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Lcr circuit

In an L-C - R series circuit, R=150 , L=0.750 H and C=0.0180

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Solving time: 2 mins

In an $L - C$ - $R$ series circuit, $R = 150Ω, L = 0.750 H$ and $C = 0.0180 μ F$ . The source has voltage amplitude $V = 150 V$ and a frequency equal to the resonance frequency of the circuit.
(a) What is the power factor?
(b) What is the average power delivered by the source?
(c) The capacitor is replaced by one with $C = 0.0360 μ F$ and the source frequency is adjusted to the new resonance value. Then, what is the average power delivered by the source?

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Electricity and Magnetism (DC Pandey)

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Harper Discussed

In an

L - C

R

series circuit,

R = 150Ω, L = 0.750 H

and

C = 0.0180 μ F

. The source has voltage amplitude

V = 150 V

and a frequency equal to the resonance frequency of the circuit.
(a) What is the power factor?
(b) What is the average power delivered by the source?
(c) The capacitor is replaced by one with

C = 0.0360 μ F

and the source frequency is adjusted to the new resonance value. Then, what is the average power delivered by the source?

6 mins ago

Discuss this question LIVE

6 mins ago

Text solutionVerified

(a) At resonance frequency,

X_{L} = X_{C}, Z = R and power factor cos ϕ = \frac{R}{Z} = 1.0

(b)

P = \frac{V _{rms}^{2}}{R} = \frac{( 150/ 2 ) ^{2}}{150} = 75 W

P = \frac{V _{rms}^{2}}{R} = 75 W

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Question Text	In an $L - C$ - $R$ series circuit, $R = 150Ω, L = 0.750 H$ and $C = 0.0180 μ F$ . The source has voltage amplitude $V = 150 V$ and a frequency equal to the resonance frequency of the circuit. (a) What is the power factor? (b) What is the average power delivered by the source? (c) The capacitor is replaced by one with $C = 0.0360 μ F$ and the source frequency is adjusted to the new resonance value. Then, what is the average power delivered by the source?
Topic	Alternating Current
Subject	Physics
Class	Class 12
Answer Type	Text solution:1
Upvotes	141

Text solutionVerified

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