Question 7.4: Figure 7.8(a) shows a rectangular magnetic core with an air-...
Figure 7.8(a) shows a rectangular magnetic core with an air-gap. Find the exciting current needed to establish a flux density of B_{g} = 1.2 T in the air-gap. Given N = 400 turns and μ_{r} (iron) = 4000.
It is a simple series magnetic circuit with its analog shown in Fig. 7.8(b).
Core length l_{c} = 2 [(20 – 4) + (16 – 4)] – 0.2 = 55.8 cm
Cross-sectional area of core A_{c} = 16 cm²
Core reluctance \mathcal{R}_{c} = \frac{55.8 × 10^{-2} }{4000 × 4π × 10^{-7} × 16 × 10^{-4}} = 0.694 × 10^{5} AT/Wb
= 0.694 × 10^{5} AT/Wb
Air-gap length l_{g} = 0.2 cm
Area of air-gap A_{g} = 16 cm²
Air gap reluctance \mathcal{R}_{g} = \frac{0.2 × 10^{-2} }{ 4π × 10^{-7} × 16 × 10^{-4}}
= 9.95 × 10^{5} AT/Wb
\mathcal{R}(total) = \mathcal{R}_{c} + \mathcal{R}_{g}
= 0.694 × 10^{5} + 9.95 × 10^{5} = 10.64 × 10^{5} AT/Wb
Flux in the magnetic circuit, \phi = B A = 1.2 × 16 × 10^{-4} = 1.92 mWb
Now Ni = \mathcal{F} = \phi \mathcal{R}
= 1.92 × 10^{-3} × 10.64 × 10^{5}
= 2043 AT
∴ Exciting current i = \frac{2043}{400} = 5.11 A
It is seen above that
\mathcal{R} _{g} / \mathcal{R} _{c} = 14.34
Therefore for simplicity of computation, \mathcal{R} _{c} (magnetic core reluctance) may be altogether neglected. Then
i = 1.2 × 16 × 10^{-4} × 9.95 × 10^{5} /400
= 4.77 A
This simplification has caused an error of 6.6% which can be easily tolerated in magnetic circuit calculations.
