**Flux Linkage of n Parallel Conductors Carrying Current:**

As shown in Fig. 2.5, consider a group of n parallel conductors carrying current I_{1}, I_{2 },…, I_{n} whose sum equals zero. Distances of these conductors from a remote point P are indicated as D_{1}, D_{2 },…, D_{n. }Let us obtain an expression for the total Flux Linkages of the ith conductor of the group considering flux up to the point P only.

The flux linkages of ith conductor due to its own current I_{i} (self linkages) are given by

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

where D_{ij} is the distance of ith conductor from jth conductor carrying current I_{j}. From Eq. (2.27) and by repeated use of Eq. (2.28), the total flux linkages of parallel conductors carrying current i due to flux up to point P are

The above equation can be reorganized as

But,

__Substituting for I___{n} 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 D_{1}/D_{n}, etc. approach In 1 = 0. Also for the sake of symmetry, denoting r′_{i} as D_{ii}, we have