and connected to the three-phase power supply with symmetrical line voltage of 380V. If the motor is running at rated power, the resistance of each phase is 6 ohm and the inductive reactance is 8 ohm, the phase current flowing into each phase winding of the motor and the current of each line are calculated

Because the load is star connected, phase voltage = 1 / line voltage = 380 / √ 3 = 220. --- phase current = 220 / impedance of each phase is √ (6 ^ 2 + 8 ^ 2) = 220 / 10 = 22
Because the load is star connected, phase current of load = line current of load, phase I = line I = 22