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 $\omega =1000s^{-1}$ ?

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