Question

What is the value of inductance L for which the current is a maximum in a series LCR circuit with $ C=10\,\mu F $ and $ \omega =1000\,{{s}^{-1}} $ ?

• A. 100 mH
• B. 1 mH
• C. Cannot be calculated unless R is known
• D. 10 mH

Answer 1

Answer: (A)

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 =100{{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 $