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Chapter 2

Q. 2.4

The magnetic circuit of Fig.2.10 has cast steel core. The cross-sectional area of the central limb is 800 mm^{2} and that of each outer limb is 600mm^{2}. Calculate the exciting current needed to set up a flux of 0.8 mWb in the air gap.Neglect magnetic leakage and fringing.The magnetization characteristic of cast steel is given in Fig. 2.16.

Step-by-Step

Verified Solution

Air gap

B_{g}=\frac{0\cdot 8}{800}\times \frac{10^{-3} }{10^{-6} } =1 T and H_{g} =\frac{1}{4\pi\times 10^{-7} } AT/ m
F_{g}=\frac{1}{4\pi \times 10^{-7} }\times 1\times 10^{-3} =796 AT
Central limb

B_{c}=B_{g}=1 T

From Fig. 2.16
H_{c} =1000 AT/ m
F_{c} =1000\times 160\times 10^{-3} =160 AT

Because of symmetry, flux divides equally between the two outer limbs. So

\phi(outer limb)=0\cdot 8/ 2=0\cdot 4mWb
B(outer limb)=\frac{0\cdot 4\times 10^{-3} }{600\times 10^{-6}  }=0\cdot 667AT
F(outer limb)=375\times 400\times 10^{-3} =150 AT
F (total)=796+ 160+ 150=1106 AT
Exciting current =1106/ 500=2\cdot 21 A