Given: After fault, (\Phi_actual = 0.8\ \textmWb) at (NI=250). So total reluctance = (250 / 0.8\times10^-3 = 312.5 \ \textkA-t/Wb). Core reluctance alone = (497.4 \ \textkA-t/Wb). If total reluctance is lower than iron alone, that’s impossible. Therefore: The original core for design purposes. The fault increased the gap.
The center limb carries (\Phi_c). That flux splits into two paths, each with total reluctance (\mathcalR_branch = \mathcalR_o + 2\mathcalR_y). The center limb reluctance is in series with the parallel combination of the two branch reluctances. magnetic circuits problems and solutions pdf
Center limb: [ \mathcalR_c = \frac0.1(4\pi\times 10^-7)(1000)(6\times 10^-4) \approx 132.6 \ \textkA-t/Wb ] Each outer limb: [ \mathcalR_o = \frac0.2(4\pi\times 10^-7)(1000)(3\times 10^-4) \approx 530.5 \ \textkA-t/Wb ] Yoke (each, two yokes in series effectively for each flux path): [ \mathcalR y = \frac0.05(4\pi\times 10^-7)(1000)(6\times 10^-4) \approx 66.3 \ \textkA-t/Wb ] Total for one outer path (center → yoke → outer limb → yoke → center): [ \mathcalR outer, total = \mathcalR_c + 2\mathcalR_y + \mathcalR_o ] [ = 132.6 + 2(66.3) + 530.5 = 795.7 \ \textkA-t/Wb ] But careful: The two outer paths are after the center limb. Given: After fault, (\Phi_actual = 0
Reluctance without gap: [ \mathcalR c,iron = \frac0.15(4\pi\times 10^-7)(600)(4\times 10^-4) \approx 497.4 \ \textkA-t/Wb ] MMF = (\Phi \mathcalR) → (250 = (1.2\times 10^-3) \times \mathcalR total,des ) So (\mathcalR_total,des \approx 208.3 \ \textkA-t/Wb) – but that’s than iron reluctance alone? That’s impossible. If total reluctance is lower than iron alone,
Hint: By symmetry, the two outer limbs carry equal flux. A DC relay has a magnetic circuit that should produce (\Phi = 1.2 \ \textmWb) at (I = 0.5 \ \textA) with (N = 500). After years of use, the measured flux is only (0.8 \ \textmWb) at the same current. You suspect an unexpected air gap has developed (e.g., due to corrosion or mechanical wear).
Flux density in yokes = same as center limb area? Yokes have (A=6\ \textcm^2), but they carry (\Phi_c)? No – yokes carry the outer branch flux? Actually each yoke segment carries (\Phi_o) if symmetric. Check: At top yoke, flux from center splits: half to left outer, half to right outer. So yoke carries (\Phi_o). [ B_yoke = \frac0.4845\times 10^-36\times 10^-4 = 0.8075 \ \textT ] Desired flux (\Phi_des = 1.2 \ \textmWb) with (NI = 250 \ \textA-turns) (since (0.5 \times 500)).