In an open circuit test full load current does not flow, hence you wont get copper loss.
Because a short-circuit test is done at low voltage so there is very small power loss in the magnetic core. That is because there is very magnetic flux.
copper losses Comment: Advantages: Their operation is silent They have a high efficiency They are relatively safe Disadvantages Expensive Copper losses and hysteress losses
In an alternator, the load current is supplied by the stator and the excitation is applied to the rotor. When the power factor is low (lagging), more excitation is required to maintain rated output voltage at rated current. More excitation is also required to maintain rated output voltage with increased output current. Increased excitation current means increased rotor losses that must be dissipated as heat. (akash)
A transformer's excitation current can be resolved into two components. The first is in phase with the primary voltage, and is responsible for the losses. The second lags the supply voltage by 90 degrees, and is responsible for magnetising the core.
The input power, Pin, is reduced by different loss sources in the system. These reductions are the difference between input power & output power. The losses are: PSCL: Stator copper losses, or I2R losses Pcore: Core losses PRCL: Rotor copper losses PF&W: Friction & windage losses Pmisc: miscellaneous losses All of these losses reduce the input power. The output power is the input power minus all of the losses. Pout = Pin - PSCL - Pcore - PRCL - PF&W - Pmisc
Because a short-circuit test is done at low voltage so there is very small power loss in the magnetic core. That is because there is very magnetic flux.
how to reduce copper losses in a transformer Copper losses are due to the resistance of the copper (or aluminum) windings. To reduce copper losses the transformer would have to be rewound with heavier gage wire.
types of magnetic losses
copper losses are power losses due to flow of current in the wires or resistances,if the resistance is R, current is I then copper losses are I2R. for a 3-phase system; copper losses are same but for a single line, total losses are 3I2R.
Copper is not something that is calculated. The amount of copper might be, or copper losses / load losses, might be, but "copper" is not calculated.
Copper losses are also referred to as I^2 R losses. Copper loss is due to heating due to the current passing through the copper windings.
Copper losses are purely voltage-drop losses (I squared R) caused by the resistance of the windings, as opposed to hysteresis losses and eddy current losses (so-called iron losses), which are magnetic in nature. They are called copper losses whether the winding conductors are made of copper or not, by the way.
Power transformers have both no load and full load losses. The key is copper wiring, as copper varies with the square inches of the secondary and primary currents.
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)
The field excitation could have been lost. Check the output from the voltage regulator.
Copper losses are energy losses from the windings, due to the currents passing through them. During an open-circuit test, there is no secondary current (so no secondary copper losses) and the primary current is very low (so the primary copper losses are minimum).
stray losses,armature copper losses,iron losses(Hysteresis and eddy current losses),mechanical losses(friction and windage losses)