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)

100 mH

Answer 2

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, $L=\frac{1}{{{(1000)}^{2}}\times 10\times {{10}^{-6}}}$ $=0.1\,\,H$ $=100\,\,mH$