An L-R-C series circuit L = 0.124 H , R = 244 , and C = 7.27 F carries an rms cu
ID: 1494112 • Letter: A
Question
An L-R-C series circuit L = 0.124 H , R = 244 , and C = 7.27 F carries an rms current of 0.454 A with a frequency of 408 Hz .
Part A
What is the phase angle?
0.825
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Part B
What is the power factor for this circuit?
0.678
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Part C
What is the impedance of the circuit?
360
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Correct
Part D
What is the rms voltage of the source?
163
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Correct
Part E
What average power is delivered by the source?
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Part F
What is the average rate at which electrical energy is converted to thermal energy in the resistor?
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Part G
What is the average rate at which electrical energy is dissipated (converted to other forms) in the capacitor?
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Part H
What is the average rate at which electrical energy is dissipated (converted to other forms) in the inductor?
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An L-R-C series circuit L = 0.124 H , R = 244 , and C = 7.27 F carries an rms current of 0.454 A with a frequency of 408 Hz .
Part A
What is the phase angle?
=0.825
radiansSubmitMy AnswersGive Up
Answer Requested
Part B
What is the power factor for this circuit?
0.678
SubmitMy AnswersGive Up
Answer Requested
Part C
What is the impedance of the circuit?
Z =360
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Correct
Part D
What is the rms voltage of the source?
Vrms =163
VSubmitMy AnswersGive Up
Correct
Part E
What average power is delivered by the source?
P = WSubmitMy AnswersGive Up
Incorrect; Try Again; 3 attempts remaining
Part F
What is the average rate at which electrical energy is converted to thermal energy in the resistor?
PR = WSubmitMy AnswersGive Up
Part G
What is the average rate at which electrical energy is dissipated (converted to other forms) in the capacitor?
WSubmitMy AnswersGive Up
Part H
What is the average rate at which electrical energy is dissipated (converted to other forms) in the inductor?
WSubmitMy AnswersGive Up
Explanation / Answer
Part A) Phase angle = cos -1 (R/Z) = cos -1 (244/360) = 0.826 radians
Part b) power factor = R/Z = 244/360 =0.678
PART E) Average power = Irms*Erms *POWER FACTOR = 0.454 A * 0.454*360*244/360
= 50.29 W
part f) Rate t which electrical energy is converted to thermal energy in the resistor
= Irms^2*R=0.454 A * 0.454*244= 50.29W
PART g ) Rate is zero as capicitors do not dissipate power
PART h ) Rate is zero as Inductors do not dissipate power
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