Question 2 of 6: Two-Window Magnetic Circuit with Air Gaps
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: Closed-book written exam, 3 hours. Six questions of equal value (20 marks each); five constitute a complete paper. All six are solved here as a study resource. AC quantities are rms; three-phase voltages are line-to-line and power is total power unless noted.
Reference texts: S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (McGraw-Hill); B. S. Guru & H. R. Hiziroglu, Electric Machinery and Transformers, 3rd ed. (Oxford); P. C. Sen, Principles of Electric Machines and Power Electronics; T. Wildi, Electrical Machines, Drives, and Power Systems.
Question 2: Two-Window Magnetic Circuit with Air Gaps (20 marks)
Given. A three-limb (two-window) silicon-iron core. The coil ($N_1=100$) sits on the left limb and drives $\Phi_1$; the centre limb carries air gap $\delta_1$ at the top and the right limb carries air gap $\delta_2$ at the bottom. Each member is 50 mm wide (equal to $l$) by 75 mm deep, so all cross-sections are equal.
Given data
Depth
δ1
δ2
Window w × h
l
N1
Φ2
75 mm
3 mm
2 mm
125 × 150 mm
50 mm
100
1 mWb
Find. The left-limb flux $\Phi_1$ and the coil current $I_1$.
[Figure not reproduced: Figure 2 (redrawn): coil on the left limb; the centre and right limbs form two parallel magnetic branches between the top and bottom yoke nodes, each containing one air gap. See the official exam paper.]
Check: The figure dimensions $w,h$ are the window (inner) dimensions; each core member is read from the drawing as 50 mm wide, giving a uniform cross-section $A = 0.050\times0.075 = 3.75\times10^{-3}\ \text{m}^2$. Mean magnetic path lengths are taken along member centre-lines. Iron field intensities are read from the 1% silicon-iron curve of Figure 3.
Approach. Treat the centre and right limbs as two parallel branches sharing the same magnetic potential drop $U$ between the top and bottom yoke junctions. From the known $\Phi_2$ compute $U$ across the right branch; equate it to the centre branch to find its flux; add to get $\Phi_1$; finally close the outer loop through the coil.
Right-branch flux density. With $A=3.75\times10^{-3}\ \text{m}^2$,
$$B_2=\frac{\Phi_2}{A}=\frac{1\times10^{-3}}{3.75\times10^{-3}}=0.267\ \text{T}.$$
From Figure 3, $H_{2}\approx 54\ \text{A/m}$ in the iron.
Magnetic potential across the parallel section. The right branch is iron path $l_r=0.548$ m (top-right yoke + right limb + bottom-right yoke, less the gap) plus gap $\delta_2$:
$$U = H_2\,l_r + \frac{B_2}{\mu_0}\delta_2 = 54(0.548) + \frac{0.267}{4\pi\times10^{-7}}(0.002)$$
$$U = 29.7 + 424.4 = \boxed{454\ \text{A}\!\cdot\!\text{t}.}$$
The 2 mm gap dominates the drop.
Centre-branch flux. The same $U$ drives the centre branch (iron $l_c=0.197$ m + gap $\delta_1=3$ mm). Solving $U=H_{si}(B_c)\,l_c + (B_c/\mu_0)\delta_1$ gives
$$B_c = 0.187\ \text{T}\ \Rightarrow\ \Phi_c = B_c A = 0.702\ \text{mWb}.$$
Left-limb flux. By continuity at the top yoke node, $\Phi_1=\Phi_c+\Phi_2$:
$$\boxed{\Phi_1 = 0.702 + 1.000 = 1.70\ \text{mWb}.}$$
Coil current. Around the outer loop, $N_1I_1 = H_{si}(B_1)\,l_{left} + U$ with $B_1=\Phi_1/A=0.454$ T, $H_1\approx101\ \text{A/m}$, and left-branch iron length $l_{left}=0.55$ m:
$$N_1 I_1 = 101(0.55) + 454 = 55.7 + 454 = 510\ \text{A}\!\cdot\!\text{t}.$$
$$\boxed{I_1 = \frac{510}{100} = 5.1\ \text{A}.}$$
