Answer :
Answer:
530.5 Hz
Explanation:
Given :
C=0.75 micro
L=120 mH
Reactance capacitor can be determined by
=[tex]\frac{1}{ wC}[/tex]
where c =capacitor
= [tex]\frac{1}{w*0.75*10^-6}[/tex]
Reactance inductor can be determined by
[tex]= w*L\\ = w*120*10^-3[/tex]
As they are in series so ,
[tex]\frac{1}{w*0.75*10^-6} =w*120*10^-3[/tex]
On solving these value we get
[tex]w=3333.33[/tex]
As we know that
f = [tex]\frac{w}{2*3.14}[/tex] =
530.5 Hz
The 530.5 Hz of frequency should be used to create a resonance condition.
Reactance capacitor is defined as the opposition to the voltage change across the element. It is inversely proportional to the angular frequency and the capacitance.
[tex]\bold {X_C = \dfrac 1{\omega C}}[/tex]
Where,
Xc = Reactance
[tex]\bold {\omega }[/tex] - angular frequency
C - capacitance.
so,
[tex]\bold {Xc = \dfrac 1 {\omega \times 0.75x10^-^6}}[/tex]
Reactance of the inductor,
[tex]\bold {X_L = \omega \times L}[/tex]
[tex]\bold {X_L = \omega \times 120x 10^-^3}[/tex]
Since, both are in series,
So,
[tex]\bold { \dfrac 1 {\omega \times 0.75x10^-^6} = \omega \times 120x 10^-^3} }}\\\\\bold {\omega = 3333.33}[/tex]
So, frequency
[tex]\bold {f = \dfrac w{2\times 3.14} = 530.5\ hz }[/tex]
Therefore, the 530.5 Hz of frequency should be used to create a resonance condition.
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