Inductance of ThreePhase Lines with Symmetrical Spacing
Consider the threephase line shown in Fig. 1.6. Each of the conductors has a radius of r and their centers form an equilateral triangle with a distance D between them. Assuming that the currents are balanced, we have

(1.23) 
Consider a point P external to the conductors. The distance of the point from the phases a, b and c are denoted by D_{pa}, D_{pb}and D_{pc}respectively.
Fig. 1.6 Threephase symmetrically spaced conductors and an external point P.
Let us assume that the flux linked by the conductor of phasea due to a current I_{a} includes the internal flux linkages but excludes the flux linkages beyond the point P .
Then from (1.18) we get

(1.24) 
The flux linkage with the conductor of phasea due to the current I_{b} , excluding all flux beyond the point P , is given by (1.17) as

(1.25) 
Similarly the flux due to the current I_{c} is

(1.26) 
Therefore the total flux in the phasea conductor is
The above expression can be expanded as

(1.27) 
From (1.22) we get
Substituting the above expression in (1.27) we get

(1.28) 
Now if we move the point P far away, then we can approximate D_{pa} » D_{pb} » D_{pc}.. Therefore their logarithmic ratios will vanish and we can write (1.28) as

(1.29) 
Hence the inductance of phasea is given as

(1.30) 
Note that due to symmetry, the inductances of phases b and c will be the same as that of phasea given above, i.e., L_{b}= L_{c} = L a . 