Consider a composite conductor which is made up of two or more strands which are in parallel. For simplicity let us assume that all the strands are identical and share the current equally. The sum of the currents in all the conductors is zero. Such a group of conductors is shown in the Fig. 1.

Fig. 1 |

The conductors 1, 2, 3, ... n carry the currents , , .... n carry the currents I

_{1}, I_{2}, I_{3 }... I_{n}. Let the distances of the conductors from a point P be D_{1p}, D_{2p}, D_{3p}... D_{np }respectively. Let ψ_{1p1 }be flux linkages of conductor due to its own current I_{1 }due to internal and external flux. The flux beyond point P is excluded. Now ψ

_{1p2 }is the flux linkages of conductor 1 due to current is equal to flux produced by I_{2 }between the point P and conductor 1. Again flux beyond P is neglected Similarly the flux linkages ψ

_{1p }with conductor 1 due to all the conductors in the group but the flux beyond point P is neglected. By expanding the logarithmic terms and rearranging the terms

The sum of all currents is zero.

Substituting this value in above equation

If point P is at infinite distance so that ln (D

_{1p}/D_{np}) ln (D_{2p}/D_{np}) etc will approach to zero (since ln 1 = 0) then we have, All the flux linkages of conductor 1 are included in the above derivation. The above expression is valid only only if sum of the currents is zero.

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