Flux Linkage of one conductor in a Group:

As shown in Fig. 2.5, consider a group of n parallel round conductors carrying phasor currents I1, I2 ,…, In whose sum equals zero. Distances of these conductors from a remote point P are indicated as D1, D2 ,…, Dn. Let us obtain an expression for the total Flux Linkage of one conductor in a Group of the group considering flux up to the point P only.

Flux Linkage of one conductor in a Group

The flux linkages of ith conductor due to its own current Ii (self linkages) are given by

The Flux Linkage of one conductor in a Group i due to current in conductor j is

where Dij is the distance of ith conductor from jth conductor carrying current Ij. From Eq. (2.27) and by repeated use of Eq. (2.28), the total flux linkages of conductor i due to flux up to point P are

Flux Linkage of one conductor in a Group

The above equation can be reorganized as

Substituting for In in the second term of Eq. (2.29) and simplifying, we have

Flux Linkages of one Conductor in a Group

In order to account for total flux linkages of conductor i, let the point P now recede to infinity. The terms such as In D1/Dn etc. approach In 1 = 0. Also for the sake of symmetry, denoting r′i as Dii, we have

Flux Linkage of one conductor in a Group