Building electrical low voltage electrical equipment current voltage and power problems

First of all, understand the relationship between voltage, current and power in AC circuits.

In AC circuits, because of the relationship between capacitance and inductance, voltage and current do not change at the same time, and there is an angle between them

Assuming that in a sinusoidal steady-state circuit, the maximum voltage is um and the maximum current is im, then the voltage rms value U=um 2 and the RMS current value i=IM 2

Its average power p=ui*cos, unit w;

Reactive power q=ui*sin, unit: var

Apparent power s=ui= (p 2 + q 2), unit va

Generally, the nameplate of our electrical is marked with apparent power, so generally you can see that the expression of power is generally ***va, not ***w, usually the civil voltage value we call is generally also the effective value of voltage, and single-phase 380V refers to the effective voltage value of U=380V, the maximum value of UM=380*2, so:

1. Single-phase 380V, 60A equipment, the effective voltage value U=380V, the effective current value 60A, the apparent power S=UI=380*60=22800 VA (the power of general electrical equipment is expressed in this way).

Average power p = ui * cos = 22800 cos w

Reactive power q=ui*sin =22800sin var

And so on.

Supplement: three-phase power supply line, the three-phase voltage in the general electrician refers to the line voltage in the three-phase line (the voltage between phases), and the phase voltage refers to the voltage at each relative neutral point.

Let the RMS voltage of the three-phase line line be u, then the relationship is .

Three-phase line phase voltage U phase = U line 3

The apparent power of a three-phase consumer can still be calculated according to the definition of apparent power.

The problem is low-level, go to the textbook to go to school, there are example questions, others write formulas for you here, no matter how detailed it is, it can't compare to you reading a book!

Your question is not comprehensive, first of all, you have to determine what kind of load you have (inductive, resistive), and you don't say AC or DC.

How to calculate the current of a 380V, 6KW device? It is best to have a formula.

How to calculate the power of single-phase 380V, 60A equipment? It is best to have a formula.

How to calculate the power of three-phase 220V and current 40A equipment? It's best to have a formula (this question is important.)

Three-phase 220V, power 20KW, what is the current? It is best to have a formula.

Single-phase 220V, power 10kw, what is the current?

What is the three-phase four-wire 220V, power, and current? There are formulas most.

In the power supply system, there is no total current of three-phase electricity, and you generally don't need to care. When the three-phase load is equal, the three-phase currents add up to 0A

Phase A 10a, phase B 12a, phase C 13a

Then I will calculate by 13 and select the square line of three separate loads The formula for calculating the power of each phase three-phase motor is: p=. From the power formula of the three-phase motor, the line current formula can be deduced: i=p

where: p is the motor power.

U is the line voltage, which is generally 380V

COS is the motor power factor, which is generally taken.

Line current for your 45kw motor:

i=p Because the starting current of the motor is very large, it is 4 to 7 times of the working current, so the starting current of the motor should also be considered, but the time of the starting current is not very long, and it is generally only considered according to the coefficient when selecting the wire. If you take this, then the current is 137A, and you can select wires, air openings, contactors, thermal relays and other equipment according to this current. Therefore, the step of calculating the current cannot be omitted.

What is the minimum power factor of electrical equipment? High and low pressure.

Hello, the industrial users of high-voltage power supply and the power users of high-voltage power supply equipped with medium blind rent load adjustment power sales megavoltage device, the power factor is 0 90 or more; Other power users of 100 kVA (kW) and above and large and medium-sized power drainage and irrigation stations, with a power factor of 0 85 or more; Wholesale and agricultural electricity with a power factor of 0 80. If the power factor cannot meet the above requirements, the power supply bureau may refuse to connect the electricity. Existing users who do not meet the above requirements should be in the .

Within two or three years, reactive power compensation equipment will be added to meet the above requirements. For users who do not add reactive power compensation equipment for a long time and do not state the reason, the power supply bureau may stop or restrict the power supply. The power supply bureau should supervise and help users to take measures to improve the power factor.

The power factor adjustment electricity fee shall be implemented in accordance with the provisions of the "Electricity and Heat" approved by the state.