## 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. 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 The above equation can be reorganized as Substituting for In in the second term of Eq. (2.29) and simplifying, we have 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