What is the value of inductance $$L$$ for which the current is maximum in a series $$LCR$$  circuit with $$C = 10\,\mu F$$   and $$\omega = 1000\,{s^{ - 1}}$$  ?

Solution :
Current in $$LCR$$  series circuit,
$$i = \frac{V}{{\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} }}$$
where, $$V$$ is $$rms$$  value of voltage $$R$$ 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, i.e. during the resonance of series $$LCR$$  circuit
$$\eqalign{ & {X_L} = {X_C}\,\,{\text{i}}{\text{.e}}{\text{.}}\,\,\omega L = \frac{1}{{\omega C}} \cr & {\text{or}}\,\,L = \frac{1}{{{\omega ^2}C}}\,\,......\left( {\text{i}} \right) \cr} $$
$$\eqalign{ & {\text{Given,}}\,\,\omega = 1000\,{s^{ - 1}},C = 10\,\mu F = 10 \times {10^{ - 6}}F \cr & {\text{Hence,}}\,\,L = \frac{1}{{{{\left( {1000} \right)}^2} \times 10 \times {{10}^{ - 6}}}} \cr & = 0.1\,H = 100\,mH \cr} $$