Solutions Pdf — Magnetic Circuits Problems And

Percent change from Problem 2: [ \frac0.232 - 0.2010.201 \times 100 \approx +15.4% ] Fringing reduces reluctance → increases flux. Ignoring fringing underestimates performance. Solution 4 – Series-Parallel Circuit Step 1 – Reluctances (all (\mu = 1000 \mu_0))

Mistake: Desired flux is (1.2\ \textmWb) – that’s higher than actual? No, problem says: after fault, measured flux = 0.8 mWb at same current. So with fault: [ \mathcalR total,fault = \frac2500.8\times 10^-3 = 312.5 \ \textkA-t/Wb ] Without fault, if no gap: (\mathcalR iron \approx 497\ \textkA-t/Wb) – but that would give even lower flux? Contradiction. magnetic circuits problems and solutions pdf

Given: Core length (l_c = 0.15 \ \textm), area (A = 4 \ \textcm^2), (\mu_r = 600) (still valid). What is the effective air gap length that explains the reduced flux? (Ignore fringing first, then discuss if fringing would make the gap larger or smaller.) 3. Complete Solutions Solution 1 – Toroidal Core (a) Reluctance of core: [ \mathcalR_c = \fracl_c\mu_0 \mu_r A = \frac0.4(4\pi \times 10^-7)(800)(5\times 10^-4) ] [ \mathcalR_c = \frac0.4(1.0053 \times 10^-3) \approx 398 \ \textkA-turns/Wb ] Percent change from Problem 2: [ \frac0

Flux: [ \Phi = \frac4001.725\times 10^6 \approx 0.232 \ \textmWb ] No, problem says: after fault, measured flux = 0

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.

Let’s find gap length that gives (\mathcalR total = 312.5\ \textkA-t/Wb): [ \mathcalR g = \mathcalR total - \mathcalR iron = 312.5 - 497.4 = -184.9 \ \text(negative → impossible) ] Conclusion: The core is saturating or the permeability has dropped. A better problem would give (\Phi_healthy) first.