The applied voltage V divides into three parts, V_{L} (across L), V_{C} (across C) and V_{R} (across R) such that

** **We knowBy KVL, V = V_{R} + j (V_{L} – V_{C})

V = √ (V²_{R} + (V_{L} – V_{C})²)

V_{L} Voltage across Inductor

V_{R} Voltage across Resistor

V_{C} Voltage across Capacitor

The impedance of the circuit is

V_{R} = IR; V_{C} =\(\frac{1}{\omega C}\); V_{L} = IωL;

V = IZ

\(IZ=\sqrt{{{\left( IR \right)}^{2}}+{{I}^{2}}{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}\).

\(Z=\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}\).

But at some particular frequency both Inductive effect and Capacitor effect cancels each other and the circuit start oscillating. This is called resonance and the resonating frequency is

\(\omega L=\frac{1}{\omega C}\).

\({{\omega }^{2}}=\frac{1}{LC}~\).

\(\omega =\frac{1}{\sqrt{LC}}\Rightarrow f=\frac{i}{2\pi \sqrt{LC}}~\).

At this frequency the current in the circuit is maximum as impedance (Z) is minimum.