Question

What is a resonant electrical circuit? What are its types? Find out an expression for resonant frequency for series LCR circuit.

Answer 1

Hint: LCR circuit consists of three elements: Inductor, Capacitor, and Resistor. These three elements can either be connected in series or in parallel. We will find the equation of current through the LCR circuit and the expression for resonant frequency using the phenomenon of maximum current passing through the circuit at the resonant frequency.

Formula used:
$Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-\dfrac{1}{{{X}_{C}}} \right)}^{2}}}$

Complete step by step answer:
Electrical resonance occurs in an electric circuit at a particular resonant frequency when the impedances or admittances of circuit elements cancel each other. Resonance is the phenomenon in the electrical circuit, where the output of the electrical circuit is maximum at one particular frequency. That particular frequency is known as the resonant frequency. At the resonant frequency, the capacitive reactance and inductive reactance are equal.
In an alternating current, if the phase of the applied potential voltage difference and the current flowing in the circuit are the same, then the circuit is called a resonance circuit. The phenomenon shown by these circuits is called resonance.

Types of LCR circuits:
Series resonance circuit: The three circuit elements; Resistor, Inductor, and Capacitor are connected in series in a circuit.
Parallel resonance circuit: The three circuit elements; Resistor, Inductor, and Capacitor are connected in parallel in a circuit.

Series LCR circuit:

Resonant frequency for Series LCR circuit:
We know that impedance of circuit is given by,
$Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-\dfrac{1}{{{X}_{C}}} \right)}^{2}}}$
Where,
$R$ is the resistance
${{X}_{L}}$ is the inductive reactance
${{X}_{C}}$ is the capacitive reactance
If $V$ is the potential difference and $I$ is the current and $\phi $ is the phase difference, then,
$\tan \phi =\dfrac{{{X}_{L}}-{{X}_{C}}}{R}$
Putting,
$\begin{align}
  & {{X}_{L}}=\omega L \\
 & {{X}_{C}}=\dfrac{1}{\omega C} \\
\end{align}$
$\tan \phi =\dfrac{\omega L-\dfrac{1}{\omega C}}{R}$
Or,
$\phi ={{\tan }^{-1}}\left( \dfrac{\omega L-\dfrac{1}{\omega C}}{R} \right)$
In the LCR circuit impedance, the current flowing in the circuit is given as,
$I={{I}_{o}}\sin \left( \omega t-\phi \right)$
Where,
${{I}_{o}}=\dfrac{{{V}_{o}}}{Z}$
Putting value of $Z$in the above equation,
${{I}_{o}}=\dfrac{{{V}_{o}}}{\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}}$
Putting,
$\begin{align}
  & {{X}_{L}}=\omega L \\
 & {{X}_{C}}=\dfrac{1}{\omega C} \\
\end{align}$
${{I}_{o}}=\dfrac{{{V}_{o}}}{\sqrt{{{R}^{2}}+{{\left( \omega L-\dfrac{1}{\omega C} \right)}^{2}}}}$
For ${{I}_{o}}$to be maximum, its denominator should be minimum,
That is,
The value of$\sqrt{{{R}^{2}}+{{\left( \omega L-\dfrac{1}{\omega C} \right)}^{2}}}$should be minimum.
Therefore,
$\omega L-\dfrac{1}{\omega C}=0$
$\omega L=\dfrac{1}{\omega C}$
And,
$Z=R$
Also,
${{I}_{o}}=\dfrac{{{V}_{o}}}{R}$
In the resonance condition, the value of impedance in the circuit is minimum and the current flowing through the circuit is maximum.
In the condition of resonance,
$\omega L=\dfrac{1}{\omega C}$
It means,
$\begin{align}
  & {{\omega }^{2}}=\dfrac{1}{LC} \\
 & \omega =\dfrac{1}{\sqrt{LC}} \\
\end{align}$
We know,
$\omega =2\pi f$
Therefore,
$f=\dfrac{1}{2\pi \sqrt{LC}}$
This is the expression for resonant frequency for series LCR circuits.

Note: Series LCR circuit means that the three elements: Resistor, Inductor and Capacitor are connected end to end in a circuit. It should be remembered that the resistance in an electrical circuit does not affect its resonant frequency. The resistor reduces the lag produced by the capacitor from $90$ degrees and it reduces the lead produced by the inductor from $90$ degrees. On the two sides, the effects are almost equal. This results in a cancelling out, and eventually the overall effect is the same as a pure LC tank resonance.