1、中文 5580 字，3500英文单词，17500英文字符 The Transformer on load Introduction to DC Machines The Transformer on load It has been shown that a primary input voltage 1V can be transformed to any desired open-circuit secondary voltage 2E by a suitable choice of turns ratio. 2E is available for c
2、irculating a load current impedance. For the moment, a lagging power factor will be considered. The secondary current and the resulting ampere-turns 22NI will change the flux, tending to demagnetize the core, reduce m and with it 1E . Because the primary leakage impedance drop is so low,
3、 a small alteration to 1E will cause an appreciable increase of primary current from 0I to a new value of 1I equal to ijXREV 111 / . The extra primary current and ampere-turns nearly cancel the whole of the secondary ampere-turns. This being so , the mutual flux suffers only a slig
4、ht modification and requires practically the same net ampere-turns 10NI as on no load. The total primary ampere-turns are increased by an amount 22NI necessary to neutralize the same amount of secondary ampere-turns. In the vector equation , 102211 NININI ； alternatively, 221011 NININI .
5、 At full load, the current 0I is only about 5% of the full-load current and so 1I is nearly equal to 122 / NNI . Because in mind that 2121 / NNEE , the input kVA which is approximately 11IE is also approximately equal to the output kVA, 22IE . The physical current has increased, an
6、d with in the primary leakage flux to which it is proportional. The total flux linking the primary ,111 mp, is shown unchanged because the total back e.m.f.,（ dtdNE /111 ） is still equal and opposite to 1V . However, there has been a redistribution of flux and the mutual component has fallen due to
7、the increase of 1 with 1I . Although the change is small, the secondary demand could not be met without a mutual flux and e.m.f. alteration to permit primary current to change. The net flux s linking the secondary winding has been further reduced by the establishment of secondary leakage flux
8、due to 2I , and this opposes m . Although m and 2 are indicated separately , they combine to one resultant in the core which will be downwards at the instant shown. Thus the secondary terminal voltage is reduced to dtdNV S /22 which can be considered in two components, i.e. dtdNdtd
9、NV m / 2222 or vectorially 2222 IjXEV . As for the primary, 2 is responsible for a substantially constant secondary leakage inductance 222222 / NiN . It will be noticed that the primary leakage flux is responsible for part of the change in the secondary terminal voltage due to its effects on t
10、he mutual flux. The two leakage fluxes are closely related； 2 , for example, by its demagnetizing action on m has caused the changes on the primary side which led to the establishment of primary leakage flux. If a low enough leading power factor is considered, the total secondary flux and the
11、mutual flux are increased causing the secondary terminal voltage to rise with load. p is unchanged in magnitude from the no load condition since, neglecting resistance, it still has to provide a total back e.m.f. equal to 1V . It is virtually the same as 11 , though now produced by the combined effe
12、ct of primary and secondary ampere-turns. The mutual flux must still change with load to give a change of 1E and permit more primary current to flow. 1E has increased this time but due to the vector combination with 1V there is still an increase of primary current. Two more points
13、should be made about the figures. Firstly, a unity turns ratio has been assumed for convenience so that '21 EE . Secondly, the physical picture is drawn for a different instant of time from the vector diagrams which show 0m , if the horizontal axis is taken as usual, to be the zero time referenc
14、e. There are instants in the cycle when primary leakage flux is zero, when the secondary leakage flux is zero, and when primary and secondary leakage flux is zero, and when primary and secondary leakage fluxes are in the same sense. The equivalent circuit already derived for the transformer with the
15、 secondary terminals open, can easily be extended to cover the loaded secondary by the addition of the secondary resistance and leakage reactance. Practically all transformers have a turns ratio different from unity although such an arrangement is sometimes employed for the purposes of electrically
16、isolating one circuit from another operating at the same voltage. To explain the case where 21 NN the reaction of the secondary will be viewed from the primary winding. The reaction is experienced only in terms of the magnetizing force due to the secondary ampere-turns. There is no way of dete
17、cting from the primary side whether 2I is large and 2N small or vice versa, it is the product of current and turns which causes the reaction. Consequently, a secondary winding can be replaced by any number of different equivalent windings and load circuits which will give rise to an iden
18、tical reaction on the primary .It is clearly convenient to change the secondary winding to an equivalent winding having the same number of turns 1N as the primary. With 2N changes to 1N , since the e.m.f.s are proportional to turns, 2212 )/(' ENNE which is the same as 1E
19、. For current, since the reaction ampere turns must be unchanged 1222 ''' NINI must be equal to 22NI .i.e. 2122 )/( INNI . For impedance , since any secondary voltage V becomes VNN )/( 21 , and secondary current I becomes INN )/( 12 , then any secondary impedance, including l
20、oad impedance, must become IVNNIV /)/('/' 221 . Consequently, 22212 )/(' RNNR and 22212 )/(' XNNX . If the primary turns are taken as reference turns, the process is called referring to the primary side. There are a few checks which can be made to see if the procedure out
21、lined is valid. For example, the copper loss in the referred secondary winding must be the same as in the original secondary otherwise the primary would have to supply a different loss power. '' 222 RI must be equal to 222RI . ）222122122 /()/( NNRNNI does in fact reduce to 222RI . Similarly the stored magnetic energy in the leakage field )2/1( 2LI which is proportional to 22'XI will be found to check as '' 22 XI . The referred secondary 2212221222 )/()/('' IENNINNEIEk V A .
