Flux Linkage of n Parallel Conductors Carrying Current:

As shown in Fig. 2.5, consider a group of n parallel conductors carrying current 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 Linkages of the ith conductor of the group considering flux up to the point P only.

Flux Linkage of n Parallel Conductors Carrying Current

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

Flux Linkage of n Parallel Conductors Carrying Current

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

Flux Linkage of n Parallel Conductors Carrying Current

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 parallel conductors carrying current i due to flux up to point P are

The above equation can be reorganized as

But,

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

Flux Linkage of n Parallel Conductors Carrying Current

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 n Parallel Conductors Carrying Current

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