A) 100 mH
B) 1 mH
C) Cannot be calculated unless R is known
D) 10 mH
Correct Answer: A
Solution :
In resonance condition, maximum current flows in the circuit. Current in LCR series circuit. \[i=\frac{V}{\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}}\] Where V is mis value of current, R is resistance, \[{{X}_{L}}\]is inductive reactance and\[{{X}_{C}}\]is capacitive reactance. For current to be maximum, denominator should be minimum which can be done, if \[{{X}_{L}}={{X}_{C}}\] This happens in resonance state of the circuit, i.e., \[\omega L=\frac{1}{\omega C}\] or \[L=\frac{1}{{{\omega }^{2}}C}\] ?(i) Given, \[\omega =1000\,{{s}^{-1}},C=10\,\mu F=10\times {{10}^{-6}}F\] Hence, \[L=\frac{1}{{{(1000)}^{2}}\times 10\times {{10}^{-6}}}=0.1\,H=100\,mH\]
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