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When AC is applied to a coil, a varying magnetic field will
be produced around it. When another coil is placed within
that magnetic field, it will induce a current flowing in
that coil. This principle is called MUTUAL INDUCTANCE
The amount of mutual inductance between the two coils
depends on the distance between the two coils, and the
angle between the two coils. When two coils are linked
together via mutual inductance in this manner, we say that
the coils are inductively COUPLED. When the mutually
inductive coils are close to each other, we say that they
are closely, or tightly coupled. When they are far apart,
we say that they are loosely coupled. The greatest amount
of coupling occurs when the coils are wound one directly
over the other and on a closed iron core. The quantity of
coupling between two coils is sometimes referred to as the
Coefficient of Coupling. The formula for Coefficient of
Coupling is:
K=M/ROOT OF(L1*L2)
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We have drive feeding an output contactor for a 5000 hp motor @ 5kV application. The motor FLA is 629amps x 1.25% service factor = 786.25 amps The cable that we current have in place between the drive and the output contactor is 350kcmil- MV-105 (parallel cables). Per the NEC table 310.60 (C) (69). This show the cable ampacity is at 615 amps per cable for a rating of 1230 amps. We have derated the cable for the cable tray and the multiple cables application and our calculations show that after the correction factor factor we are at 922amps. 1230 amps x .75% correction factor factor = 922.5 amps. The customer is driving us to then use another safety factor of .25% off of the already derated cable. Can you please provide a longhand calculation for your solution on this application?
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