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 i.e., 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 \[{{\phi }_{{{B}_{2}}}}={{\phi }_{{{B}_{2}}}}\] Since, \[{{M}_{12}}={{M}_{21}}\,=M,\] hence we have \[{{M}_{12}}\,{{M}_{21}}={{M}^{2}}=\frac{{{N}_{1}}\,{{N}_{2}}\,{{\phi }_{{{B}_{1}}}}\,{{\phi }_{{{B}_{2}}}}}{{{i}_{1}}\,{{i}_{2}}}={{L}_{1}}\,{{L}_{2}}\] \[\therefore {{M}_{mas}}=\sqrt{{{L}_{1}}\,{{L}_{2}}}\] Given, \[{{L}_{1}}=2\,mH,\,\,{{L}_{2}}=8\,mH\] \[\therefore {{M}_{\max }}=\sqrt{2\times 8}=\sqrt{16}=4\,mH\]
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