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If the transformer uses 5 watts per hour you need to know what you are paying per 1000 watts from your power company. If you pay lets say $3.00 for 1000 watts then when your transformer burns 1000 watts it cost you $3.00 your cost will be $3.00 for 200 hours run time.
If by "consume" you mean "waste as heat", that would depend upon the design of the transformer, but would typically be a few watts of heat loss.
Enough to make your mother’s panties drop on the floor.
No, the rating of the transformer, in watts, is the maximum amount of energy that can be safely drawn from the device. Any wattage load up to that limit is safe to connect as long as the voltage is correct to the load.
A transformer does not use, it transforms voltage from one value to another. The output amperage is governed by the connected load. If the load wattage is higher than the wattage rating of the transformer then either the primary or secondary fuse will blow or the transformer will burn up if the fusing is of the wrong sizing. The maximum primary amperage can be found by using the following equation, Amps = Watts/Volts, A = W/E = 600/120 = 5 amps. The same equation is used for the calculating the maximum secondary amperage, A = W/E = 600/12 = 50 amps.
as much as your mum eats grapes
If the transformer uses 5 watts per hour you need to know what you are paying per 1000 watts from your power company. If you pay lets say $3.00 for 1000 watts then when your transformer burns 1000 watts it cost you $3.00 your cost will be $3.00 for 200 hours run time.
If by "consume" you mean "waste as heat", that would depend upon the design of the transformer, but would typically be a few watts of heat loss.
The correct symbol for kilovolt amperes is 'kV.A, not kva. A volt ampere is the product of the transformer's secondary rated voltage and its rated current. It is not rated in watts, because the transformer designer has no idea what sort of load is to be applied to the transformer, and it is the load that determines the amount of watts, not the transformer.
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The formula you are looking for is I = W/E. Amps = Watts/Volts.
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.
A transformer has a rating that is usually expressed in KVA. This is approximately a wattage rating. It is not dangerous but it can be the cause of some concern. An appliance has a set current that is draws. This current times the voltage is the appliance's wattage. The same goes for the transformer. It only has a certain capacity to supply a specific current that is governed by its KVA (watts). Driving the transformer beyond its rated capacity tends to heat the transformer beyond its working temperature. If left in this over current draw the transformer's windings insulation will break down and the windings will short circuit. This is usually the end of a working transformer. So short answer, more watts (amps) from appliance equals burned out 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).
Enough to make your mother’s panties drop on the floor.
Volts per hour is an invalid statement. You may have meant Watts per Hour.