show that the resonant frequency wo of
RLC series circuit is the geometric mean
of omega1,omega2 the lower and upper half power frequencies respectively
Answers
Explanation:
RLC Parallel Circuit Problems with Solutions
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Concept:
In the case of resonant frequency, Z= R, i.e, in the when series RLC circuit, the impedance, and resistance of the circuit are equal.
Given:
The resonant frequency.
Find:
The resonant frequency is the square root of the geometric mean of ω₁ and ω₂.
Solution:
When R-L-C are in series,
The impedance can be written as
Z² = R² + (Xc-Xi)²
In the case of resonant frequency, Z=R,
Xc-Xi = 0
Xc = Xi
(1/ω₀C) = ω₀L
ω₀² = 1/(LC)
ω₀= 1/(√LC)
Let, ω₁ be the angular frequency of the capacitor and ω₂ be the angular frequency of the inductor.
Xc = (1/ω₁C)
C = 1/(ω₁Xc)
Similarly, Xi= ω₂L
L = Xi/ ω₂
Substituting the value of L and C,
ω₀= 1/(√(1/(ω₁Xc)×Xi/ ω₂))
As Xc = Xi,
ω₀= 1/(√(1/(ω₁ω₂)) = √(ω₁ω₂)
Hence, the resonant frequency ω₀ = √(ω₁ω₂).
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