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Resonance In Series LCR Circuit - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Resonance in Series LCR circuit is considered one the most difficult concept.

  • 35 Questions around this concept.

Solve by difficulty

The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of

In a  LCR circuit capacitance is changed from C to 2C . For the resonant frequency to remain unchanged, the inductance should be changed from L to

In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 k \Omega with C = 2 \mu F. The resonant frequency \omega is 200 rad/s. At resonance the voltage across L is

Given below are two statements :
Statements I: An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor.
Statement II: An AC circuit containing a pure capacitor or a pure inductor consumes high power due to its non-zero power factor.

In the light of above statements, choose the correct answer from the options given below :

Concepts Covered - 1

Resonance in Series LCR circuit

Resonance in Series LCR circuit-

As we have discusssed that when- 

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

then tanφ is zero i.e. phase angle (φ) 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{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 applied frequency is lesser to resonant frequency f0,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.

                                     X_{L}=X_{c}= \omega _{0}L= \frac{1}{\omega _{0}c}

                                                    \nu _{0}= \frac{1}{2\pi \sqrt{Lc}}\left ( Hz \right )

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Resonance in Series LCR circuit

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