Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, December 2016 — 04-BS-4 Electric Circuits and Power. Three hours duration, closed book (one aid sheet permitted). Seven questions are printed; any five constitute a complete paper, but all seven are solved below as a complete study resource.
Reference texts: Sadiku, Fundamentals of Electric Circuits, 7th ed. (circuit analysis, transients, AC power, rectifiers – Questions 1–6); Chapman, Electric Machinery Fundamentals, 5th ed. (magnetic circuits – Question 7).
Given. A double-window (3-leg) core carrying the excitation coil on the left leg, with a center leg and a right leg (containing a 2 mm air gap) forming two parallel return paths between the top and bottom yokes.
Given data
μr
2000
N
100 turns
I
1 A
Cross-section A
100 mm²
Leg height
10 cm
Window half-width
5 cm
Air gap (right leg)
2 mm
Find. (a) the MMF; (b) the reluctance of each segment; (c) the analog (reluctance) circuit; (d) the flux, flux density, and field intensity in the air gap.
Figure 7 — reluctance analog circuit (also the answer to part c): the MMF source F = NI, in series with the left leg's reluctance, drives flux through two parallel paths — the center leg and the right leg (iron + air gap) — between the top (T) and bottom (Bm) yoke junctions.
Approach. Take the mean path along each leg plus its associated half of the top and bottom yokes as one reluctance "segment": the left leg (which carries the coil) is in series with the MMF source; at the top and bottom yoke junctions this feeds two parallel return paths — the center leg, and the right leg (iron, in series with the air gap). This is exactly analogous to an electrical circuit with a source and series resistance driving two parallel branches, so total flux and its current-divider split follow the same algebra as Question 1/2.
(b) Reluctance of each segment. Using $\mathcal{R}=\ell/(\mu_0\mu_rA)$ for iron and $\mathcal{R}=\ell/(\mu_0A)$ for the air gap ($\mu_r=1$), with mean lengths: left leg + its two yoke halves $=10+5+5=20\text{ cm}$; center leg $=10\text{ cm}$; right leg iron $=10-0.2=9.8\text{ cm}$ plus its two yoke halves $=9.8+5+5=19.8\text{ cm}$; gap $=0.2\text{ cm}$:
$$ \mathcal{R}_{left}=\frac{0.20}{\mu_0(2000)(10^{-4})}=\boxed{7.958\times10^{5}\text{ A}\cdot\text{t/Wb}} $$
$$ \mathcal{R}_{center}=\frac{0.10}{\mu_0(2000)(10^{-4})}=\boxed{3.979\times10^{5}\text{ A}\cdot\text{t/Wb}} $$
$$ \mathcal{R}_{right,iron}=\frac{0.198}{\mu_0(2000)(10^{-4})}=\boxed{7.878\times10^{5}\text{ A}\cdot\text{t/Wb}} $$
$$ \mathcal{R}_{gap}=\frac{0.002}{\mu_0(1)(10^{-4})}=\boxed{1.592\times10^{7}\text{ A}\cdot\text{t/Wb}} $$
(the air gap's reluctance is roughly 20× the entire iron path combined, exactly as expected since air's permeability is $\mu_r=2000$ times smaller.)
(c) Analog circuit. See Figure 7: the MMF source $F=NI$ in series with $\mathcal{R}_{left}$ drives the parallel combination of $\mathcal{R}_{center}$ and $(\mathcal{R}_{right,iron}+\mathcal{R}_{gap})$, exactly mirroring an electrical source-plus-two-parallel-branches network.
(d) Flux, flux density, and field intensity in the gap. With $\mathcal{R}_{right}=\mathcal{R}_{right,iron}+\mathcal{R}_{gap}=1.671\times10^7\text{ A}\cdot\text{t/Wb}$, the parallel combination and total reluctance are
$$ \mathcal{R}_{par}=\left(\frac{1}{\mathcal{R}_{center}}+\frac{1}{\mathcal{R}_{right}}\right)^{-1}=3.886\times10^5,\qquad \mathcal{R}_{tot}=\mathcal{R}_{left}+\mathcal{R}_{par}=1.184\times10^6\text{ A}\cdot\text{t/Wb} $$
The flux leaving the source (left leg) is $\phi_{left}=F/\mathcal{R}_{tot}=84.43\ \mu\text{Wb}$; by the reluctance current-divider rule the flux through the right (gapped) branch is
$$ \phi_{gap}=\phi_{left}\,\frac{\mathcal{R}_{center}}{\mathcal{R}_{center}+\mathcal{R}_{right}}=\boxed{1.964\ \mu\text{Wb}} $$
$$ B_{gap}=\frac{\phi_{gap}}{A}=\frac{1.964\times10^{-6}}{10^{-4}}=\boxed{0.01964\text{ T}} $$
$$ H_{gap}=\frac{B_{gap}}{\mu_0}=\boxed{15{,}632.3\text{ A}\cdot\text{t/m}} $$
