Inductance of a Single-phase Line

Consider two solid round conductors with radii of r₁ and r₂ as shown in Fig. 1.5. One conductor is the return circuit for the other. This implies that if the current in conductor 1 is I then the current in conductor 2 is -I . First let us consider conductor 1. The current flowing in the conductor will set up flux lines. However, the flux beyond a distance D + r₂ from the center of the conductor links a net current of zero and therefore does not contribute to the flux linkage of the circuit. Also at a distance less than D - r₂ from the center of conductor 1 the current flowing through this conductor links the flux. Moreover since D >> r₂ we can make the following approximations

Fig. 1.5 A single-phase line with two conductors.

Therefore from (1.12) and (1.17) we can specify the inductance of conductor 1 due to internal and external flux as

H/m

(1.18)

We can rearrange L₁ given in (1.18) as follows

Substituting r₁¢ = r₁ e^-1/4 in the above expression we get

H/m

(1.19)

The radius r₁¢ can be assumed to be that of a fictitious conductor that has no internal flux but with the same inductance as that of a conductor with radius r₁ .

In a similar way the inductance due current in the conductor 2 is given by

H/m

(1.20)

Therefore the inductance of the complete circuit is

H/m

(1.21)

If we assume r₁¢ = r₂¢ = r¢ , then the total inductance becomes

H/m

(1.22)

where r¢ = re^-1/4.