Question

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.

Short answer

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

Step-by-step solution

Given

• 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.

Determining the concept

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.

Determining that the series combination of the circuit has the same resonant frequency as separate circuits

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.