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Chapter 2

Q. 2.1

The magnetic circuit of Fig. 2.4(a) has dimensions:A_{c} =4\times 4cm^{2} ,l_{g} =0\cdot 06cm, l_{c} =40 cm; N=600  turns. Assume the value of\mu _{r} =6000 for iron. Find the exciting current forB_{c} =1\cdot 2T and the corresponding flux and flux linkages.

Step-by-Step

Verified Solution

From Eq. (2.9),

Ni= H_{c} l_{c}+ H_{g}l_{g} ,

Ni=\frac{B_{c} }{\mu _{c} }l_{c} + \frac{B_{g} }{\mu _{0} }l_{g}

the ampere-turns for the circuit are given by Ni=\frac{B_{c} }{\mu _{o}\mu _{r} }l_{c} + \frac{B_{g} }{\mu _{0} }l_{g}

Neglecting fringingA_{c}= A_{g} therefore B_{c}= B_{g}

Then i=\frac{B_{c} }{\mu _{0}N } \left\lgroup\frac{l_{c} }{\mu _{r} }+ l_{g} \right\rgroup =\frac{1\cdot 2}{4\pi \times 10^{- 7}\times 600 } \left\lgroup\frac{40}{6000}+ 0\cdot 06 \right\rgroup \times 10^{- 2} =1\cdot 06A

The reader should note that the reluctance of the iron path of 40 cm is only \left\lgroup\frac{2/ 3}{6} \right\rgroup =0\cdot 11 of the reluctance of the 0\cdot 06 cm air-gap.

\phi =B_{c}A_{c}=1\cdot 2\times 16\times 10^{-4} =19\cdot 2\times 10^{-4} Wb

Flux linkages, \lambda =N\phi =600\times 19 \cdot 2\times 10^{-4} =1\cdot 152Wb- truns

If fringing is to be taken into account, one gap length is added to each dimension of the air-gap constituting the area.
Then A_{g} =\left(4+ 0\cdot 06\right) \left(4+ 0\cdot 06\right) =16\cdot 484 cm^{2}

Effective A_{g} \gt A_{c} reduces the air-gap reluctance. Now

B_{g} =\frac{19\cdot 2\times 10^{-4} }{16\cdot 484\times 10^{-4} } =1\cdot 165T

From Eq. (i)

Ni=\frac{B_{c} }{\mu _{o}\mu _{r} }l_{c} + \frac{B_{g} }{\mu _{0} }l_{g}

i=\frac{1}{\mu _{0}N } \left\lgroup\frac{B_{c}l_{c} }{\mu _{r} }+ B_{g}l_{g} \right\rgroup =\frac{1}{4\pi \times 10^{-7}\times 600 } \left\lgroup\frac{1\cdot 2\times 40\times 10^{-2} }{6000}+ 1\cdot 165\times 0\cdot 06\times 10^{-2} \right\rgroup =1\cdot 0332A