Divide the VA by the voltage. But this transformer is probably a 3-phase one, maybe working on a 11,000 volt system, and in this case the live-to-neutral voltage is 6351 v and each phase supplies 3333 kVA. In this case the current supplied is 3,333,333 / 6351 or 525 amps on each phase. This is a common type of transformer in electricity supply.
Another Answer
Assuming you are talking about the secondary line current of a three-phase transformer, you simply divide the rated apparent power of the transformer by (1.732 x rated secondary line voltage).
Don't confuse line currents with phase currents. Line currents travel in the lines that connect the transformer with the load, whereas phase currents circulate within the windings.
For example, let's assume you have a 10 MVA primary distribution transformer, with a rated secondary line voltage of 11 kV. The full load current would be:
secondary line current = 10 000 000 / (1.732 x 11 000) = 525 A
It is the rated maximum current that can be taken from the transformer. This is equal to the VA rating divided by the output voltage. So a 6 kVA 240 v transformer would have a maximum current rating of 6000/240 or 25 amps.
No load current depends on the design of the transformer, and what voltage it is energized at. It will typically be below 1% of full load, and can be significantly below 1% for utility sized transformers.
The ratings state the limits on voltage and current for operating the transformer at full load. The rated voltage times the rated current gives the rated VA of the transformer. Transormers are not usually rated directly for power because this depends on the power factor of the load applied.
Winding copper losses of a transformer can be measured in a short circuit test of a transformer. Impedance voltage is given to the primary and the secondary is often shortcircuited. (some times the reverse is done of this). Full load currents are made to flow in both primary and secondary circuits. This current flow heats up the 2 windings of the transformer. Power consumed at this time gives the transformer copper losses.
typically the ratio is on the current transformer it will say something like 100:5 witch means 5a meter becomes 100A full scale the 5 instead of a 1 gives you a hint about the capacity of the transformer its 20:1 so you could use it to make a 1A meter into a 20A but making a 10A meter into a 200A may lead to problems (accuracy or power)
is it primary current ?
2 to 5% of full load current
CT=========current transformer PT=========potential transformer these are the instrumental transformers.
First find the ratio of the transformer. 6600/220 volts. Second find the secondary current, I = W/E, 99000/220. Third divide the secondary current by the transformer ratio. The answer will be the primary current. To check your answer (W (or VA) = V x A) multiply the primary current times the primary voltage and the secondary current times the secondary voltage and they should both equal the transformer's kVA.
By calculate do you mean calculate the connections required or phasor diagrams? full question would be helpful
It is the rated maximum current that can be taken from the transformer. This is equal to the VA rating divided by the output voltage. So a 6 kVA 240 v transformer would have a maximum current rating of 6000/240 or 25 amps.
No load current, in a transformer for example, is the current necessary for exciting the transformer. If you wish to keep it energized, and you need to keep it energized at full voltage, there is nothing you can do to reduce this other than replace the transformer with one that has lower no load current. If you are referring to a different piece of equipment, you may need to specify what you are meaning by "no load current".
No load current depends on the design of the transformer, and what voltage it is energized at. It will typically be below 1% of full load, and can be significantly below 1% for utility sized transformers.
The ratings state the limits on voltage and current for operating the transformer at full load. The rated voltage times the rated current gives the rated VA of the transformer. Transormers are not usually rated directly for power because this depends on the power factor of the load applied.
It depends on the rated voltage. Take 1600 KVA and divide by KV, and you will get A.
Transformer short circuit tests are used to determine the impedances (positive and zero sequence) of the transformer. A simple explanation: to do this one winding is shorted, and voltage is applied to another winding to circulate the normal full load current of the transformer. The impedance of the transformer is the applied voltage divided by the induced current. If one winding was not shorted, the voltage divided by induced current would not give the impedance of the transformer - the induced current would be much lower, giving a much higher impedance measurement that would be essentially meaningless.
A short-circuit test is done to determine the power lost in the resistance of the primary and secondary windings of the transformer. It is done at full load current but with only enough voltage to give the required current with the secondary short circuited. An open-circuit test is done at full load voltage but no current is taken from the secondary, and this enables the power lost in the magnetic core of the transformer to be measured. As well a power, the tests also allow the inductances to be measured as well as the resistances, in order ot characterise the transformer fully.