It depends on what you mean by '5 mva'. Because you haven't used the correct symbol, no-one will know whether you mean '5 millivolt amperes' or '5 megavolt amperes' -there's a HUGE difference! The correct symbol for 'millivolt ampere' is 'mV.A' and the correct symbol for 'megavolt ampere' is 'MV.A'. I hope you can now see the importance of using the correct symbol.
To answer your question, the load volt amperes should not exceed the transformer's volt amperes. So, in your case, the load shouldn't exceed 5 mV.A or 5 MV.A -whichever you mean.
To determine the amperage output of a transformer, you need to know the voltage it operates at. Assuming a standard voltage of 120 volts for a household transformer, you can use the formula Amperage = VA / Voltage. In this case, a 60-VA transformer operating at 120 volts would output 0.5 amperes (A) of current.
All magnetic materials require some energy before they will operate. You can think of it as the price of admission. In an electrical transformer, the excite current is actually a composite of the energy which is lost due to eddy currents in the magnetic material as well as hysteresis losses of the material. Excite current is both frequency and voltage dependent. Instead of saying, "before my husband will do anything I have to bake him a cake", you could say"before my transformer runs at 115Vinput, I have to supply x milliAmps of excite current just to get the core going." A special class of transformers known as CT's or Current Transformer's works in much the same way, except their voltage is reflected off the load instead of coming directly in through the primary. This energy is dissipated as heat. In power transformers, the concern is energy lost with no load connected on the output. In current transformers, the concern is the varying loss of accuracy due to the highly non-linear nature of excite current, especially at low excite levels.
Its a common card so its cheap. Go to eBay there should be dozens of new mint blue eyes for sale at any given time.
Depending on their physical size, turns ratio, core material, design frequency of operation, and power handling capability, transformers typically cost somewhere between pennies and thousands of dollars.
TOO MUCH VOLTAGE, INCLIMATE WEATHER (Snow, heavy Rain, ice etc) down powerlines and trees...something touches it. DO NOT TOUCH TRANSFORMER THOUGH!
It tells us how much is the transformer utilised in a given process. For a rectifier,TUF =(D.c.power delivered to the load)/(power rating of transformer secondary)
25 kV.A for a permanent load; more if the load is variable.
..the questions does not say wether the 30kVA transformer is 3 phase transformer or single phase transformer..but it is implying that a single phase welding load at 16A per phase is to be connect to it, it is assumed then that the transformer is 3 phase transformer..we assume load is rated 240V.. ..though not much details is given about the transformer voltage specs. but if is rated 3 phase 415/240V, the approx full load current per phase is given by 30KVAx1.3912=41.736A.. ..but if the transformer was rated single phase 240V say, the approx full load current FLC=30000/240V gives 125ampers.. ..if you further devide the FLC by the intended load current.. ..Recommended no of welders per for a 3 phase 415V transformer and assuming single welding sets =41.736/16=2.6..so you can connect max two welding sets to this transformer per phase,..max 6 weld sets can connect at an approx load factor of 76%.. ..for the single phase transformer, FLC/16=125/16, gives 7.8 but is also recommended not to connect more than 6 welding sets to such transformer for the same reasons..
It depends on the load
No because the load is 638 VA which is too much for the transformer.
The inductance of the transformer is much higher than the resistance of the transformer, resulting in very low real power losses (in watts), but some reactive power (vars).
75 kV.A is the rated apparent power of the transformer which it can supply, continuously, to a load without overheating. When the transformer is not supplying a load, the primary current is (a) very small and, (b) lagging the supply voltage by practically 90 electrical degrees. Bear in mind that energy losses only occur for the component of current that is actually in phase with the supply voltage. So the energy consumed to the transformer is very small and is due to the resistance of the primary winding (copper loss) and a relatively small loss in the core (iron loss). Just how much energy this accounts for and, therefore, how much it costs to run the off-load transformer, is not possible to tell without knowing the full specification of the transformer.
If the load current is too high, the power lost in the transformer windings will be too high and it will overheat. If the voltage is excessive, the power lost in eddy currents in the magnetic core will be too high and it will overheat.
A transformer has two coils coupled via the magnetic field, and when it has no load all you see is the inductance of the primary coil, which has to be fairly high to create the necessary amount of magnetic flux. When the transformer supplies a resistive load, the input looks much more resistive.
The current drawn by a transformer on no load is very small, typically less than 5% of full load current. For a 50 kVA transformer, this would be around 1-2 amps per phase. At full load, the current will depend on the efficiency of the transformer but could be around 15-20 amps per phase. There are losses in transformers, including copper losses (I^2R) and core losses, which lead to efficiency being less than 100%. The losses are generally small in well-designed transformers, but they do contribute to heating and decrease the overall efficiency.
Measuring No-LoadIn theory the no-load current of a transformer is zero. But in practice there is iron loss and core loss in the transformer, so there is power loss. Connect an ammeter in series with the stabilizer to measure the no-load current. Check your energy meter at no-load to see how much power is consumed. Ohms law: I(Amps) = E(voltage) divided by R(resistance). In the case of coils (transformer), the resistance of the coil would simply be the total impedance (Z). If I am remembering this correctly, you get, I=E/(R+Z)
The dc is supplied via a centre-tap on the transformer. This ensures zero dc net flux in the transformer core, which enables a much smaller core to be used.