Loop inductance of a single-phase two-wire line Considered a single-phase line consisting of two conductors (phase and neutral) a and b of equal radius r. They are situated at a distance D meters. The cross sections of conductors are shown in the diagram below.
Let the current flow in the conductors are opposite in direction so that one becomes the return path for the other.
The flux linkages of conductor ‘a’ is given by the formula:
λa=2×10−7[Ia‌ln‌+Ib‌ln‌] Ia=+I,Ib=−I,Daa=r′,Dab=D λa=2×10−7[I[ln‌]−I[ln‌]] =2×10−7I[ln‌] La=λaI =2×10−7[ln‌] =Lb=Lc Loop inductance
=La+Lb=4×10−7[ln‌] ........(1)
Inductance per conductor in a three-phase line (symmetrical)
Let the spacing between the conductors be D and the radius of each conductor, r. The flux linkages of conductor a is given by the equation:
The flux linkages of the conductor ‘a’ is given by the formula:
λa=2×10−7[Ia‌ln‌+Ib‌ln‌+Ic‌ln‌] Where
Daa=r′,
Dab=Dbc=Dac=D λa=2×10−7[Ia‌ln‌+Ib‌ln‌+Ic‌ln‌] Now from
3−ϕ,3 wire system
Ia+Ib+Ic=0 Ia=−(Ib+Ic) λa=2×10−7[Ia‌ln‌+(Ib+Ic)‌ln‌] λa=2×10−7[Ia‌ln‌−(Ia)‌ln‌] λa=2×10−7[Ia‌ln‌] La==2×10−7[ln‌]=Lb=Lc ....(2)
Where
La is the inductance per conductor in a three-phase line
From equations (1) and (2),
Inductance per conductor in a 3-phase line =
() times loop inductance in 1 phase.