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In an L-R-C series circuit, 305 , 0.401 , and 5.98×10-8 . When the ac source operates at the resonance frequency of the circuit, the current
Question
In an L-R-C series circuit, 305 , 0.401 , and 5.98×10-8 . When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.491A
What is the voltage amplitude of the source?
What is the amplitude of the voltage across the resistor?
What is the amplitude of the voltage across the inductor?
What is the amplitude of the voltage across the capacitor?
What is the average power supplied by the source?
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Physics
3 years
2021-08-31T18:34:37+00:00
2021-08-31T18:34:37+00:00 1 Answers
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Answers ( )
Answer:
Explanation:
Given;
inductance, L = 305 H
Resistance , R = 0.401 Ω
Capacitance, C = 5.98 x 10⁻⁸ F
Amplitude current, I₀ = 0.491A
resonance frequency (F₀) = ?
Part (a) What is the voltage amplitude of the source?
V₀ = I₀ Z
Z is the impedance, since the circuit is in resonance Z = R
V₀ = I₀ Z = I₀ R = 0.491 x 0.401 = 0.197 V
Part (b) What is the amplitude of the voltage across the resistor?
V₀ = I₀ R = 0.491 x 0.401 = 0.197 V
Part (c) What is the amplitude of the voltage across the inductor?
Where XL is the inductive reactance
Part (d) What is the amplitude of the voltage across the capacitor?
Where XC is the capacitive reactance
Part (e) What is the average power supplied by the source?