# Calculating maximum wattage

If we get 220V from outlet, would it be possible to tell the maximum power (wattage)?

Since,

Power = Voltage (V) x Current (I)


Also, Is it possible to get the maximum amount from current (I) from 220V? Assuming any type of standard wire with resistance used?

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Syria, I think that you can learn things in a more organic way reading a book (or also from Wiki) that starts from the basic concepts. –  clabacchio Jan 22 '12 at 11:29
@clabacchio - note name = Sryia,noy Syria. Message will reach them as on end of question but may not if on an answer. –  Russell McMahon Jan 22 '12 at 15:34
@clabacchio - You called him/her "Syria". The name is Sryia. When you post a comment to a question the questioner sees your post reqardless of what you call them. If you post a comment to an answer they may not if you spell their name wrong. –  Russell McMahon Jan 23 '12 at 0:39
@clabacchio - (1) Yes - all ok in this case. (2) You cannot edit old comments because "the system" decrees so. ie the reason is "BECAUSE!!!!". –  Russell McMahon Jan 23 '12 at 11:03
(3) You can bypass the edit limitation to some extent by copying your old comment, deleting the old comment (which you can always do) and then starting a new comment, pasting the old text and then editing the text as desired. This has the disadvantage of losing the time priority of the comment. If there have been responses to your comment then the new one would be after them and so out of order. You can also leave the old one in place, copy it , paste to a new comment and say eg "Updating comment above ..." . –  Russell McMahon Jan 23 '12 at 11:03

Assuming the wiring is up to standards then you can multiply the voltage (220V) with the current written on the circuit breaker. You will get the apparent power, measured in VA. How much real power you get depends on the device (load) you are trying to connect - a purely resistive load (like a heater or an incandescent light bulb (not entirely resistive, but close enough) then the real power will be equal to the apparent power. Otherwise you have to multiply the apparent power with the power factor.

Real power = apparent power * PF

The power factor may be written on the device or it may not. PC power supplies without PFC (Power Factor Correction) have power factor around 0.6, power supplies with PFC have power factor around 0.99. Purely resistive load has power factor of 1, purely reactive (ideal inductor or capacitor) load has power factor of 0.

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One more thought on this: Circuit breakers usually will allow a certain amount of overcurrent over a specific time. Here in Germany, most circuit breakers are the "B16" type, which allow a (over-)current of 18-23A for about one hour until the thermal protection will trip them. To draw this amount of power in a safe way, I would use a room heater. I would not push the envelop with this kind of experiment, unless you have good fire insurance. –  0x6d64 Jan 22 '12 at 11:14

The amount of power or "Wattage" that you can get at 230 VAC depends almost completely on what the supply authorities allow you to get at a given point in the distribution system.

50,000 Watt single phase 230 VAC transformers are available in NZ.
Larger would be possible.

Your question may refer to power from a wall socket or from a phase coming into a domestic premises or to an industrial circuit or a pole transformer, but you did not make it clear which of these you intended.

A domestic or household power plug is often rated at 10 A = 230 x 10 = 2300 Watt nominal.

The circuit that it is fed from may have several 10A outlets on it and may be fused at say 25 a = 230 x 25 = 5750 Watt - say ~= 6 kW.

A domestic hot water hater fast recovery element may be rated at 6 kW or maybe even 10 kW.

A single phase 230 VAC supply to a house will have a "pile fuse" which limits the current able to be fed to the house. This may be in the 60 A - 150 A range = ~ 13 to 34 kW.

This company Transformer specialties in my country will provide transformers up to 50 kVA = 50000/230 = 217 A single phase 230 VAC.

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