A) 5H
B) 1 H
C) 10 H
D) \[4\centerdot 5H\]
Correct Answer: B
Solution :
The current in the inductor coil is given by \[{{I}_{0}}=\frac{{{E}_{0}}}{{{X}_{L}}}=\frac{\sqrt{2}{{E}_{v}}}{2\pi vL}\] \[L=\frac{\sqrt{2}{{E}_{v}}}{2\pi v{{I}_{0}}}=\frac{1.414\times 200}{2\times 3.14\times 50\times 0.9}=1\,H\]
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