Find the expression for the electrons resonance frequency using F=ma in equation the polarization p=-ner^ and the depolarization field would be E=-4πP/3
Answers
Answer:
The resonant frequency is the frequency of a circuit under resonant. A Resonant circuit is also known as the LC circuit or tank circuit. This circuit contains an inductor and capacitor attached parallel to each other. The resonant circuits are used to create a particular frequency or to select a particular frequency from a complex circuit.
The Formula for Resonant Frequency:
So, the resonant frequency formula is:
f0=12πLC√
Where f0 is the the resonant frequency is denoted as, the inductance is L and the capacitance is C
Derivation:
Let us consider a series connection of R, L and C. This series connection is excited by an AC source.
Let us first calculate the impedance Z of the circuit.
Z=R+jωL–jωC
Z=R+j(ωL–1ωC)
With the condition of resonance, the circuit is purely resistive. This means the imaginary part of the impedance Z will be zero during resonance condition or at a resonant frequency. You should always keep this in your mind while calculating the resonant frequency for a given circuit.
This means,
ωL–1ωC=0
ωL=1ωC
so,
ω2=1LC
Also
ω=12πf
i.e. f = 12πω
Substituting the values , we get
f0=12π(√LC)
Therefore, Resonant Frequency f_0 for the Resonance Circuit, will be
f0=12πLC√
Solved Examples
Q.1: An electrical circuit is given. Determine the the resonant frequency of this circuit. It has inductance of 25 mH , and capacitance as 5μF?
Answer:
The given parameters in the problem are:
L=50mH=50×10−3H
C=5μF=5×10−6F
The formula for resonant frequency is:
f0=12πLC√
Substituting the values in the above formula,
f0=12π50×10−3×5×10−6√
f0=12π×5×10−4
f0=12×3.14×5×10−4
f0 = 318.47 Hz
Therefore resonance frequency will be 318.47 Hz
Q.2: Determine the resonant frequency of a circuit whose value of inductance is 40 mH and capacitance is 8μF.
Solution:
L = 40 mH
L = 40×10−3H
C = 8μF=8×10−6F
To calculate the resonant frequency we will apply the formula as below:
f0=12πLC√
Substituting the values in the above formula,
f0=12×3.1440×10−3×8×10−6√
Thus after solving we get,
f0=281.35Hz
Therefore resonance frequency will be 281.35 Hz