# How can I determine the maximum wattage that I can hook up to a breaker?

All of my heating is on one breaker, 240V on a double pole breaker, each marked 20A (it's a small condo). I wonder how I can determine the maximum wattage I can hook up to that breaker.

Power (Wattage) = Current x Voltage (P=I*V)

20A X 240V = 4800W

• If a double pole breaker is marked as 20A then its 20A. Its not doubled because its a double pole.
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Commented Jan 25, 2011 at 2:33
• That formula is for DC. For AC you also need to consider power factor, although for heating alone it's usually not a real concern. Just be aware that as a general rule it's not the correct formula for AC. Commented Jan 25, 2011 at 8:00
• +1, Although that will give an estimate only - the breaker will not have exactly 20A cutoff, just very close to that, so it might happen that it breaks the circuit on slightly lower or slightly higher current. Commented Jan 25, 2011 at 9:43
• not really 4800W. The reality it's that P=U*I and this gives you VA, in order to get W you need to multiply this with cosPhi, that you can take arround 0.8 to be safe (so it shoud be just a shade under 4000W) Commented Jan 25, 2011 at 11:26
• @sharptoorh: the circuitbreaker works at 20A without uninterupting the circuit for an unlimited ammount of time, but when you get over it will trip according to it's tripping characteristics. the higher the curret the faster it trips. so you could run let's say... 30A for a couple of minutes through it before it trips. Commented Jan 25, 2011 at 11:27

Calculating the total wattage using Ohm's law is quite simple.

`Watts = Volts (V or E) * Current (A or I)`

Therefore

`Watts = 240 Volts * 20 Amperes = 4800 Watts`

However, if you live in the US (and possibly Canada has similar rules) and follow National Electrical Code, you're not quite done yet. 424.3(B) says that fixed electric space-heating equipment shall be considered a continuous load. 210.19(A)(1) says that branch circuit conductors shall be sized at 100% of noncontinuous loads, plus 125% of continuous load. 210.20(A) says the same for overcurrent protection.

This all means that if you install a 4800 Watt heater, you'll actually have to increase the conductor and overcurrent protection. Instead of 12 AWG copper and 20 amperes breakers, you'll need 10 AWG and 30 ampere breakers. So if you want to use the 12 AWG copper conductors and 20 ampere breakers, you'll have to install a 3840 Watt or smaller heater.

`4800 Watts * 80% = 3840 Watts`

If you start getting in to inductive loads, you'll have to take power factor into account. But I think that's beyond the scope of this question.

In Canada and in the United States (If i am not mistaken) total wattage is (A x W) x 80%.This goes to say, a double pole bridged 20amp breaker would be consider 40AMP. So 40AMPS x 120(Or corresponding voltage) x 80% = 4800 X .80 = 3840 would be the maximum voltage.

In order to find out how many fixtures you can put on that circuit, divide the total save load by the wattage of the fixtures.

• I don't even know where to begin, this answer is just wrong. Commented Oct 22, 2014 at 11:01
• In all of the universe, Wattage (W) is Voltage (V) * Current (A or I). Nowhere in the universe does a 20 ampere breaker give you 40 amperes. In the US (and maybe Canada), branch-circuit conductors and overcurrent devices for fixed electric space-heating equipment must have an ampacity not less than 125% of the total heating load. So the calculation should be `240 Volts * 20 Amperes * 80% = 4800 * 80% = 3840 Watts`. Commented Oct 22, 2014 at 11:15