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. $10\text{ }mH$

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, $\omega =1000\,{{S}^{-1}},C=10\,\mu F=10\times {{10}^{-6}}\,F$ Hence, $L=\frac{1}{{{(1000)}^{2}}\times 10\times {{10}^{-6}}}$ $=0.1\,H$ $=100\,mH$