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.
Calculating the total wattage using Ohm's law is quite simple.
Watts = Volts (V or E) * Current (A or I)
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.