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Chapter 2

Q. 2.7

For the magnetic circuit of Fig.2.17 find the self and mutual inductances between the two coils.Core permeability = 1600.

Step-by-Step

Verified Solution

l_{1} =\left(6+ 0\cdot 5+ 1\right)\times 2+ \left(4+ 2\right)=21 cm
l_{2} =\left(3+ 0\cdot 5+ 1\right) \times 2+ \left(4+ 2\right)=15 cm
l_{0} =4+ 2=6 cm
R_{1} =\frac{21\times 10^{-2} }{4\pi \times 10^{-7}\times 1600\times 2\times 2\times 10^{-4} } =0\cdot 261\times 10^{6}
R_{2}=\frac{15\times 10^{-2} }{4\pi \times 10^{-7}\times 1600\times 2\times 2\times 10^{-4} } =0\cdot 187\times 10^{6}
R_{0}=\frac{6\times 10^{-2} }{4\pi \times 10^{-7}\times 1600\times 1\times 2\times 10^{-4} } =0\cdot 149\times 10^{6}

(i) Coil 1 excited with 1A

R=R_{1}+ R_{0} \parallel R_{2} =0\cdot 261+ 0\cdot 1871\parallel 0\cdot 149=0\cdot 344\times 10^{6}
\phi _{1} =\left(500\times 1\right) / \left(0\cdot 344\times 10^{6} \right) =1\cdot 453 mWb
\phi _{21}=\phi _{2} =1\cdot 453\times 0\cdot 149/ \left(0\cdot 149+ 0\cdot 187\right) =0\cdot 64 mWb
L_{11} =N_{1} \phi _{1} =500\times 1\cdot 453\times 10^{-3} =0\cdot 7265 H
M_{21}=N_{2} \phi _{21} =1000\times 0\cdot 649\times 10^{-3} =0\cdot 64 H

(ii) Coil 2 excited with 1A

R=R_{2} + \left(R_{0}R_{1} \right) / \left(R_{0}+ R_{1} \right) =\left[0\cdot 187+ \left(0\cdot 149\times 0\cdot 281\right)/ \left(0\cdot 149+ 0\cdot 281\right) \right]\times 10^{6} =0\cdot 284\times 10^{6}
\phi _{2} =\left(1000\times 1\right) / \left(0\cdot 284\times 10^{6} \right) =3\cdot 52 mWb
L_{22} =N_{2} \phi _{2} =1000\times 3\cdot 52\times 10^{-3} =3\cdot 52H
M_{12}= M_{12} (bilateral)=0\cdot 65H