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The maximum power of a square power supply can be over how much current, and how much power. 2.5 square wires, for example. How many wires should be used in the construction of the project.

For 1.5, 2.5, 4, 6, and 10mm2 wires, the number of the cross section can be multiplied by 5 times. 7 x! G2?! Q0 h%]

For 16 and 25mm2 wires, the number of cross sections can be multiplied by 4 times.

For 35 and 50mm2 wires, the number of cross sections can be multiplied by 3 times.

For 70 and 95mm2 wires, the number of cross sections can be multiplied by 2.5 times. 9 D. o- F$O% U. / O

For wire 120, 150 and 185mm2 can be multiplied by 2 times the cross-sectional area of 2 h& o: V7 G. R/ _. an F

The working temperature of 30 DEG C, 90% load under long-term continuous flow is as follows: W K. K'y: L# `3 # P, y

1.5 square millimeters -- 18A

2.5 mm - 26A) v& O'C8 o _2 x0 / I Q4)

4 square millimeters -- 26A* L; H. G8 D. E "H

6 square millimeters -- 47A3 M5 v& V "s; {0 P/ T" b${, E (D6 Y)

10 square millimeters -- 66A

16 mm - 92A (|; Y "s" Z4 u T7 p- s* O!

25 square millimeters -- 120A

Square millimeter -- 150A

Power P= voltage U x current I = 220 volts 18 An = 3960 watts

The load current value (part) of the electric wire specified by the standard GB4706.1-1992/1998

Copper core wire:.. copper core line intercepting area.. permissible long current..2.5 square millimeter (16A to 25A).. 8 V3 k& ^6 V9 A; J1 L

4 mm (25A ~ 32A)..6 mm (32A ~ 40A) + a$t- B7 B (}2 |, A: G8 W

Aluminum wire, aluminum core wire cross-sectional area. Allow long-term current..2.5 mm (13A ~ 20A). K j'~3: D9 u% H3 {_ _

4 square millimeter (20A to 25A).. 6 square millimeter (25A to 32A) + S: Z! F9? 'V& H

S: U1 k/ U6 o (z)

For example: "B7 / / / / / / / J3 D- M# F7 n; W

1, the power consumption of each computer is about 200 ~ 300W (about 1 to 1.5A), so 10 computers will need a 2.5 mm2 copper core wire to supply electricity, otherwise, there may be a fire.

2. 3 large air conditioners consume about 3000W (about 14A), and then 1 air conditioners need a separate 2.5 square millimeter copper wire. C8 u u S1 a/ x# # B

3, now housing into the line is generally 4 square mm copper cable, therefore, at the same time open the household appliances shall not exceed 25A (5500 watts), it will be housing the wires replaced 6 square mm copper cable is useless, because the wire into the meter is 4 square millimeter. *}9 {1 I (W6 J* X0] "c& \ & U

4. The early housing (15 years ago) is usually a 2.5 square millimeter aluminum wire, so the home appliance that is opened at the same time must not exceed 13A (2800 watts).

5, the relatively large power consumption of household appliances are: air conditioning 5A (1.2), electric water heater 10A, microwave oven 4A, rice cooker 4A, dishwasher 8A, washing machine 10A with drying function, electric water heater 4A% K% u9 l: Z$V (HP!}.

90% of the fire caused by the power is caused by the heating of the joints. Therefore, all joints must be welded. The contactor can not be welded for 5~10 years, such as sockets, air switches and so on. L+ M$A7 N0 _2 L8 C! B; E& H

And. Long term current allowed by the national standard

4 square is 25-32A

The square is 32-40A

7}5 D W8 a+ L9 "N's& D* M |.

In fact, these are the theory of safety value is greater than the limit value of $a/ x1 K/ W K$l+ g2,5; the maximum power of the square wire allows the use of 8000W 5500W. is: 4 square, 6 square 9000W no problem.40A digital meter normal 9000W is absolutely no problem. The mechanical 12000W will not be burned.

Copper core cable ampacity standard cable ampacity lips: # H* p& K1 V2 Y

The estimation formula:

Two point five times nine, go up and go down a cis number. P0 X- z+ N+ u# q$"

Thirty-five multiplied by three point five, and pairs of pairs are reduced by five. P4 P$u t. ^7 S8) |; J

Conditions are converted and converted, high temperature ten percent off copper upgrades. ) y! X. K)? * n$M'C9 E

Intubation root number two three four, 87 forty percent off full flow. 4 G! S) B "T/ J