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Q.
What is the value of inductance $L$ for which the current is maximum in a series $L C R$ circuit with $C=10\, \mu F$ and $\omega=1000 \,s ^{-1} ?$
Alternating Current
Solution:
In resonance condition, maximum current flows in the circuit.
Current in $L C R$ 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 $L=\frac{1}{\omega^{2} C} \ldots$ (ii)
Given, $\omega=1000 \,s ^{-1}, C=10 \,\mu F =10 \times 10^{-6} F$
Hence, $L=\frac{1}{(1000)^{2} \times 10 \times 10^{-6}}$
$=0.1\, H =100 \,mH$