A series circuit with a resistor–inductor-capacitor combination R₁, L₁, C₁has the same resonant frequency as a second circuit with a different combination R₂, L₂, C₂. You now connect the two combinations in series. Show that this new circuit has the same resonant frequency as the separate circuits.

• The first series circuit with the resistor-inductor-capacitor combination is R₁, L₁, C₁
• The second series circuit with the resistor-inductor-capacitor combination is R₂, L₂, C₂
• Two circuits have the same resonant frequency.

Use the concept of resonance. The resonance is the current amplitude in a RLCseries circuit driven by a maximum sinusoidal external emf when the driving angular frequency ${\mathit{\omega }}_{\mathbf{d}}$equals the natural angular frequency$\mathit{\omega }$of the circuit.

The formulae are as follows:

$$\begin{gathered} X_C=\frac{1}{{\omega }_{d}C} \\ {X}_L={\omega }_{d}L \end{gathered}$$

Where, $\omega$ is angular frequency, Cis capacitance.

The series combination of the circuit has the same resonant frequency as the separate circuits:

According to the resonance condition, the inductive reactance equals the capacitive reactance. Then,

X_C = X_L

The capacitive reactance is,

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

The inductive reactance is$X_L=\omega L$

For the first series circuit,

$\omega L_1=\frac{1}{\omega C_1}$

For the second series circuit,

$\omega L_2=\frac{1}{\omega C_2}$

The resonance $\omega$values are the same for both circuits.

For the series combination of the resonance, add these equations as

$$\begin{gathered} \omega L_1+\omega L_2=\frac{1}{\omega C_1}+\frac{1}{\omega C_2} \\ \omega \left(L_1+L_2\right)=\frac{1}{\omega }\left(\frac{1}{C_1}+\frac{1}{C_2}\right)...........\left(1\right) \end{gathered}$$

According to the series combination of the inductance,

$L_{eq}=L_1+L_2$

And the series combination of the capacitor,

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$

The equation (1) becomes

$\omega L_{eq}=\frac{1}{\omega }\frac{1}{C_{eq}}$

This is the resonance of the combined circuit.

Hence, it is proved that the new circuit has the same resonant frequency as the separate circuit.