# Series Resonant Circuit

A resonant circuit (series or parallel) must have an inductive and a capacitive element. A resistive element will always be present due to the internal resistance of the source ($R_s$), the internal resistance of the inductor ($R_l$), and any added resistance to control the shape of the response curve (Rdesign ). The basic configuration for the series resonant circuit appears in Fig. 1(a) with the resistive elements listed above. The “cleaner” appearance of Fig. 1(b) is a result of combining the series resistive elements into one total value.
Fig. 1: Series resonant circuit.
That is, $$\bbox[10px,border:1px solid grey]{R = R_s + R_l + R_d} \tag{1}$$ The total impedance of this network at any frequency is determined by $$Z_T = R + j X_L - j X_C = R + j (X_L - X_C)$$ The resonant conditions described in the introduction will occur when $$\bbox[10px,border:1px solid grey]{X_L = X_C} \tag{2}$$ removing the reactive component from the total impedance equation. The total impedance at resonance is then simply $$\bbox[10px,border:1px solid grey]{Z_Ts =R } \tag{3}$$ representing the minimum value of $Z_T$ at any frequency. The subscript s will be employed to indicate series resonant conditions.
The resonant frequency can be determined in terms of the inductance and capacitance by examining the defining equation for resonance eq. 2: $$X_L = X_C$$ Substituting yields $$wL = { 1\over wC}$$ and $$w^2 = { 1\over LC}$$ and $$\bbox[10px,border:1px solid grey]{w = { 1\over \sqrt{LC}}} \tag{4}$$ or $$\bbox[10px,border:1px solid grey]{f_s = { 1\over 2 \pi \sqrt{LC}}} \tag{5}$$ The current through the circuit at resonance is $$I = {E \angle 0^\circ \over R \angle 0^\circ} = { E \over R} \angle 0^\circ$$ which you will note is the maximum current for the circuit of Fig. 1 for an applied voltage $E$ since $Z_T$ is a minimum value. Consider also that the input voltage and current are in phase at resonance.
Since the current is the same through the capacitor and inductor, the voltage across each is equal in magnitude but 180° out of phase at resonance: $$V_L = (I \angle 0^\circ)(X_L \angle 90^\circ) = IX_L \angle 90^\circ$$ $$V_C = (I \angle 0^\circ)(X_C \angle -90^\circ) = IX_L \angle -90^\circ$$ and, since $X_L = X_C$, the magnitude of $V_L$ equals $V_C$ at resonance; that is, $$\bbox[10px,border:1px solid grey]{V_{L_{s}} = V_{C_{s}}} \tag{5}$$
Fig. 2: Phasor diagram for the series resonant circuit at resonance.
Figure 2, a phasor diagram of the voltages and current, clearly indicates that the voltage across the resistor at resonance is the input voltage, and $E$, and $I$, and $V_R$ are in phase at resonance.
The average power to the resistor at resonance is equal to $I^2R$, and the reactive power to the capacitor and inductor are $I^2X_C$ and $I^2X_L$, respectively.
Fig. 3: Power triangle for the series resonant circuit at resonance.
The power triangle at resonance (Fig. 3) shows that the total apparent power is equal to the average power dissipated by the resistor since $Q_L = Q_C$. The power factor of the circuit at resonance is $$F_p = \cos \theta = { P \over S}$$ and $$F_p = 1$$ Plotting the power curves of each element on the same set of axes (Fig. 4), we note that, even though the total reactive power at any instant is equal to zero (note that $t = t1$), energy is still being absorbed and released by the inductor and capacitor at resonance.
Fig. 4: Power curves at resonance for the series resonant circuit
A closer examination reveals that the energy absorbed by the inductor from time 0 to t1 is the same as the energy released by the capacitor from 0 to t1. The reverse occurs from t1 to t2, and so on. Therefore, the total apparent power continues to be equal to the average power, even though the inductor and capacitor are absorbing and releasing energy. This condition occurs only at resonance. The slightest change in frequency introduces a reactive component into the power triangle, which will increase the apparent power of the system above the average power dissipation, and resonance will no longer exist.