Determine total magnetic flux, Electrical Engineering

Determine Total magnetic flux:

A ring is composed of three sections. The cross-sectional area is 0.001 m2 for each section. The mean lengths of each section are la = 0.3 m, lb = 0.2m, lc = 0.1 m. An air-gap length of 0.1 mm is cut in the ring. Relative permeabilities for sections a, b, c are 5000, 1000 and 10000 respectively. Flux in the air gap is 7.5 × 10- 4 Wb and the coil has 100 turns. Determine (a) Total m.m.f, and (b) exciting current.

Solution

Reluctance of section a of ring

Sa  =( la/ μ0 μra A )=  0.3 / (4π× 10- 7 × 5000 × 0.001)

= 47746.37 AT/Wb

Reluctance of section b of ring

Sb  = lb/ μ0 μrb A =0.2 / (4π× 10- 7 × 1000 × 0.001)

    = 159154.57 AT/Wb

Reluctance of section c of ring

Sc  = lc/ μ0 μrc A =0.1 / 4π× 10- 7 × 10000 × 0.001 = 7957.73 AT/Wb

Reluctance of air-gap

S g  = lg / μ0 Ag =  0.1 × 10- 3/4π× 10- 7 × 0.001

                            = 79577.73 AT/Wb

Total Reluctance

S = Sa  + Sb  + Sc  + Sg  = 294436.4 AT/Wb

(a) Total m.m.f = Flux × Reluctance = 7.5 × 10- 4 Wb × 294436.4 AT/Wb = 220.83 AT.

(b) Exciting Current = m.m.f/ No. of turns  = 220.83/100  = 2.21 Amp

Posted Date: 2/4/2013 5:47:37 AM | Location : United States







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