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., |
| \[\omega L=\frac{1}{\omega C}\] |
| or \[L=\frac{1}{{{\omega }^{2}}C}\] (i) |
| Given, \[\omega =1000\,{{s}^{-1}},\,C=10\,\mu F=10\times {{10}^{-6}}\,F\] |
| Hence, \[L=\frac{1}{{{(1000)}^{2}}\times 10\times {{10}^{-6}}}\] |
| \[=0.1\text{ }H\] |
| \[=100\text{ }mH\] |
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