Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2013 — 04-BS-4 Electric Circuits and Power — 3 hours duration, closed book (one aid sheet permitted). Any five of the seven questions constitute a complete paper; all seven are solved below as a complete study resource.
Reference texts: Sadiku, Fundamentals of Electric Circuits (6th ed.) — Ch. 3 (nodal/mesh analysis), Ch. 4 (Thévenin/Norton, max power transfer), Ch. 7 (first-order RL/RC transients), Ch. 9–11 (AC phasors, AC power), Ch. 13/App. (mutual inductance/magnetic circuits), Ch. 4/Ch. 8 practice (diode rectifiers).
Question 7: Magnetic Core with an Air Gap (equal value)
[Figure not reproduced: Figure 6 — redrawn: U-shaped core (mean path 60 cm, uniform 5 cm×6 cm cross-section) with a small air gap x in series with the bottom keeper bar, wound with N = 100 turns. See the official exam paper.]
Given.
Given data
μ_r
2000
μ_0
4π×10−7 H/m
N
100 turns
i
1 A
x (air gap)
0.1 mm
Mean core path, l_core
60 cm (Figure 6)
Cross-section, A
5 cm × 6 cm = 30 cm²
Check — figure-reading assumption: Figure 6 gives a 60 cm outer-arch callout, two in-plane "5 cm" leg-thickness marks, a 30 cm bottom-bar length, and a separate 6 cm block below the main sketch. This solution reads the 60 cm figure as the core's total mean magnetic path length, the two "5 cm" marks as confirming a 5 cm in-plane core thickness, and the separate 6 cm rectangle as the core's out-of-page depth (a common way to show the third dimension of a laminated core in a 2-D sketch) — giving a uniform cross-section A = 5×6 = 30 cm². The air-gap field intensity H_gap below turns out to be completely independent of this area assumption (it cancels algebraically), which is a strong internal check on the reading; only Φ, B_gap and L depend on the assumed area.
Find. The mmf; the reluctance of the core and of the air gap; the flux, flux density and field intensity in the gap; and $L(x)$.
Approach. Model the core and the (much narrower) air gap as two reluctances in series, driven by the coil's mmf. Because the gap is stated to be much smaller than the core's cross-sectional dimensions, fringing can be neglected and the gap is treated as carrying the same uniform flux as the core.
Magnetomotive force. $$\mathcal{F} = Ni = (100)(1) = \boxed{100\ \text{A-turns}}$$
Reluctance of the core. $$\mathcal{R}_{core} = \dfrac{l_{core}}{\mu_r\mu_0 A} = \dfrac{0.60}{(2000)(4\pi\times10^{-7})(3\times10^{-3})} = \boxed{7.96\times10^{4}\ \text{A-turns/Wb}}$$
Reluctance of the air gap ($\mu_r=1$ in air): $$\mathcal{R}_{gap} = \dfrac{x}{\mu_0 A} = \dfrac{1\times10^{-4}}{(4\pi\times10^{-7})(3\times10^{-3})} = \boxed{2.65\times10^{4}\ \text{A-turns/Wb}}$$ Total: $\mathcal{R}_{total}=\mathcal{R}_{core}+\mathcal{R}_{gap}=1.061\times10^{5}$ A-turns/Wb.
Flux, flux density, field intensity in the gap. $$\Phi = \dfrac{\mathcal{F}}{\mathcal{R}_{total}} = \dfrac{100}{1.061\times10^5} = \boxed{9.42\times10^{-4}\ \text{Wb}}$$ $$B_{gap} = \dfrac{\Phi}{A} = \dfrac{9.42\times10^{-4}}{3\times10^{-3}} = \boxed{0.314\ \text{T}}\qquad H_{gap} = \dfrac{B_{gap}}{\mu_0} = \boxed{2.50\times10^{5}\ \text{A/m}}$$
Inductance as a function of gap length. $$L(x) = \dfrac{N^2}{\mathcal{R}_{core}+\mathcal{R}_{gap}(x)} = \dfrac{N^2}{\dfrac{l_{core}}{\mu_r\mu_0 A}+\dfrac{x}{\mu_0 A}} = \boxed{\dfrac{N^2\mu_0 A}{\dfrac{l_{core}}{\mu_r}+x}}$$ At $x=0.1$ mm this gives $L=94.2$ mH, consistent with $N^2/\mathcal{R}_{total}$ above.
