Circuit containing Inductor and Capacitor in Series (L-C Series Circuit )

Mathematical Analysis of L-C Series Circuit :
Alternating Current Circuit Containing Inductor and Capacitor in series
Let us consider, a circuit containing inductor $L$ capacitor $C$ and these are connected in series. If an alternating voltage source is applied across it then the resultant voltage of the L-C circuit

$V=V_{L} - V_{C} \qquad(1)$

We know that:

$V_{L} = iX_{L}$
$V_{C} = iX_{C}$

So from equation $(1)$

$V= iX_{L} - iX_{C} $

$V=i \left(X_{L} - X_{C} \right) $

$\frac{V}{i}=\left(X_{L} - X_{C} \right) $

$Z=\left(X_{L} - X_{C} \right) \qquad(2)$

$Z \rightarrow$ Impedance of L-C circuit.
$X_{L} \rightarrow$ Inductive Reactance which has value $\omega L$
$X_{C} \rightarrow$ Capacitive Reactance which has value $\frac{1}{\omega C}$

So from equation $(2)$, we get

$Z=\left( \omega L - \frac{1}{\omega C} \right) \qquad(3)$

The phase of resultant voltage:
Phasor Diagram For L-C Circuit
The phase of resultant voltage from current is $90^{\circ}$ as shown in the figure above.

The Impedance and Phase at Resonance Condition:($X_{L} = X_{C}$):

At resonance $X_{L} = X_{C} \qquad(5)$

$\omega L = \frac{1}{\omega C}$

$\omega^{2} = \frac{1}{L C}$

$\omega = \sqrt{\frac{1}{L C}}$

$2 \pi f = \sqrt{\frac{1}{L C}}$

$ f = \frac{1}{2 \pi}\sqrt{\frac{1}{L C}}$

Where $f \rightarrow$ Natural frequency of the circuit.

1.) The Impedance of the circuit at resonance condition:

Substitute the resonance condition i.e. $X_{L} = X_{C}$ in equation $(2)$ then the impedance of the L-C Circuit


The impedance of the L-C circuit at resonance condition is zero.

2.) The Phase of resultant voltage at resonance condition:

There is not any change in the phase of resultant voltage at resonance condition i.e. that will be the same $90^{\circ}$.