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A series RLC circuit is driven by an ac source with a phasor voltage Vs=10∠30° V. If the circuit resonates at 10 3 rad/s and the average power absorbed by the resistor at resonance is 2.5W, determine that values of R, L, and C, given that Q =5.

Sagot :

Nuhulawal20idn

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

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