A) 0.044 H
B) 0.065 H
C) 80 H
D) 0.08 H
Correct Answer: B
Solution :
[b] Here \[i\,=\frac{e}{\sqrt{{{R}^{2}}+X_{L}^{2}}}=\frac{e}{\sqrt{{{R}^{2}}+{{\omega }^{2}}{{L}^{2}}}}=\frac{e}{\sqrt{{{R}^{2}}+4{{\pi }^{2}}{{v}^{2}}\,{{L}^{2}}}}\] \[10\,=\frac{220}{\sqrt{64+4{{\pi }^{2}}\,{{(50)}^{2}}\,L}}\] \[\left[ \because \,\,R=\frac{V}{I}=\frac{80}{10}=8 \right]\] On solving we get L = 0.065 H
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