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Usually for this type of installation there is a control transformer installed in the starter enclosure. The primary of the transformer is the motors line voltage and the secondary is the control voltage which is usually 120 volts. In Canada the control transformer needs fusing on the secondary side of the transformer. In the US the transformer needs fusing on the primary and secondary sides of the transformer. The secondary side of the transformer's terminals are labeled X1 and X2. The X2 lead is grounded to the starter's enclosure. From this point out the wire is then termed the circuit's neutral wire. This wire then connects to one side of the magnetic starters draw in coil. Yes, 120 volt coil would require a neutral, this could be derived from the control transformer if it has one or from an external source.
Inductors are considered to be a load for reactive power, meaning that they will draw reactive power from the system. Capacitors are considered to be sourced of reactive power, they feed reactive power into the system. If you have a circuit that is at unity (balanced with inductors and capacitors) no reactive power will be drawn from the source. You will have unity power factor. If your circuit is more inductive than capacitive it will be drawing reactive power from the source. The opposite is also true for capacitors.
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?
No, not directly. The supply voltage has to rise or the resistance has to fall to get over-current. If there was a secondary control voltage that was part of a voltage control circuit for a higher voltage, it is conceivable that a voltage drop in control circuit could cause an over-voltage in the supply. Motors are constant power devices, so this could be true for a motor. If you have a 1hp motor (loaded at 1hp), it will want to draw 1hp of power no matter the supply voltage. If the voltage dips, the motor will require more current to keep it spinning at it's normal speed.
Resistance ideal transformer is the one having no core losses, infinite permeability no mmf needed to set up flux), windings are having no resistances or reactances.
Usually for this type of installation there is a control transformer installed in the starter enclosure. The primary of the transformer is the motors line voltage and the secondary is the control voltage which is usually 120 volts. In Canada the control transformer needs fusing on the secondary side of the transformer. In the US the transformer needs fusing on the primary and secondary sides of the transformer. The secondary side of the transformer's terminals are labeled X1 and X2. The X2 lead is grounded to the starter's enclosure. From this point out the wire is then termed the circuit's neutral wire. This wire then connects to one side of the magnetic starters draw in coil. Yes, 120 volt coil would require a neutral, this could be derived from the control transformer if it has one or from an external source.
It depends on the load. A good transformer has over 90% (some as high as 99%) efficiency. So the power drawn by it is a function of the power in the load, plus a small amount due to losses in the transformer.
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Yes, use a ladder diagram.
Yes, a plugged in transformer uses power with no load on it.No transformer (or any electrical device, for that matter) is ideal, so there are always losses. There is parasitic capacitance, etc. that leads to power draw, particularly between the laminations. Plus, while the laminations are supposed to be insulated from each other, there is leakage current, and this causes power draw due to the tertiary transformer winding, that is partially shorted, which the laminations represent.Another AnswerThis simple answer is practically none.The primary winding of a transformer draws a very small current (
Depending on your application... Use a voltage dividor (two resistors) if it doesn't draw a lot of current. You could also try a voltage regulator or using an op amp as a buffer, with the input the voltage you want. That voltage could come from a voltage divider, since op amps draw extremely low current.
A: If the transformer is connected to a power input of course it will draw current. The primary is a long wire it has own resistance wrap around an iron core. Of course there will be primary current whether there is a load on the secondary or not.
Yes. The Va numbers indicate the rating of the transformer (12VA indicates that you can draw upto 12/24 = 1/2 Amps at 24 volts and 20VA indicates that you can draw about 20/24 = 0.83 Amps). Usually 20va transformer will cost more than the 12va transformer. So, unless a higher power output is required, it is adequate to use the lower rating transformer. However, you cannot go the other way - if you use a 12va transformer in-lieu of the 20va one, if the application needs more power, then you have a reliability problem. The transformer will have more loss, heat-up and then eventually burn out.
True
Inductors are considered to be a load for reactive power, meaning that they will draw reactive power from the system. Capacitors are considered to be sourced of reactive power, they feed reactive power into the system. If you have a circuit that is at unity (balanced with inductors and capacitors) no reactive power will be drawn from the source. You will have unity power factor. If your circuit is more inductive than capacitive it will be drawing reactive power from the source. The opposite is also true for capacitors.
Watts are power. If the lights were mostly or totally switched off, you'd have a circuit generating 600W of heat somewhere if the transformer still took 600W, not only that, but when you switched on, the 600W that the transformer was consuming, would not disappear, so the total drain would be 1.2kW. ---- Don't understand the above answer. The 600 watts on the transformer nameplate is the maximum amount of wattage that the transformer can produce and still be within its safety limits. It doesn't draw that wattage all the time. If you had two 50 watt lamps connected to the transformer then the transformer has the capacity of 500 watts left. The transformer will only produce the wattage that the load requests. The transformer has the ability to supply twelve 50 watt bulbs. 12 x 50 = 600. Any more bulbs than 12 and the transformer is in an overload condition.
12/240, 0.05 amps, maybe about 30% more because of the poor power factor of small transformers.