Question

What is the value of inductance $L$ for which the current is 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)

In resonance condition, maximum current flows in the circuit.
Current in $LCR$ series circuit,
$i=\frac{V}{\sqrt{R^2+{\left(X_L-X_C\right)}^2}}$
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,
$X_L=X_C$
This happens in resonance state of the circuit, i.e.,
$\omega L=\frac{1}{\omega C}$
or $L=\frac{1}{{\omega }^{2}C}\dots$ (ii)
Given, $\omega =1000s^{-1},C=10\mu F=10\times 10^{-6}F$
Hence, $L=\frac{1}{(1000{)}^2\times 10\times 10^{-6}}$
$=0.1H=100mH$