The copper losses, because they vary as the square of the secondary/primary currents.
To calculate the no load current from transformer & core loss is also calculated.
major component of power loss in a transformer is secondary resistance.when transformer is operated under no load,no current flows through the secondary.so under no load conditions transformer has just very small megnetic losses.
The no load losses are the losses caused by energizing the transformer. These are constant losses, regardless of loading. This in effect tells you the efficiency of the transformer. (Power in) - (no load losses) = (Power out)
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)
there are several losses in a transformer that prevent it from attaining 100% efficiency. One is core loss, which can be divided into Hysteresis losses, Eddy currents and Magnetostriction loses. see for more details http://en.wikipedia.org/wiki/Transformer#Energy_losses
Copper loss varies with the load.
the efficiency is maximum in a transformer when no load loss is equal to load loss.
To calculate the no load current from transformer & core loss is also calculated.
major component of power loss in a transformer is secondary resistance.when transformer is operated under no load,no current flows through the secondary.so under no load conditions transformer has just very small megnetic losses.
The no load losses are the losses caused by energizing the transformer. These are constant losses, regardless of loading. This in effect tells you the efficiency of the transformer. (Power in) - (no load losses) = (Power out)
A transformer is fundamentally a set of coils; therefore, a transformer is an inductive load. However, by "transformer load", you seem to mean "the load that is connected to a transformer". Whether that load is inductive or capacitive depends mostly on what is hooked up to the transformer.
That type of transformer normally has about 99% efficiency so the full-load loss would be 1% or 6 kW.
Yes because the transformer heating (power losses) depend on the load current and the load voltage. It can be assumed that the voltage stays more or less constant, therefore the iron loss is also constant. The copper loss depends on the square of the load current. So it is the VA of the load that determines the power loss and any heating.
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)
Hope this helpsAn "OFF-Load tap transformer" can only have it's tap adjusted when it is De-energized,while the "On-Load tap transformer" can adjust its tap under load conditions.Kind RegardsHammad KhanUniversity of Western AustraliaAnswerAn 'off load' transformer is one whose secondary is open circuited, and not supplying a load. An 'on load' (not 'load') transformer is one that is connected to a load.
there are several losses in a transformer that prevent it from attaining 100% efficiency. One is core loss, which can be divided into Hysteresis losses, Eddy currents and Magnetostriction loses. see for more details http://en.wikipedia.org/wiki/Transformer#Energy_losses
eddy current loss in the transformer core is reduced by