A) \[100\,mH\]
B) \[1\,mH\]
C) Cannot be calculated unless R is known
D) \[10\text{ }mH\]
Correct Answer: A
Solution :
Key Idea 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 rms 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 ie, \[\omega L=\frac{1}{\omega C}\] 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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