**Inductance of a Single-phase Line **
Consider two solid round conductors with radii of *r*_{1} and *r*_{2} 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 *+ *r*_{2} 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 *- *r*_{2} from the center of conductor 1 the current flowing through this conductor links the flux. Moreover since *D *>> *r*_{2} 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 *L*_{1} given in (1.18) as follows
Substituting *r*_{1}¢ = *r*_{1} *e*^{-1/4} in the above expression we get
H/m |
(1.19) |
The radius *r*_{1}¢ 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 *r*_{1} .
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 *r*_{1}¢ = *r*_{2}¢ = *r*¢ , then the total inductance becomes
H/m |
(1.22) |
where *r*¢ = *re*^{-}^{1/4}. |