The value of inductance per conductor in a three-phase line is _______ time(s) the loop inductance for the single-phase lines.

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:
\(λ_a=2\times10^{-7}[I_aln\frac{1}{D_{aa}}+I_bln\frac{1}{D_{ab}}]\)

I_a =+I , I_b = -I, D_aa = r', D_ab = D

 \(λ_a=2\times10^{-7}[I[ln\frac{1}{D_{aa}}]-I[ln\frac{1}{D_{ab}}]] =2\times10^{-7}I[ln\frac{D_{ab}}{D_{aa}}]\)

\(\large{L_a=\frac{λ_a}{I}=2\times10^{-7}[ln\frac{D}{r'}]=L_b=L_c}\)

 \(Loop\ inductance = L_a+L_b=4\times10^{-7}[ln\frac{D}{r'}]......(1)\)

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:

\(\large{λ_a=2\times10^{-7}[I_aln\frac{1}{D_{aa}}+I_bln\frac{1}{D_{ab}}+I_cln\frac{1}{D_{ac}}]}\)

Where D_aa = r', 
D_ab = D_bc =  D_ac = D

\(\large{λ_a=2\times10^{-7}[I_aln\frac{1}{r'}+I_bln\frac{1}{D}+I_cln\frac{1}{D}]}\)

Now from 3 -ϕ,3 wire system

I_a + I_b + I_c = 0
I_a = - (Ib + Ic)

,\(\large{λ_a=2\times10^{-7}[I_aln\frac{1}{r'}+(I_b+I_c)ln\frac{1}{D}}]\)

\(\large{λ_a=2\times10^{-7}[I_aln\frac{1}{r'}-(I_a)ln\frac{1}{D}}]\)

\(\large{λ_a=2\times10^{-7}[I_aln\frac{D}{r'}}]\)

\(\large{L_a=\frac{λ_a}{I}=2\times10^{-7}[ln\frac{D}{r'}]=L_b=L_c}......(2)\)

Where L_a is the inductance per conductor in a three-phase line 

From equations (1) and (2),

Inductance per conductor in a 3-phase line = (1 / 2) times loop inductance in 1 phase.