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In a series R-L-C circuit, L = 0.190 H , C = 84.5 μF and the...

In a series R-L-C circuit, L = 0.190 H , C = 84.5 μF and the voltage amplitude of the source is 290 V .

Part A

What is the resonance angular frequency of the circuit?
ω0 =   rad/s  

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Part B

When the source operates at the resonance angular frequency, the current amplitude in the circuit is 0.610 A . What is the resistance R of the resistor?
R =   Ω  

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Part C

At the resonance frequency, what are the peak voltages across the inductor, the capacitor, and the resistor?

Enter your answer as three numbers separated with commas.
VL, VC, VR =   V  
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Answer #1

Solution Part A The condition for the resonance is inductive reactance is equal to capacitive reactance as T: o LC OA- LC 0.1290 0.610 475.4Ω Part C When the source operates at the resonance angular frequency the voltage across the Resistance R=- 290Capacitance is を-X 0.610 429.57x84.5x10 =168v

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