Determine self inductance of coil, Electrical Engineering

Determine self inductance of coil:

A flux of 0.5 m Wb is generated by coil A of 600 turns wound on a ring with a current of 2 A in it. Determine (a) self inductance of coil A, (b) the e.m.f induced in coil A when a current of 6 A passing through it is switched off, supposing the current to fall to zero in 2 millisecond, and (c) the mutual inductance between the coils, if a second coil B of 400 turns is uniformly wound over the first coil A.

Solution

 (a)       Self inductance of coil  A = N1  ( φ 1/ i1 )= (600 × 0.5 × 10- 3 )/2= 0.15 H

 (b) e = L (di / dt) = 0.15 × (6 - 0) /2 × 10- 3  = 450 V

 (c) M = N2  ( φ 1 / i1  )= (400 × 0.5 × 10- 3) /2= 0.1 H .

Posted Date: 2/4/2013 6:07:53 AM | Location : United States







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