A) 1/8
B) ¼
C) 1/2
D) 1
Correct Answer: B
Solution :
Given: Self-inductance of first coil \[{{L}_{1}}=4\,mH\] Self-inductance of second coil \[{{L}_{2}}=1\,mH\] From the relation induced emf \[e=L\frac{di}{dt}\] Hence, \[{{e}_{1}}={{L}_{1}}\frac{d{{i}_{1}}}{dt}\]and \[{{e}_{2}}={{L}_{2}}\frac{d{{i}_{2}}}{dt}\] As the power given are same \[{{e}_{1}}={{e}_{2}}\] or \[{{L}_{1}}=\frac{d{{I}_{1}}}{dt}={{L}_{2}}\frac{d{{I}_{2}}}{dt}\] or \[\frac{{{L}_{1}}}{{{L}_{2}}}=\frac{d{{I}_{2}}}{d{{I}_{1}}}=\frac{{{I}_{2}}}{{{I}_{1}}}\] or \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{L}_{2}}}{{{L}_{1}}}=\frac{1}{4}\]You need to login to perform this action.
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