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

Question 7 of 7: Series–Parallel Magnetic Circuit

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, Spring 2016. Closed book (formula sheets supplied); do any five of seven questions, all of equal value (20 marks each). Three hours. All ac quantities are rms; three-phase voltages are line-to-line and power is total real power unless noted. All seven questions are solved here as a complete study resource.

Reference texts. S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (DC machines ch. 8, transformers ch. 2, induction motors ch. 7, synchronous machines ch. 4–5, parallel operation ch. 5); J. D. Glover et al., Power System Analysis and Design, 6th ed. (per-phase and power-factor analysis ch. 2); T. Wildi, Electrical Machines, Drives, and Power Systems. Canadian frame: EGBC/PEO national exam; SI units.

Question 7: Series–Parallel Magnetic Circuit (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 drives flux that splits into a shunt centre leg (branch 2) and a right branch (segments 3, 4 plus the air gap) carrying the required gap flux Φ_g = 2 mWb.

Given data — Question 7
Coil turns / gap flux	N = 500, Φ_g = 2 mWb
Areas (cm²)	A₁=40, A₂=12, A₃=A₄=25, A_g=26
Iron path lengths	l₁=40, l₂=24, l₃=l₄=26 cm
Air-gap length	l_g = 0.25 mm
Magnetization curve	read H(B) off Figure 3

Find. The coil current I that establishes 2 mWb in the air gap.

[Figure not reproduced: Figure 3 (redrawn) — the coil leg (Φ 1 ) feeds a shunt centre leg (Φ 2 ) in parallel with the right branch (Φ g through segments 3, 4 and the air gap). Flux: Φ 1 = Φ 2 + Φ g . See the official exam paper.]

Approach. This is a series–parallel magnetic circuit (magnetic-Ohm's-law analogue). Start from the known gap flux to get the magnetic potential across the parallel section, use that potential to find the shunt-leg flux, add the two for the coil-leg flux, and finally sum the coil-leg mmf and the parallel-section potential to get $NI$.

Air-gap field. $B_g=\dfrac{\Phi_g}{A_g}=\dfrac{2\times10^{-3}}{26\times10^{-4}}=0.769$ T, so $H_g=\dfrac{B_g}{\mu_0}=\dfrac{0.769}{4\pi\times10^{-7}}=6.12\times10^5$ A/m and the gap mmf is $H_gl_g=6.12\times10^5(0.25\times10^{-3})=153$ A·t.
Right-branch iron. $B_3=B_4=\dfrac{\Phi_g}{A_3}=\dfrac{2\times10^{-3}}{25\times10^{-4}}=0.80$ T; from the curve $H_3\approx 128$ A/m, so the two segments contribute $H_3(l_3+l_4)=128(0.52)=66.6$ A·t.
Magnetic potential across the parallel section. $$U=H_3l_3+H_gl_g+H_4l_4=66.6+153.0=219.6\text{ A}\cdot\text{t}.$$ This same potential drives the centre (shunt) leg.
Shunt-leg (branch 2) flux. $H_2=\dfrac{U}{l_2}=\dfrac{219.6}{0.24}=915$ A/m; from the curve $B_2\approx 1.326$ T, so $\Phi_2=B_2A_2=1.326(12\times10^{-4})=1.591\text{ mWb}.$
Coil-leg flux and field. By continuity $\Phi_1=\Phi_g+\Phi_2=2.0+1.591=3.591$ mWb, giving $B_1=\dfrac{\Phi_1}{A_1}=\dfrac{3.591\times10^{-3}}{40\times10^{-4}}=0.898$ T and, from the curve, $H_1\approx 160$ A/m.
Required current. The coil mmf equals the leg drop plus the parallel-section potential: $$NI=H_1l_1+U=160.3(0.40)+219.6=64.1+219.6=283.7\text{ A}\cdot\text{t},$$ hence $$\boxed{I=\frac{NI}{N}=\frac{283.7}{500}=0.567\text{ A}.}$$

Final results — Question 7
Quantity	Value
Gap flux density / field	B_g = 0.769 T, H_g = 6.12×10⁵ A/m
Parallel-section potential U	219.6 A·t
Shunt-leg flux Φ₂	1.591 mWb
Coil-leg flux Φ₁	3.591 mWb
Total mmf NI	283.7 A·t
Required coil current	0.567 A

Check: Iron field intensities H₃ ≈ 128 A/m (B=0.80 T), H₂ = 915 A/m → B ≈ 1.326 T, and H₁ ≈ 160 A/m (B=0.898 T) are read from the supplied magnetization curve (points on the printed curve: B = 0.5/0.8/0.9/1.0/1.1 T at H ≈ 74/128/161/213/295 A/m; H = 500/700/900/1000 A/m at B ≈ 1.236/1.296/1.325/1.333 T). The 153 A·t air gap is the largest single drop (about 54% of NI), but the iron is not negligible: misreading every H by ±5 A/m moves I by less than 0.02 A, while reading the knee of the curve as much steeper than printed understates I by roughly 15%.

Topic: Series–parallel magnetic circuit with an air gap, solved by the magnetic-Ohm's-law analogue.

Key relations: $B=\Phi/A$; $H=B/\mu_0$ in air (nonlinear $H(B)$ from the curve in iron); mmf $=\!\int\! H\,dl$; magnetic KCL $\Phi_1=\Phi_2+\Phi_g$; magnetic KVL $NI=H_1l_1+U$.

Why this works: Flux plays the role of current, mmf the role of voltage, and reluctance the role of resistance, so Kirchhoff-type laws apply: fluxes sum at a node and mmf drops sum around a loop. Because the iron is nonlinear, we work from the known gap flux (which fixes the branch fields directly) rather than by inverting reluctances, and read $H$ from the B–H curve at each computed flux density. The air gap, though only 0.25 mm long, dominates the required mmf because air has an enormously higher reluctance than iron.

Common pitfall: Treating the centre and right branches as in series (they are parallel across the same magnetic potential), or reading H at the wrong flux density — the flux density is fixed by geometry ($B=\Phi/A$), and H then follows from the curve, not the reverse.

Source: Chapman, Electric Machinery Fundamentals §1.3–1.5 (magnetic circuits, air-gap fringing neglected).

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