Resonance

Series Resonant Circuits

Consider the series resonant RLC 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:

$Q_{S}={\frac {\omega _{0}L}{R_{L}}}={\frac {1}{\omega _{0}R_{L}C}}$

(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 $180^{\circ }$ out of phase with the capacitor voltage, thus cancelling each other out.

Parallel Resonant Circuits

For the parallel resonant RLC circuit in Fig. 2, the total admittance seen by the current source is:

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

(8)

The admittance becomes purely real at the resonant frequency, $\omega _{0}={\tfrac {1}{\sqrt {LC}}}$. Once again, we can calculate the voltage across $R_{L}$ as:

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

(9)

Thus, the currents through the inductor and capacitor can be expressed as:

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

(9)

(10)

Where the parallel quality factor, $Q_{P}={\tfrac {R}{\omega _{0}L}}=\omega _{0}R_{C}C$. Once again, we see that even though the currents across the reactances cancel out, the individual currents could be very large.