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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