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本文提出一种解电网的方法,用这种方法来解电网,可以避免解联立方程之麻烦,以节省劳动力。本文分下列两部分:1.一般电网解法——凡是一切直线律电网,不论其中含有若干电动势,都可应用这解法。这解法是首先将电网中的电动势代换以等值的电源电流,其次将电网中的结点逐点化去,使电网化简为一等值支路,以求该支路两端点的电势差,然后再返原为原来的电网,以求其中各结点的电势,如此,最后就很易计算出各支路的电流。2.只含一个电动势电网的特殊解法——凡是一切直线律电网,若其中只含有一个电动势,则用这解法更为简便。这解法与一般电网解法所不同的地方就是可将电网返原的这一步骤省去。为了要省去这一步骤,只须在电网化简前,电网各结点上均虚设一定量的电源电流,如此,当电网化简为一等值支路时,就能直接求得原来电网中各结点的电势。再者,本文所提出的解电网方法对于交流电网及直流电网均能适用,而本文是按解交流电网的方式叙述。In this paper a method for solving electric network is suggested. By this method, the solution of the network may be found conveniently without solving simultaneous equations.This paper consists of two parts. The first part deals with the method for solving the general linear network. The procedure of which is, firstly, to replace the e.m.fs. in the network by an equivalent source of currents; secondly, to eliminate the joints of the network one by one until the network transforms into an equivalent branch for finding the potential difference between the terminals of this branch; thirdly, to restore the equivalent branch to original network for finding the potentials of all joints of the network; finally, we can calculate the currents of the branches of the network easily. The second part deals with the special method for solving the linear network which contains only one e.m.f. The difference between this method and the former one is that the proceeding of restoring the equivalent branch to original network in this method may be omitted. In order to omit this proceeding, it is necessary to assume a certain amount of current source on every joint of the network before the transformation of the network. Therefore, when the network transforms into an equivalent branch, we may find the potentials of all joints of a network which contains only one e.m.f.
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