A) 100 mH
B) 1 mh
C) Cannot be calculated unless R is known
D) 10 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 i,e., \[\omega L=\frac{1}{\omega C}\] \[orL=\frac{1}{{{\omega }^{2}}C}....(i)\] 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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