Resonance In Series LCR Circuit MCQ - Practice Questions with Answers

Resonance in Series LCR circuit-

As we have discussed that when- 

     

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

then $\tan \phi$ is zero i.e. phase angle $(\phi )$ 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 impedance become -

$$Z=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}=R$$

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,

$$\omega =\frac{1}{\sqrt{LC}}$$

As, $\omega =2\pi f_o$. Where $f_o$ is the frequency of applied voltage.

                                   

                                                                So,

                                                                           

$$f_o=\frac{1}{2\pi \sqrt{LC}}$$

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 i₀) 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 the applied frequency is lesser to resonant frequency f₀,the current is high otherwise

2. If R is large, the curve is broadsided which means that those is limited change in current for resonance and non-resonance conditions

                  

Note -                                

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

$$\begin{array}{rl}{X}_L=X_c& ={\omega }_{0}L=\frac{1}{{\omega }_{0}c}\\ {\nu }_0& =\frac{1}{2\pi \sqrt{Lc}}(Hz)\end{array}$$