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Chapter 7

Q. 7.4

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.

Step-by-Step

Verified Solution

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.

7.8