A) 1 : 2
B) 2 : 1
C) 4 : 1
D) 1 : 4
Correct Answer: D
Solution :
\[E=-L\frac{dI}{dt};W=\frac{1}{2}L{{I}^{2}}\] We know that \[E=-\frac{LdI}{dt}\]As\[\frac{dI}{dt}\] is same, Hence, \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{L}_{1}}}{{{L}_{2}}}=4\] Power is constant so \[{{E}_{1}}{{I}_{1}}={{E}_{2}}{{I}_{2}};\] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{I}_{2}}}{{{I}_{1}}}=4\]We know that \[W=\frac{1}{2}L{{I}^{2}};\] \[\therefore \]\[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{{{L}_{1}}I_{1}^{2}}{{{L}_{2}}I_{1}^{2}}=4\times \frac{1}{16}=\frac{1}{4}\]You need to login to perform this action.
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