An R L C circuit with R = 24.5 , L = 393 mH , and C = 45.5 F is connected to an
ID: 1648888 • Letter: A
Question
An RLC circuit with R = 24.5 , L = 393 mH , and C = 45.5 F is connected to an ac generator with an rms voltage of 30 V .
Part A
Determine the average power delivered to this circuit when the frequency of the generator is equal to the resonance frequency.
Express your answer using two significant figures.
Part B
Determine the average power delivered to this circuit when the frequency of the generator is twice the resonance frequency.
Express your answer using two significant figures.
Part C
Determine the average power delivered to this circuit when the frequency of the generator is half the resonance frequency.
Explanation / Answer
Given
RLC circuit with R = 24.5 , L = 393 mH , and C = 45.5 F , rms voltage is Vrms = 30 V
Part A
The average power delivered to this circuit when the frequency of the generator is equal to the resonance frequency.
that is at resonance reactance Z = R ===> P_avg = Vrms^2*R/Z^2 = Vrms^2*R/R^2 = Vrms^2/R = 30^2/24.5 W = 36.73 W
ParT B
The average power delivered to this circuit when the frequency of the generator is twice the resonance frequency.
that is W = 2(1/sqrt(LC)) = 2/sqrt(LC)
P_avg = Vrms^2*R/Z^2 , we know that the reactance of capacitor and inductor are
Xc = 1/(W*C) , XL = W*L and Z = sqrt(R^2+(XL-XC)^2)
substituting all these
P_avg = (Vrms^2*R)/(R^2+(W*L-1/W*C)^2)
= (Vrms^2*R)/(R^2+(2*L/sqrt(LC)-(sqrt(LC)/2*C))^2)
= (Vrms^2*R)/(R^2+(2sqrt(L/C)-(0.5*sqrt(L/C))^2)
= (Vrms^2*R)/(R^2+(1.5sqrt(L/C))^2)
= (30^2*24.5)/(24.5^2+(1.5 sqrt(0.393/(44.5*10^-6))))^2 W
=0.040 W
Part C
The average power delivered to this circuit when the frequency of the generator is half the resonance frequency
that is W = 0.5(1/sqrt(LC))
P_avg = (Vrms^2*R)/(R^2+(W*L-1/W*C)^2)
= (Vrms^2*R)/(R^2+(L/2*sqrt(LC)-2*(sqrt(LC)/C))^2
= (Vrms^2*R)/(R^2+(0.5*sqrt(L/C)-(2/sqrt(L/c)(c))^2)
= (30^2*24.5)/(24.5^2+(0.5 sqrt(0.393/(44.5*10^-6))-(2/(sqrt(0.393*44.5*10^-6)*44.5*10^-6)))^2) W
= 1.90*10^-10 W
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