What Wire Should I Use?

i will not hire an electrician because. I need to find it out. and if you can't answer my question? i think you better not to reply

Then try British Standard 7671 instead.

you have asked a very elementary question. This shows that you do not have the experience to be doing any electrical work. Yes, I could answer the question but I don't want you to install this incorrectly and cause a problem.

I admit i dont have exp. on it, that's why I ask anyone could answer this. So if you know this?then leave an answer.

What voltage; what length? If you can't answer, then neither can anyone else.

Want to know the gauge of conductor to carry 21 X 3800 W loads at some unknown voltage which may be a foot or ten miles away from source? Send more information and show a little more decorum.

at 220 volts

estimated lenght would be 25 mtrs. please show me the calculations. on how do you get the size of wire.

In the USA, wire gauges are specified by National Electrical Code or International Electrical Code, or some other such code.

Length is required to be considered for voltage drop from source to load. For 25 meters, it's not much of a factor.

3800 Watts @ 220 Volts yields ~18 Amps. AWG (American Wire Gauge) #12 is sufficient, meets code requirements, etc. #12 has a diameter of 0.030 inches = 0.78 mm, and it has a cross section of 6530 cir mils = 3.31 mm².

Total current for 21 X 3800W @ 220 volts = 363 amps. Mains conductor to panel feeding 21 units should be AWG 4/0 copper type THHN (or equivalent) or larger. Other insulation types may require larger diameters, such as type TW or UF (underground feeder0, which requires that the conductor be 300 kcmils.

AWG 4/0 diameter is 0.528 inches = 13.41 mm

300 kcmils diameter is 0.630 inches = 16 mm

These sizes do not represent any temperature compensation, etc., and data came from NEC 2002 code for copper conductor ampacity table, and it is for information only. This post does not represent any suggestion that the data herein contained is suitable for your application.

You wanted to know how to arrive at a number, so here it is.

I would agree with Wareagle. Unless you have experience - which apparently you don't - this is a dangerous idea that could cause shock, fire or death. Do yourself and family a favour by finding someone locally who knows what he's doing. There's a lot more to this then just wire size.

- what do you know about grounding?

- do you know how to wire up the receptacles?

- do you know what type of receptacles are to be used?

- what type of wire will you be using?

- how are the wires and receptacles attached to the walls?

- etc. etc.

Grae

At what voltage?

at 220 volts

And how long are these circuits?

Knock yourself out.

Do-it-Yourself Repairs & Basic Home Wiring Projects Articles

If you have just a regular building, use white wire. But, if you really have 21 rooms, that's different: use yellow.

The first calculation would be 3800 W ÷ 220 V ≈ 17.3 A. In the U.S. that would call for AWG 12 wire. Also you may need to check the voltage loss between the supply panel and the load, especially if the wiring runs are long.

It depends on local electrical standards and regulations which are specific to your country and state, it also depends on cable length, existing circuit protection rating (fuse/MCCB protecting the cable), etc. I am assuming this is relating to your previous question on your new apartment/house (but 21 rooms and only 5 appliances?)

http://cr4.globalspec.com/thread/47391/Can-You-Calculate-the-Wattage

What country are you in? Can you provide more information, otherwise all you will get from me is a reference to AS/NZS3000 (our Australasian wiring standard) giving 3 core flat TPS 1.5mm2 (or alternatively 2.5mm2) building cable.

http://www.generalcable.co.nz/newzealand/Products/Energy/TPS/1.1.5.3.6.1.pdf

yah this was from 21 rooms. And im from philippines. Coz assume in every room they ahave the same loads. And it was range to 3700watts to 3800 watts

One room with 3800KW - and 21 such rooms? What is it supposed to be? Even if I consider a hotel, then still 3.8KW per room is just a bit much.

With AC also one is likely to go for Central rather than distributed.

Could you describe the load type in this room? Let says: water heater, electric stove (4 coils, a bigger coil is 15 kW may not be include in this room), aircon, and lighting lamps.

I amperes = 3800/(0.8*1.732*220) = 21.59 Amperes. So, the main cable goes to each room distributin box (or meter panel) to supply 3800 watt is 4 mm2.

Note: Local rule must be followed.

Revision:

I amperes = 3800/(0.8*1.732*220) shall be I amperes = 3800/(0.8*220). Omitted 1.732.