Resonance

Series Resonant Circuits

Consider the series resonant circuit shown in Fig. 1. We can calculate the total impedance seen by the source $v_{S}$, as:

$Z=R_{L}+j\omega L+{\frac {1}{j\omega C}}$

(1)

Combining the imaginary terms of the impedance, we get:

$Z=R_{L}+j\left(\omega L-{\frac {1}{\omega C}}\right)$

(2)

We can see that the imaginary component of the impedance becomes zero at the resonant frequency, $\omega _{0}$, equal to:

$\omega _{0}={\frac {1}{\sqrt {LC}}}$

(3)

Note that the cancellation is narrowband, since perfect cancellation occurs only at a single frequency. Thus, from the point of view of the voltage source $v_{S}$, the impedance $Z$ is purely real. We can then calculate the current, $i_{S}$ as:

$i_{S}={\frac {v_{S}}{R_{L}}}$

(4)

We then use this current to calculate the voltage across the inductor and capacitor:

$v_{L}=i_{S}\cdot j\omega _{0}L={\frac {v_{S}}{R_{L}}}\cdot j\omega _{0}L=v_{S}\cdot jQ_{S}$

(5)

$v_{C}=i_{S}\cdot {\frac {1}{j\omega _{0}C}}=-{\frac {v_{S}}{R_{L}}}\cdot j{\frac {1}{\omega _{0}C}}=-v_{S}\cdot jQ_{S}$

(6)

Where $Q_{S}$ is the series circuit quality factor:

(7)

Note that:

1. For small values of $R_{L}$ results in large values of $Q$, leading to large voltages across the inductor and capacitor.
2. The voltage across the inductor is exactly out of phase with the capacitor voltage, thus cancelling each other out.