A) 10 mH
B) 6 mH
C) 4 mH
D) 16 mH
Correct Answer: C
Solution :
When the total flux associated with one coil links with the other ie, a case of maximum flux linkage, then \[{{M}_{12}}=\frac{{{N}_{2}}{{\phi }_{{{B}_{2}}}}}{{{i}_{1}}}\] and \[{{M}_{21}}=\frac{{{N}_{1}}{{\phi }_{{{B}_{1}}}}}{{{i}_{2}}}\] Similarly, \[{{L}_{1}}=\frac{{{N}_{1}}{{\phi }_{{{B}_{1}}}}}{{{i}_{1}}}\] and \[{{L}_{2}}=\frac{{{N}_{2}}{{\phi }_{{{B}_{2}}}}}{{{i}_{2}}}\] If all the flux of coil 2 links coil 1 and vice-versa then \[{{\phi }_{{{B}_{2}}}}={{\phi }_{{{B}_{1}}}}\] Since, \[{{M}_{12}}={{M}_{21}}=M\], hence we have \[{{M}_{12}}{{M}_{21}}={{M}^{2}}=\frac{{{N}_{1}}{{N}_{2}}{{\phi }_{{{B}_{1}}}}{{\phi }_{B2}}}{{{i}_{1}}{{i}_{2}}}={{L}_{1}}{{L}_{2}}\] \[\therefore \] \[{{M}_{\max }}=\sqrt{{{L}_{1}}{{L}_{2}}}\] Given, \[{{L}_{1}}=2mH,\,\,{{L}_{2}}=8mH\] \[\therefore \] \[{{M}_{\max }}=\sqrt{2\times 8}=\sqrt{16}=4mH\]
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