100 is to 800 as 75 is to X. Cross multiply, 100 x X and 800 x 75, 100X = 60000. To get rid of the 100 divide it into both sides of the equation. X = 600.
If the copper loss is linear, at 75 percent the copper loss will be 600 watts.
The transformer can be tested on open and short circuit to find the iron losses and copper losses separately, which uses a fraction of the power than having to run the transformer on full-load.
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.
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
2 to 5% of full load current
No load current is excitation current, and is usually specified as a certain percent of base KVA rating at a specific voltage (often tested at 90, 100, and 110% rated voltage). This can be found in the test report for the specific transformer in question, or should be supplied by the manufacturer. Note it is specified as a percent of base, so if heated up from the high side, the current will be less than from the lowside - the transformer's excitation takes the same amount of power regardless of energizing voltage. You could test this roughly yourself by applying rated voltage to the lowside, and measuring the induced current. The load current for a single phase 11kVA transformer would be: 11K / (L-N voltage).
Winding copper losses of a transformer can be measured in a short circuit test of a transformer. Impedance voltage is given to the primary and the secondary is often shortcircuited. (some times the reverse is done of this). Full load currents are made to flow in both primary and secondary circuits. This current flow heats up the 2 windings of the transformer. Power consumed at this time gives the transformer copper losses.
The transformer can be tested on open and short circuit to find the iron losses and copper losses separately, which uses a fraction of the power than having to run the transformer on full-load.
Copper losses are directly related to loading of the transformer. Iron (core) losses are a result of magnetizing of the core of the transformer, and are relatively constant from no load to full load. With this in mind, it should be clear that the above statement is false. Maximum efficiency results with low core losses, and low copper losses. Copper losses cannot be helped, so it is important to minimize core losses to increase the efficiency of a transformer.AnswerYes, it is perfectly correct -well, with the proviso that transformers normally operate somewhat below full load and, so, are designed to achieve maximum efficiency somewhat below full load. A transformer's maximum efficiency does indeed occur when the copper losses and iron losses are equal. Unfortunately, the mathematical proof of this is too complicated to reproduce here, I suggest that you check out any reputable electrical engineering textbook.
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.
CT=========current transformer PT=========potential transformer these are the instrumental transformers.
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
güzel
the internet..... ._.
An open-circuit test is done with the transformer running at its rated voltage but with no load. This measures the power lost in the magnetic core. (IR Losses) A short-circuit test is done with the transformer running at its full rated current in all windings but at a low voltage. The secondary is shorted and the primary voltage is adjusted to give the rated current. This measures the power lost in the copper windings. (Copper losses)
The change in output voltage from no load to full load defines the voltage regulation of that transformer.
Fully loaded - 2.62 amps at 11kV. The no load depends on the transformer design, but it will usually be significantly less than the full load amps (not sure on this size, but on larger transformers it is typically ~.05 - .1% full load, so you'd be looking at ~2.5 mA RMS). The connection type is not important. Transformers are very efficient, thus there is not a whole lot of loss in the "average" transformer. The actual loss will depend on the design criteria of the transformer.
is it primary current ?