To answer this question the voltage must be known. Watts = Amps x Volts.
The equation that you are looking for is I = E/R. Amps = 120/20 = 6 amps. This is one of the basic equations of Ohm's Law.
in one ton that is 8amps more or less for the full load amprs. for no load amprs around 4amprs
The formula you are looking for is Watts = Amps x Volts. Amps = Watts/Volts. This comes to 4 amps load. Minimum size fuse would be 5 amps.
One can purchase a load balancing switch at many retail locations, both in person and online. Examples of such locations that offer the product online include Amazon and eBay.
One 150 watt inverter reports 0.2 Amps (=approx 2.4watts) Another inverter (180 watts?) reported 0.4 Amps If you find a fairly complete spec sheet, it may tell you the Amps or Watts that it uses under "No Load". I'm the originator of the question, and I discovered the "additional" specs.
It is the terminology given to differentiate the dimmer switch from a three way dimmer switch. It is a replacement for an ordinary lighting load switch when you want to dim the lighting load for aesthetic value.
To wire an up-down switch, you would typically connect the power source to one terminal of the switch and the load (such as a motor) to the other terminal. When the switch is in the up position, it sends power to the load to move it in one direction. When the switch is in the down position, it reverses the polarity of the power to the load to move it in the opposite direction. Make sure to follow the wiring diagram provided with the switch to ensure proper installation.
Watts = volts x amps x Power Factor. Assume a PF of one for a resistive load. Wall outlets in US are typically 120 volts. 240 / 120 = 2 amps. At 80 % rated load a typical 20 amp circuit souls handle 16 amps. Therefore, 8 wall outlets each with a single 240 watt device.
If you have a 200 amp main breaker, it will trip when any one of three phases exceeds 200 amps. This does not give you 600 amps total, only 200 amps total, but on each of three phases. Your three phase machinery and equipment will likely load all phases evenly, and a good electrical design will load your lights and receptacles evenly on all three phases. Doing this reduces the possibility that two phases are loaded at 100 amps, but one is loaded at 205 amps, so the main trips for example.
481 amps if the load has a power factor of 1, but if the power factor is less than one (e.g. if it's a motor) you also have to divide by the power factor.
When you have a 110 v or 220 v supply, you can draw some amps by putting a load on. If the load draws little power, like a 40 watt lamp, the current is small. If the load is a heavy one like a 1500 watt convector, there is more current. To find the current, just divide the watts by the volts using a calculator.
Insufficient data. Current drawn depends on voltage applied and resistance/impedance of the load. You can't apply 5.5 Amps unless you do that from a current source (or from too high a voltage) and it's unlikely that you have one unless you are in an electrical/electronics lab and know what you are talking about. On the other hand, if you manage to stuff 5.5 Amps through a device that is only rated at 5 Amps, then you are forcing too much power into it. It will be electrically stressed and it may (or may not) fail. Back to first statement - insufficient data.