The total load in watts would be W = A x V. 20 x 120 = 2400 watts. Any wattage higher than this will trip the breaker and shut the circuit off.
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The theoretical resistive load is V/I = 120/20 = 6 Ohms. The lower the resistance the higher the current. Usually you don't want to operate above the 80% point so the number would be 120/16 = 7.5 Ohms.
Question is incorrect. in a 240 Volt single phase circuit, how can you have A phase and B phase?
If you are talking about a 6 volt coil, yes, so long as the contacts are rated for the 230 volt circuit. If you are talking about 6 volt contacts, no, absolutely not.
It all depends on the load. The formula for calculating amps, volts or ohms (resistance of load) is E=IR, where E is the voltage, I is the current and R is the load or circuit resistance. So, if you know the resistance in ohms and the current in amps, you multiply them together to get the voltage of the circuit. Again, it depends on the load, so a 12 volt car battery can deliver 1.5 amps if the load is 8 ohms whereas a 120 volt circuit will deliver 1.5 amps if the load is 80 ohms. This is all simplified and is based on a resistive load. If the load is capacitive or inductive, then phase angles come into play and the math is more complicated using imaginary numbers and J-operators.
1 volt
A volt meter will do the job.
A 'volt ampere' (not 'volt amp'!) is the unit for theapparent power of a load in an a.c. circuit. It is simply the product of the supply voltage and the load current.
The current in a 220 volt circuit depends on the resistance of the load connected to it. Ohm's Law (I = V/R) states that current (I) is equal to voltage (V) divided by resistance (R). So, the current will vary based on the resistance of the circuit.
For a 220 volt circuit with a 40 amp load, the appropriate wire size would be 8 AWG (American Wire Gauge) copper wire.
Question is incorrect. in a 240 Volt single phase circuit, how can you have A phase and B phase?
A volt can not be connected to a circuit.
AWG # 10 wire on 30 amp circuit.
When you mean simple circuit, I assume a source of one volt across a load of one ohm, which, according to Ohm's Law, equals one amp.
resistance = volt / current . 440 volt across a parallel circuit means the same 440 volt across both resistance s. hence resistance r = volt / current . 440 / 20 amp = 27.5 ohms total resistance
Usually a volt meter is placed across a component to measure the voltage drop across that component. Doing this places the volt meter resistance in parallel with that component's resistance, which will always lower the total resistance. Since the volt meter resistance is usually very large relative to the resistance of the element being measured, the total resistance does not change significantly. The formula for total resistance of two parallel elements is: Rtot = (R1*R2)/(R1+R2), as R1 (the volt meter) >> R2, Rtot ~= (R1*R2) / (R1) = R2 If a volt meter is placed into a circuit instead of around an element of that circuit, it will raise the resistance of the circuit, load the circuit with, and interrupt "normal" operation of the circuit (normal operation = how things would be without the meter in place). More importantly, the volt meter would then be measuring the voltage developped across itself (instead of an element of the circuit), which is not the point of this tool / this would be a misapplication of a volt meter.
Series circuit? Add 'em!
LxW of building x 3 volt amps per sq. foot
LxW of building x 3 volt amps per sq. foot