Question 5 of 7: Magnetic Circuit – Reluctance, Flux, and Inductance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-4 Electric Circuits and Power — December 2014 national exam. 3 hours, closed book (one aid sheet permitted). Any five of the seven questions constitute a complete paper; the first five as they appear in the answer book are marked. All seven questions are solved below as a complete study resource.
Reference texts: Sadiku & Alexander, Fundamentals of Electric Circuits (6th ed.) — DC circuit analysis, Thevenin/Norton equivalents, first-order transients, AC steady-state phasors and power, magnetic circuits, diode rectifiers (Ch. 2–4, 9–11, 13); Mano & Ciletti, Digital Design — combinational logic design (Ch. 2–4), cited inline for Question 7.
Question 5: Magnetic Circuit – Reluctance, Flux, and Inductance (20 marks)
Check: the figure gives only ONE in-plane dimension (5 cm) for the limb cross-section; the problem statement says only that the cross-section is "uniform," with no separate depth given. This solution assumes a square 5 cm × 5 cm cross-section ($A=25\text{ cm}^2$). Because the air gaps dominate the total reluctance (≈83%, computed below), the boxed results are not sensitive to reasonable alternative iron mean-path estimates – but they DO scale with this assumed area, so a different specified depth would rescale $B_{gap}$, $H_{gap}$ and $L$ proportionally.
Given. Single-window rectangular core with two air gaps in series (1 mm at the centre of the top yoke; 2 mm in the left limb). $\mu_r=2000$, $N=100$ turns (wound in series across both limbs, carrying the same current), $i=2\text{ A}$.
Given data (dimensions from Figure 5)
Segment
Mean length
Top yoke (2×35 cm, split by 1 mm gap)
70 cm iron
Right limb (2×5 cm, no gap)
10 cm iron
Left limb (2×5 cm, minus 2 mm gap)
9.8 cm iron
Bottom yoke
30 cm iron
Air gaps
1 mm (top) + 2 mm (left limb)
[Figure not reproduced. See the official exam paper.]
Find. (a) MMF; (b) reluctance of each segment; (c) $\phi$, $B_{gap}$, $H_{gap}$ in each gap; (d) coil inductance $L$.
Approach. This is a single series magnetic loop (one flux path, no parallel branches), so total reluctance is the sum of the iron-segment reluctances and the two gap reluctances; MMF drives the same flux through every segment including both gaps.
a) Magnetomotive force.
$$\mathcal{F}=Ni=(100)(2)=\boxed{200\text{ A-turns}}$$
b) Reluctance of each part. Using $\mathcal{R}=\ell/(\mu A)$ with $A=25\text{ cm}^2=2.5\times10^{-3}\text{ m}^2$:
$$\mathcal{R}_{iron}=\frac{\ell_{iron}}{\mu_0\mu_rA}=\frac{1.198}{(4\pi\times10^{-7})(2000)(2.5\times10^{-3})}=1.907\times10^{5}\text{ A-t/Wb}$$
$$\mathcal{R}_{gap,1mm}=\frac{0.001}{\mu_0A}=3.183\times10^{5}\text{ A-t/Wb}, \qquad \mathcal{R}_{gap,2mm}=\frac{0.002}{\mu_0A}=6.366\times10^{5}\text{ A-t/Wb}$$
$$\mathcal{R}_{total}=\mathcal{R}_{iron}+\mathcal{R}_{gap,1mm}+\mathcal{R}_{gap,2mm}=\boxed{1.146\times10^{6}\text{ A-t/Wb}}$$
(the two gaps alone account for ≈83% of the total, confirming the check note above.)
c) Flux, flux density, field intensity in each gap. Since this is a single series loop, the SAME flux passes through both gaps:
$$\phi=\frac{\mathcal{F}}{\mathcal{R}_{total}}=\frac{200}{1.146\times10^6}=\boxed{1.746\times10^{-4}\text{ Wb (both gaps)}}$$
$$B_{gap}=\frac{\phi}{A}=\frac{1.746\times10^{-4}}{2.5\times10^{-3}}=\boxed{0.0698\text{ T (both gaps)}}$$
$$H_{gap}=\frac{B_{gap}}{\mu_0}=\frac{0.0698}{4\pi\times10^{-7}}=\boxed{55{,}571\text{ A/m (both gaps)}}$$
