Problem 4.3.3: Series-parallel core with a gap in the centre limb

Interactive: Magnetic Circuit Solver

Problem 4.3.3

Series-parallel core with a gap in the centre limb

In the magnetic circuit below, calculate the required current to be passed in the coil having $200$ turns in order to establish a flux of $1.28 mWb$ in the air gap. Neglect fringing effect and leakage flux. The $B$ – $H$ curve of the material is given in the figure. (Permeability of air may be taken as $μ_{0} = 1/800000$ .) ( $l_{1} = l_{4} = 0.9 m$ , $l_{2} = l_{3} = 0.25 m$ , $d = 0.1 mm$ , cross-sectional areas at branches 1 and 4 $= 0.0024 m^{2}$ , cross-sectional areas at branches 2 and 3 and the air gap $= 0.0016 m^{2}$ .)

Problem 4.3.3. The B–H curve of the core material with the three readings of the solution: H=476\,\mathrm{A/m} at 0.8\,\mathrm{T}, B_4=0.375\,\mathrm{T} at H_4=336\,\mathrm{A/m}, and H_1=510\,\mathrm{A/m} at B_1=0.91\,\mathrm{T}. — **Figure 4.3.23.** Problem 4.3.3. The $B$ – $H$ curve of the core material with the three readings of the solution: $H = 476 A/m$ at $0.8 T$ , $B_{4} = 0.375 T$ at $H_{4} = 336 A/m$ , and $H_{1} = 510 A/m$ at $B_{1} = 0.91 T$ .

Try it yourself first.

Strategy

Branch 1 (the limb with the coil) carries the whole flux. At its ends the circuit splits into the centre path ( $l_{2}$ , gap, $l_{3}$ ) and the outer path ( $l_{4}$ ), which are in parallel: they span the same magnetic potential difference $U$ . Start where the data are, at the gap flux, and go upstream: $U$ from the centre path, $Φ_{4}$ from the outer path, $Φ_{1}$ from the node, $N I$ from the loop through branch 1.

(b) B_{\mathrm{gap}} against I. Problem 4.3.3. (a) Branch 1 carries everything, and the gapped centre limb and branch 4 are in parallel across the same U. (b) The gap induction as a function of the coil current, computed from the curve of Figure 4.3.23: 0.8\,\mathrm{T} needs 3.8\,\mathrm{A}. — **Figure 4.3.24.** (b) $B_{gap}$ against $I$ . Problem 4.3.3. (a) Branch 1 carries everything, and the gapped centre limb and branch 4 are in parallel across the same $U$ . (b) The gap induction as a function of the coil current, computed from the curve of Figure 4.3.23: $0.8 T$ needs $3.8 A$ .

1. The centre path

$B_{g} = Φ_{g} / S_{g} = 1.28 \times 1 0^{- 3} /0.0016 = 0.80 T$ . Branches 2 and 3 have the same area as the gap and carry the same flux, so $B_{2} = B_{3} = 0.80 T$ , for which the curve gives $H_{2} = H_{3} = 476 A/m$ . In the gap $H_{g} = 0.8 \times 800000 = 640000 A/m$ . Hence

U = H_{2} l_{2} + H_{3} l_{3} + H_{g} d = 476 \times 0.5 + 640000 \times 1 0^{- 4} = 238 + 64 = 302 At .

2. The outer path

Branch 4 spans the same $U$ : $H_{4} = U / l_{4} = 302/0.9 = 336 A/m$ , and the curve gives $B_{4} = 0.375 T$ , so

Φ_{4} = B_{4} S_{4} = 0.375 \times 0.0024 = 0.90 mWb .

3. The node

$Φ_{1} = Φ_{g} + Φ_{4} = 1.28 + 0.90 = 2.18 mWb$ , so $B_{1} = 2.18 \times 1 0^{- 3} /0.0024 = 0.91 T$ and, from the curve, $H_{1} = 510 A/m$ .

4. The loop through the coil

N I = H_{1} l_{1} + U = 510 \times 0.9 + 302 = 459 + 302 = 761 At, I = \frac{761}{200} .

Answer

I \approx 3.8 A

Check the three readings and the flux split, then switch to 'Given I' and vary the current.

InteractiveLoading simulation…

Check

The other loop, branch 1 and branch 4: $H_{1} l_{1} + H_{4} l_{4} = 459 + 336 \times 0.9 = 459 + 302 = 761 At = N I$ . Here the iron dominates, not the gap: $0.1 mm$ of air costs only $64 At$ out of $761$ . The parallel branch is not a detail. It carries $0.90 mWb$ , $41 %$ of what the coil must push, and ignoring it ( $Φ_{1} = Φ_{g}$ , $B_{1} = 0.53 T$ , $H_{1} \approx 390 A/m$ ) would give $N I \approx 655 At$ , $I \approx 3.3 A$ , that is $14 %$ too little. Notice also that steps 1–4 never used a value of $μ$ : for a nonlinear material one works with $B$ and $H$ read from the curve, and $Z = l / (μ S)$ is only a bookkeeping device.