【正文】 on is applied. The behavior of a transformer as detected at both sets of terminals is the same as the behavior detected at the corresponding terminals of this circuit when the appropriate parameters are inserted. The slightly different representation showing the coils 1N and 2N side by side with a core in between is only used for convenience. On the transformer itself, the coils are, of course, wound round the same core. Very little error is introduced if the magizing branch is transferred to the primary terminals, but a few anomalies will arise. For example, the current shown flowing through the primary impedance is no longer the whole of the primary current. The error is quite small since 0I is usually such a small fraction of1I . Slightly different answers may be obtained to a particular problem depending on whether or not allowance is made for this error. With this simplified circuit, the primary and referred secondary impedances can be added to give: 6 221211 )/(Re NNRR ?? And 221211 )/( NNXXXe ?? It should be pointed out that the equivalent circuit as derived here is only valid for normal operation at power frequencies。2 RR ? and 1239。 XX ? 。 . 239。2X and 5 39。 IENNINNEIEk V A ???? . The argument is sound, though at first it may have seemed suspect. In fact, if the actual secondary winding was removed physically from the core and replaced by the equivalent winding and load circuit designed to give the parameters 1N , 39。 22 XI . The referred secondary 2212221222 )/()/(39。XI will be found to check as 39。39。 RNNR ? and 22212 )/(39。/39。39。 ENNE ? which is the same as 1E . For current, since the reaction ampere turns must be unchanged 1222 39。 2? , for example, by its demagizing 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 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 . equal to 1V . It is virtually the same as 11? , though now produced by the bined effect of primary and secondary ampereturns. 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 3 but due to the vector bination with 1V there is still an increase of primary current. Two more points should be made about the figures. Firstly, a unity turns ratio has been assumed for convenience so that 39。1 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 opencircuit secondary voltage 2E by a suitable choice of turn’s ratio. 2E is available for circulating a load current impedance. For the moment, a lagging power factor will be considered. The secondary current and the resulting ampereturns 22NI will change the flux, tending to demagize the core, reduce m? and with it 1E . Because the primary leakage impedance drop is so low, 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 ampereturns nearly cancel the whole of the secondary ampereturns. This being so, the mutual flux suffers only a slight modification and requires practically the same ampereturns 10NI as on no load. The total primary ampereturns are increased by an amount 22NI necessary to neutralize the same amount of secondary ampereturns. In the vector equation, 102211 NININI ?? 。 alternatively, 221011 NININI ?? . At full load, the current 0I is only about 5% of the fullload 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, and 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 ., （ dtdNE /111 ?? ） is still equal and 2 opposite to 1V . However, there has been a redistribution of flux and the mutual ponent has fallen due to the increase of 1? with 1I . Although the change is small, the secondary demand could not be met without a mutual flux and . alteration to permit primary current to change. The flux s? linking the secondary winding has been further reduced by the establishment of secondary leakage flux due to 2I , and this opposes m? . Although m? and 2? are indicated separately, they bine 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 ponents, . dtdNdtdNV 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 the mutual flux. The two leakage fluxes are closely related。21 EE ? . Secondly, the physical picture is drawn for a different instant of time from the vector diagrams which show 0??m , if the horizontal axis is taken as usual, to be the zero time reference. 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 secondary terminals open, can easily be extended to cover the loaded secondary by th