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Why LV side is shorted in short circuit test?
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.
How transformer get burn?
Current overload from whatever circuit draws current from the transformer? Proper fusing of its supply might protect a transformer from this cause. Or it could have developed a shorted turn fault because the insulation on a winding got old and perished? Or maybe the transformer got damaged if the appliance it is mounted in was dropped?
How much to run a 75 KVA transformer with no load?
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.
What happens when a transformer is fed with dc voltage?
a transformer require AC to function as desired it transform the AC to different levels. DC on a transformer can only see the actual primary or secondary resistance if the source is not limited in current it will burn the transformer by excessive heating since it will see only wire resistance. Answer 2 the application of an AC voltage, V on one winding of the transformer produces alternating flux that links the entire core of the transformer. The changing flux induces an emf, E that opposes the main voltage. the current through the winding in this case is {(V-E)/R}; R= winding resistance in case of application of DC, since there is no changing flux, there is no induced emf and hence the current will be V/R since the resistance of the winding is very small, the current is very high and this can burn away the windings. hence DC is not used.
Will using a transformer to convert 230V to 12V DC consume more energy than using a few capacitors and resistors without a transformer?
no, the transformer is much more efficient <><><> HOWEVER- a transformer only changes the voltage of AC current- it does not change it to DC. You will still need diodes or rectifiers. You can also use a motor/generator- a 230v AC motor turns a 12 v DC generator.
Does led lighting require a transformer per light or one per circuit?
One transformer should feed several led's in parallel, but you have to work out how much current each one will take, total it up, and then get a transformer with the correct output voltage and that can supply the total current. Don't exceed the transformer max current or it may fail through overheating
Why LV side is shorted in short circuit test?
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.
Why the no-load current of an induction motor is high compared to a transformer?
Because of the presence of the air-gap between stator and rotor in motor the magnetization current is much higher than that of a transformer in addition to the friction and windage losses due to the rotation of the rotor.
How transformer get burn?
Current overload from whatever circuit draws current from the transformer? Proper fusing of its supply might protect a transformer from this cause. Or it could have developed a shorted turn fault because the insulation on a winding got old and perished? Or maybe the transformer got damaged if the appliance it is mounted in was dropped?
Why does the magnetization current impose an upper limit on the voltage applied to a transformer?
The magnetization current imposes an upper limit on the voltage applied to a transformer core due to the ability of the respective spark gap to extinguish. If too much current supply is available the spark gap becomes overheated. This makes it unable to "switch-off" when the high or elevated supply current is flowing through it.
Why does the magnetization current impose on upper limit on the voltage applied to a transformer core?
The magnetization current imposes an upper limit on the voltage applied to a transformer core due to the ability of the respective spark gap to extinguish. If too much current supply is available the spark gap becomes overheated. This makes it unable to "switch-off" when the high or elevated supply current is flowing through it.
Is there any possibility of making a Transformer function as a CT IF Yes how could it be made. Please explain clearly?
Theoretically, yes. But, in practise, probably not! The main thing to understand about a current transformer is that its primary winding is always connected in series with a circuit's load, rather than in parallel with its supply and, so, must be capable of handling the circuit's load current which will be very much larger than the primary current of a regular 'voltage' transformer.
How much current can a 50 volt ampere 36volt transformer deliver?
I (Amps) = VA / E (Volts) I = 50 / 36 I = 1.39A Do the math!
Is there any way to convert car led strips for wall outlets?
Probably a transformer that converts voltage from 120v AC to 12v DC. Commonly found at Radio Shack or an electronics store. Depending upon how much the lights draw for current, you have to find a transformer that will supply that current. Such as .5 amp or 2 amps or as much as 5 amps-----.
How much to run a 75 KVA transformer with no load?
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.
Does magnetizing a seed affect its growth?
No. How deep you plant it, how much water you give it upon planting it, and whether it is exposed to sunlight or not does.
What happens when a transformer is fed with dc voltage?
a transformer require AC to function as desired it transform the AC to different levels. DC on a transformer can only see the actual primary or secondary resistance if the source is not limited in current it will burn the transformer by excessive heating since it will see only wire resistance. Answer 2 the application of an AC voltage, V on one winding of the transformer produces alternating flux that links the entire core of the transformer. The changing flux induces an emf, E that opposes the main voltage. the current through the winding in this case is {(V-E)/R}; R= winding resistance in case of application of DC, since there is no changing flux, there is no induced emf and hence the current will be V/R since the resistance of the winding is very small, the current is very high and this can burn away the windings. hence DC is not used.
