A) 0.1 H
B) 0.01 H
C) 0.2 H
D) 0.02 H
Correct Answer: A
Solution :
Angular velocity, \[{{\omega }_{0}}=2\pi \,n=2\pi \times 100\] \[{{\omega }_{0}}=2\times 3\times 100=600\,rad/s\]\[(\because \pi =3)\] Further \[{{\omega }_{0}}=\frac{1}{\sqrt{LC}}\] ... (i) Also \[{{X}_{C}}=\frac{1}{C{{\omega }_{0}}}=60\,\Omega \] \[\Rightarrow \] \[C=\frac{1}{{{\omega }_{0}}\times 60}=\frac{1}{600\times 60}\] \[\Rightarrow \] \[C=\frac{1}{36\times {{10}^{3}}}F\] So, put values in Eq. (i), we get \[600=\frac{1}{\sqrt{L\left( \frac{1}{36\times {{10}^{3}}} \right)}}\] \[36\times {{10}^{4}}=\frac{36\times {{10}^{3}}}{L}\] \[\Rightarrow \] \[L=\frac{36\times {{10}^{3}}}{36\times {{10}^{4}}}=\frac{1}{10}=0.1H\]
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