Magnetic Circuits Problems And Solutions Pdf -

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.

Let’s correct the fault diagnosis realistically: magnetic circuits problems and solutions pdf

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)) Given: After fault, (\Phi_actual = 0

Ah – critical insight: If the core originally had , its reluctance is 497 kA-t/Wb. Then flux would be (250/497k \approx 0.503 \ \textmWb), not 1.2 mWb. So the “desired” 1.2 mWb must have come from a different core or higher current. The problem as written is inconsistent – an excellent teaching point: always check if numbers make physical sense . If total reluctance is lower than iron alone,

Total reluctance seen by MMF: [ \mathcalR_total = \mathcalR c + \mathcalR eq,branches = 132.6 + 331.55 = 464.15 \ \textkA-t/Wb ] MMF = (300 \times 1.5 = 450 \ \textA-turns) [ \Phi_c = \frac450464.15 \times 10^3 \approx 0.969 \ \textmWb ] Then (\Phi_o = \Phi_c / 2 = 0.4845 \ \textmWb)

Flux: [ \Phi = \frac4001.725\times 10^6 \approx 0.232 \ \textmWb ]