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 \[{{\phi }_{{{B}_{2}}}}={{\phi }_{{{B}_{1}}}}\] Since,\[{{M}_{12}}={{M}_{21}}=M,\] hence we have \[{{M}_{12}}{{M}_{21}}={{M}^{2}}=\frac{{{N}_{1}}{{N}_{2}}{{\phi }_{{{B}_{1}}}}{{\phi }_{{{B}_{2}}}}}{{{i}_{1}}{{i}_{2}}}={{L}_{1}}{{L}_{2}}\] \[{{M}_{\max }}=\sqrt{{{L}_{1}}{{L}_{2}}}\] Given, \[{{L}_{1}}=2\,mH,\,{{L}_{2}}=8\,mH\] \[{{M}_{ax}}=\sqrt{2\times 8}=\sqrt{16}=4\,mH\]
You need to login to perform this action.
You will be redirected in
3 sec
