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Answer:
R = 20Ω
L = 0.1 H
C = 1 × 10⁻⁵ F
Explanation:
Given the data in the question;
Vs = 10∠30°V { peak value }
V"s[tex]_{rms[/tex] = 10/√2 ∠30° V
resonance freq w₀ = 10³ rad/s
Average Power at resonance Power[tex]_{avg[/tex] = 2.5 W
Q = 5
values of R, L, and C = ?
We know that;
Power[tex]_{avg[/tex] = |V"s[tex]_{rms[/tex]|² / R
{ resonance circuit is purely resistive }
we substitute
2.5 = (10/√2)² × 1/R
2.5 = 50 × 1/R
R = 50 / 2.5
R = 20Ω
We also know that;
Q = w₀L / R
we substitute
5 = ( 10³ × L ) / 20
5 × 20 = 10³ × L
100 = 10³ × L
L = 100 / 10³
L = 0.1 H
Also;
w₀ = 1 / √LC
square both side
w₀² = 1 / LC
w₀²LC = 1
C = 1 / w₀²L
we substitute
C = 1 / [ (10³)² × 0.1 ]
C = 1 / [ 1000000 × 0.1 ]
C = 1 / [ 100000 ]
C = 0.00001 ≈ 1 × 10⁻⁵ F
Therefore;
R = 20Ω
L = 0.1 H
C = 1 × 10⁻⁵ F