202 THE MAGNETIC CIRCUIT [ART. 62 



an unsymmetrical spacing is replaced by an equivalent equidistant 

 spacing. The exact solution for an unsymmetrical spacing is given 

 in tin 1 next article. Let the instantaneous values of the three cur- 

 in the wires A, B and C of a three-phase transmission line be 

 lit \2 and 13. The sum of the three currents at each instant is 

 zero, or 



ii +12 -f-i 3 =0 ....... (126) 



Let Qeq be the equivalent flux which links at any instant with the 

 wire A. The instantaneous e.m.f. induced in this wire is 



dt ........ (127) 



The equivalent flux consists of the actual flux outside the wire plus 

 the sum of the fluxes inside the wire, each infinitesimal tube of flues 

 being reduced in the proper ratio, according to the fraction of the 

 cross-section of the wire with which it is linked. Or, what is the 

 same thing, each wire is replaced by an equivalent hollow cylinder 

 of infinitesimal thickness, without partial linkages, as in problem 

 9 in Art. 59 (consult also the definition of equivalent permeance 

 given in Art. 58). 



In order to determine tf> eq we replace the three-phase system 

 by two superimposed single-phase systems. The current ii in the 

 wire A may be thought of as the sum of the currents 12 and 13, 

 each flowing in a separate fictitious wire, and both of these wires 

 coinciding with A. The currents +12 in B and 12 in A form 

 one single-phase loop, while the currents +13 in C and 13 in A 

 form the other loop. The flux <P eq which surrounds A is the sum 

 of the fluxes produced by these two loops. The flux per unit 

 length of the line, due to the first loop, is equal to <P iq 'i2, since 

 the number of turns is equal to one. For the same reason (P' eq = 

 L f where L' is determined by eq. (125). Hence, the flux per 

 unit length of the line, due to the first loop, is L'i^ Similarly, 

 the flux due to the second loop and linked with A is equal to 

 L'i3, the same value of Z/ being used because the spacing and 

 the sizes of all of the wires are the same. Thus, 



(^ eq = ~L'i 2 -Ui3=Ui l) .... (128) 



the last result being obtained by substituting the value of 12+13 

 from eq. (126). Thus, eq. (127) becomes simply 



. . . (129) 



