Resonance In Series LCR Circuit - Practice Questions & MCQ

Resonance in Series LCR circuit-

As we have discussed when-  

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

then $\tan \varphi$ is zero i.e. phase angle ( $\varphi$ ) is zero and voltage and current are in phase. We have called it electric resonance. So, if $X_L=X_C$, then the equation of impedence become -

$
Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}=R
$
So, we get minimum value of $Z$.
In this case impedance is purely resistive and minimum and currents has its maximum value. Now as -

$
\omega L=\frac{1}{\omega C}
$
So,

$
\omega=\frac{1}{\sqrt{L C}}
$
As, $\omega=2 \pi f_o$. Where $f_o$ is the frequency of applied voltage.

So,

$
f_o=\frac{1}{2 \pi \sqrt{L C}}
$
This frequency is called resonant frequency of the circuit.
Peak current in this case is given by -

$
i_o=\frac{V_o}{R}
$
We will now discuss about the resonance curve and its nature. We will show the variation in circuit current (peak current $i_0$ ) with change in frequency of the applied voltage -

This figure/graph shows the variation of current with the frequency. 

Conclusions from the graph -
1. If $R$ has small value, the resonance is sharp which means that if applied frequency is lesser to resonant frequency $\mathrm{f}_0$, the current is high otherwise
2. If $R$ is large, the curve is broad sided which means that those is limited change in current for resonance and non -resonance conditions

Note -
The natural or resonant frequency is Independent from resistance of the circuit.

$
\begin{aligned}
X_L=X_c & =\omega_0 L=\frac{1}{\omega_0 c} \\
\nu_0 & =\frac{1}{2 \pi \sqrt{L c}}(H z)
\end{aligned}
$