You need to specify the rating , voltage and phases of the motor to answer this question.
P=1.732*415*0.8*i
=38.9
7.5 amps
The ratio is between 30% to 40 %
Ohm's law: voltage is current times resistance 0.03 amperes times 1000 ohms = 30 volts.
It would probably be 20-30 amps. <<>> The electrical code book states a value of 18.7amps for a 3HP 208 volt motor. If the nameplate data is available it is always best to use it when calculating wire size and overload protection for the motor.
A delta connection should cause a 30 degree phase shift. It is necessary to know the configuration to tell whether this is +30 or -30 degrees.Also, if you invert the connection, you can end up with [-30 + 180] 150 or [30 + 180] 210 degree phase shifting.I've seen one transformer that had a delta connection that was phase shifted 150 degrees, but this is definitely NOT the norm.CommentSince the current is determined by the load, the answer to this question is that it depends upona. the nature of the load (resistive, R-L, R-C, R-L-C), andb. whether the load is balanced or unbalanced.In other words, it depends....
Electrical wire size is directly dependant on the load amperage that is connected to it. The higher the load amperage, the larger the cross sectional area of the wire needs to be. The connected amperage to a conductor is determined by a group of electrical experts and their results are then written into the latest addition of the electrical code book of the country in which you live.
The ratio is between 30% to 40 %
The starting current of any electrical equipment which is 6 times more than that of the full load current. ex: full load current = 5 A the starting current will be 5*6 = 30 A.
if you know the gauge of the wire you can learn its current carrying capacity
Ohm's law: voltage is current times resistance 0.03 amperes times 1000 ohms = 30 volts.
If, Ct value = 50 meter unit = 30 so, 50 x 30 = 1500 kwh
Minimum current would be 10000 divided by 240 but it might be up to 30% more if the load has a poor power factor.
This is for 30 day accounts... 100% of total debtors ledger less current * 90%
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The maximum load on a 30 amp breaker is 30 amps at 40 degrees C. If the ambient temperature is higher that 40 degrees C then the breaker will trip at a lower current. Keep in mind the breaker is a dual function device. On a short circuit the magnetic component will trip the breaker instantaneously. If you are designing a branch circuit for a load that is constantly on, code may require you to "derate" the branch by 20 percent so that the ordinary load is not at the rated load of the overcurrent protection.
The load current must not exceed 30 amps. That's the safe limit.
First, a fuse with a current rating, 30 amp, is designed to allow current to flow thru the circuit or equipment or part of your car that uses or requires 30 amps and actually a bit less. If the current in the circuit, wires etc exceeds 30 amps, the fuse will open (a metal strip that burns some call it blown but it is actually an open so the excess current cannot damage the circuit or wiring it is associated with. Now, there are basic formulas you can use to calculate wattage (actually Power P) which would be the power rating let's say for the equipment you may be asking about. If your circuit is designed to use say 30 amps and it's in a car (12volt battery) the general formula would be P=IV=30x12=360w. If the equipment has a power rating and you have 12v then I (current)=P/V. The equipmet could be considered a Load (resistance R), you can calculate the current that would be drawn or used in the circuit by P=V2/R (that's V squared not x2) or P=I2R(I squared, not times 2). You can transpose ohms law to find R (the Load) Where V=IR so R=V/I. That's the best I can do not knowing more.
It would probably be 20-30 amps. <<>> The electrical code book states a value of 18.7amps for a 3HP 208 volt motor. If the nameplate data is available it is always best to use it when calculating wire size and overload protection for the motor.