A) \[281.25\,mH\]
B) \[312.75\,mH\]
C) \[450.00\,mH\]
D) \[225.50\,mH\]
Correct Answer: A
Solution :
[a] \[\frac{1}{2}C{{V}^{2}}=\frac{1}{2}Li_{0}^{2}\] \[L=\frac{(0.5\times {{10}^{-6}}){{(150)}^{2}}{{(250)}^{2}}}{{{(50)}^{2}}}\] \[L=281.25mH\]You need to login to perform this action.
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