transformer rating is not measured based on current..it is also caliculated based on voltage.............ther main losses in a transformer are copper losses and core losses........cu losses mainly depend on voltage and core losses mainly depend on current..........thats why we measure the rating of transformer in KVA(kilo volt ampere)..................
I use a chart in the back of a High school automotives text in my garage.
I recommend you use the following google search:ELECTRICAL WIRE AMP CHARTThis will take you to several sights with good charts and articles on both AC and Automotive wiring.Current doesn't 'pass through' a transformer. There are two currents: a load (or secondary) current and a primary current. To measure these you would have to use two ammeters: one in the primary circuit and the other in the secondary circuit. If we are dealing with high voltages, then we would need to insert current Transformers into the primary and secondary circuits, and use them to supply our ammeters.
You work back from the load amperage. Load amperage x load voltage will give you VA. Then you find a transformer that has the primary and secondary voltages that you need to use. The rating size will be the VA, or KVA depending on the size of the load, that you calculated from above.
A transformer is rated in terms of its secondary apparent power -that is the product of its secondary rated voltage and rated current.
It really depends on the rating you're wishing to test. there's a full load amperage rating, a voltage withstand rating, steady state voltage ratings, etc.
Circuit breakers are used to protect the conductor that is connected to it. The sizing of the conductor is based on the current of the load. As the connected load current is increased so must the wire size to accommodate that higher current. Therefore as the wire size increases so must the breaker size to accommodate the higher load current.
The equivalent mm2 cross-sectional area of a 5.26 mm2 conductor is a # 10 AWG conductor. A # 10 AWG conductor size is protected by a 30 amp fuse.
It's not the voltage that determine the size of the conductor,it's the current. The rule of thumb is: 6A for every 1mm sq
you need the current of motor or the KW/HP rating
Without knowing the nominal voltage rating, who can tell? <<>> The sizing of cables or conductors is based on the amperage that the conductor can safely carry. The formula for amperage is I = W/E. Amps = 12000/Volts. With out a voltage stated an answer to this question can not be answered. When you calculate the amperage, re-ask your question for a conductor size for a given amperage.
Circuit breakers are used to protect the conductor that is connected to it. The sizing of the conductor is based on the current of the load. As the connected load current is increased so must the wire size to accommodate that higher current. Therefore as the wire size increases so must the breaker size to accommodate the higher load current.
The size of the conductor is in direct relation ship to its rating capacity of carrying a current. The larger the diameter of the conductor the larger the amperage rating capacity of the conductor.
for DC load and if its copper wire the cross section of the conductor wire is generally calculated as 1/4 th of the current rating . Eg for a DC load of 16 amp the copper conductor with 4 mm square is selected.
Because neutral doesn't have to carry the load current .This is either used for unbalanced current (in Y-connection) or for earthing purpose which don't require high ampere rating....
A breaker is sized by the load current. A breaker is used to protect the load feeder from an over current being applied to the conductor. If the load draws up to 30 amps, the wire size needs to be a #10 copper conductor or larger depending on the distance to the load with an insulation rating of 90 degrees C. The 460 volt source has to do with the insulation rating of the conductor. In this case the #10 conductor will need an insulation rating of 600 volts. Do not use 300 volt rated wire.
Fuse sizing is calculated by the size of the conductor. The conductor is calculated by the amperage to the load. The above answer is partially right, however in calculating fuse sizes for say a motor load of 10 hp on 208 volts 3 phase. The table in 430.250 list this motor as pulling 30.8 amps. Conductor size would be 30.8 times 1.25 or 38.5 amps. This value would lead you to wire table 310.15(B)(16) and a conductor size of number 8THWN. This conductor is listed for 50 ampacity, however due to the starting current of the motor. A fuse could be sized at 30.8 times 1.75 or 53.9 amps. The next standard size would be 60 amps. the fuse size is above what the conductor is rated for, but because of motor overload protection the conductor is protected. If breakers were used instead of fuses, a breaker would be sized at 30.8 times 2.5 or 77 amps or 80 amp next standard size.
A #14 copper conductor is rated for 15 amps. A #12 copper conductor is rated for 20 amps.
i=v/R and R=rho.L/A . by analysing these fourmulas we find the size . here i is electric current in the conductor to flow or load current to flow . v is the voltage across the conductor , R is the resistance of the conductor . L is the length of the conductor ,rho is a constant of a material called specific resistance here the the material is the material of the conductor , A is the area of cross section of the conductor .
A #1 copper conductor with an insulation rating of 90 degrees C is rated at 145 amps.
To answer a current carrying capacity question the size of the wire has to be stated.
The amount of current flowing through a conductor is governed by the amperage of the connected load. This is why there are different sizes of conductors. Each conductor size is only allowed a specific amount of amperage to flow through it. If the amperage load is higher that what the conductor is rated for then the next larger size conductor has to be used. The limiting of the amperage to specific size conductors keeps the conductor from heating beyond the conductors specifications, under full amperage conditions.
The equivalent mm2 cross-sectional area of a 5.26 mm2 conductor is a # 10 AWG conductor. A # 10 AWG conductor size is protected by a 30 amp fuse.