The magnetic circuit above represents a relay. The coilhas 500 turns and the mea
ID: 1830280 • Letter: T
Question
The magnetic circuit above represents a relay. The coilhas 500 turns and the mean core path is Lc = 360 mm. When the airgap lengths are 1.5 mm each, a flux density of 0.8 Tesla isrequired to cause the relay to operate. The core is caststeel. As a result, the field intensity (H, A-T/m) is 500 H. So Hg= B/uo = 0.8 T / 4*pi*10^-7 = 636619.7724. A) Determine the current in the coil, which should be 4.19amps.. B) Determine the values of permeability ( uc = 1.57*10^-3 ) and relative permeability of the core ( ur= 1250 ) . C) If the air gap is zero, determine the current in the coilfor the same flux density in the core, which should equal 0.368amps. This is everything the professor would give us, he said we donot need the cross sectional area of the core because MMF = NI =( Hc * Lc ) + ( Hg * Lg ) and because of thedata provided above, we should be able to arrive at the answers Ihave given here. You have the answers; Please show all steps. Thank you.Explanation / Answer
I'm in the same class and this is what I got. The teacher told mefrom the graph B=0.8 tesla and Hc = 510 At/m.. Given: N = 500 turns = 0.8 T lc = 360 mm Air gaps 1 and 2 = 1.5 mm Uo = 4pi x 10^-7 a) mmf of core: Fc = Hc x lc = (510)(360 x 10^-3) = 183.6 mmf of air gap: Fg = Hg x lg = [B/Uo] (2)(lg) =[0.8/4pi x 10^-7] (2)(1.5x10^-3) = 1909.86 mmf total: Ft = Fc + Fg = 183.6 + 1909.86 =2093.46 current i: i = Ft/N = 2093.46/500 = 4.19A b)Uc = Bc / Hc = 0.8 / 510 = 1.57x10^-3 Ur = Uc / Uo = 1.57x10^-3 / 4pix10^-7 =1250 c) I just did the same thing as part a but the air gap would be0. Fc = Hc x lc = (510)(360 x 10^-3) = 183.6 Fg = 0 Current i : i = Fc / N = 183.6 / 500 =.368A For part c I think my calculator is giving me rounding errors butplease double check the work and see if that's right. Thanks!
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