22-Elec-A6 Power Systems and Machines · December 2016

Question 6 of 6: Series-Parallel Magnetic Circuit with Air Gap

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: Professional Engineers Ontario — 07-Elec-A6 Power Systems and Machines, December 2016. Closed-book; formula sheet supplied. Six questions of equal value; candidates answer any five. All six are solved here as a study resource. All ac voltages and currents are rms; three-phase voltages are line-to-line unless noted.

Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (transformers, induction/synchronous machines, dc motors); J. D. Glover et al., Power System Analysis and Design, 6th ed. (three-phase power, two-wattmeter method, power-factor correction).

Question 6: Series-Parallel Magnetic Circuit with Air Gap (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 500-turn coil on the left leg; a centre leg in parallel with a right branch that contains the air gap and carries the specified gap flux $\Phi_g=2$ mWb.

Given data
Member	Area	Length
Coil leg (1)	A₁ = 40 cm²	l₁ = 40 cm
Centre leg (2)	A₂ = 12 cm²	l₂ = 24 cm
Right branch iron (3, 4)	A₃ = A₄ = 25 cm²	l₃ = l₄ = 26 cm
Air gap	A_g = 26 cm²	l_g = 0.25 mm

Find. The coil current $I$ (with $N=500$) needed to drive $\Phi_g=2$ mWb through the gap.

[Figure not reproduced: Figure 2 (redrawn): coil leg flux $\Phi_1$ splits into the centre leg ($\Phi_2$) and the right branch ($\Phi_g$), which are magnetically in parallel. See the official exam paper.]

Approach. Work backward from the known gap flux. Find the mmf drop across the right branch (gap + two iron members); this equals the common magnetic potential $U$ across the parallel centre leg. From $U$ get the centre-leg flux, add it to the gap flux to get the coil-leg flux, then sum the coil-leg mmf and $U$ to get $NI$. Read $H$ from the magnetization curve at each iron flux density.

Air-gap mmf. $B_g=\Phi_g/A_g=2\times10^{-3}/26\times10^{-4}=0.769$ T, so $H_g=B_g/\mu_0=6.12\times10^{5}$ A/m and $$\mathcal{F}_g=H_g l_g=(6.12\times10^{5})(0.25\times10^{-3})=153.0\ \text{A}\cdot\text{t}.$$
Right-branch iron mmf. The gap flux also flows through members 3 and 4: $B_3=B_4=\Phi_g/A_3=2\times10^{-3}/25\times10^{-4}=0.80$ T. From the curve $H\approx80$ A/m, so each iron member drops $H\,l=80(0.26)=20.8$ A·t.
Common magnetic potential (parallel node). $$\boxed{U=\mathcal{F}_g+H_3 l_3+H_4 l_4 = 153.0+20.8+20.8 = 194.6\ \text{A}\cdot\text{t}.}$$
Centre-leg flux. The centre leg sees the same $U$: $H_2=U/l_2=194.6/0.24=811$ A/m. From the curve $B_2\approx1.32$ T (the leg is in saturation, so the read is robust), giving $$\Phi_2=B_2 A_2=1.32(12\times10^{-4})=1.587\times10^{-3}\ \text{Wb}.$$
Coil-leg flux and field intensity (magnetic KCL). $$\Phi_1=\Phi_g+\Phi_2 = 2.0+1.587 = 3.587\ \text{mWb},\quad B_1=\frac{\Phi_1}{A_1}=\frac{3.587\times10^{-3}}{40\times10^{-4}}=0.897\ \text{T},$$ from which $H_1\approx100$ A/m off the curve.
Total mmf and coil current. $$NI=H_1 l_1+U = 100(0.40)+194.6 = 234.6\ \text{A}\cdot\text{t},$$ $$\boxed{I=\frac{NI}{N}=\frac{234.6}{500}=0.47\ \text{A}.}$$

Final results — Question 6
Quantity	Value
Gap flux density B_g	0.769 T
Magnetic potential U (parallel node)	194.6 A·t
Centre-leg flux Φ₂	1.587 mWb
Coil-leg flux Φ₁	3.587 mWb
Total mmf NI	234.6 A·t
Coil current I	0.47 A

Topic: Nonlinear series-parallel magnetic circuit with an air gap (magnetic analogue of Kirchhoff's laws).

Key relations: $B=\Phi/A$; $H=f(B)$ from the B–H curve (iron), $H=B/\mu_0$ (gap); mmf drop $\mathcal{F}=Hl$; magnetic KVL $\sum Hl=NI$ around a loop; magnetic KCL $\sum\Phi=0$ at a node; parallel branches share one magnetic potential $U$.

Why this works: Flux, mmf and reluctance play the roles of current, voltage and resistance. Because the gap flux is given, its $B$ (hence $H$ and $\mathcal{F}$) is fixed and the whole right branch's mmf is known—so we can march from the parallel node outward instead of iterating. The centre leg and right branch are in parallel (same two end nodes), so they share $U$; the coil leg carries their combined flux in series with that potential. The air gap, though only 0.25 mm long, dominates the mmf (153 of 235 A·t) because $\mu_0$ is thousands of times smaller than the iron's permeability.

Common pitfall: Treating the centre and right branches as being in series, or reading $H$ at the wrong flux density—always read $H$ at the $B$ fixed by geometry ($B=\Phi/A$), never invert the curve. Forgetting that $B_2$ differs from $B_g$ because the areas differ is another frequent slip.

Source: Chapman, Electric Machinery Fundamentals, 5th ed., §1 (magnetic circuits, air gaps, B–H nonlinearity).

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