AC circuit with three elements in series R= 100ohms C= 5.066microH and L= 5mH AC
ID: 2140817 • Letter: A
Question
AC circuit with three elements in series R= 100ohms C= 5.066microH and L= 5mH
AC Sourse is VtRLC=200Sin(2(pi)FT +A), where 200 is Vm and A=omega and It= Im(sin(2piFT) omega is the phase difference beteen VtRLC and current
A) For Frequency = 2kHz find Xl ,Xc ,Z, L , Im Vmr , Vml, Vmc then use the rms to find the average power in each element as wll as the total and Power factor.
B) Find the Resonance frequency of the ciruit and FInd Xl ,Xc ,Z, L , Im Vmr , Vml, Vmc at the resadance frequency.
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AC circuit with three elements in series R= 100ohms C= 5.066microH and L= 5mH AC Sourse is VtRLC=200Sin(2(pi)FT +A), where 200 is Vm and A=omega and It= Im(sin(2piFT) omega is the phase difference beteen VtRLC and currentExplanation / Answer
A)
R = 100 ohm
C = 5.066 uF
L = 5mH
f = 2kHz = 2000 Hz
Vt = 200*sin(2*pi*ft+A)
Xl = 2*pi*f*L = 2*3.14*2000*5*10^-3 = 62.8 ohm
Xc = 1/(2*pi*f*C) =1/(2*3.14*2000*5*10^-6) = 15.92 ohm
Z = sqrt(R^2 + (Xl-Xc)^2) = sqrt(100^2+(62.8-15.92)^2) = 110.4 ohm
So, Im = V/Z = 200/(110.4) = 1.81 A
Vmr = Im*R = 1.81*100 = 181 V
Vml = Im*Xl = 1.81*62.8 = 113.7 V
Vmc = Im*Xc = 1.81*15.92 = 28.8 V
avg power in R = Pr = Irms^2*R = (Im/sqrt(2))^2*R = Im^2*R/2
= 1.81^2*100/2 = 163.8 W
avg power in L = Pl = Vrms*Irms =Im*Vml/2 = 1.81*113.2/2 = 102.45 W
avg power in C = Pc = (Vrms,C)^2/Xc= Vmc^2/2Xc = 28.8^2/(2*15.92) = 26.1 W
power factor = cos(atan((Xl-Xc)/R)) = 0.91
Total power = Irms*Vrms = Im*Vm/2 = 1.81*200/2 = 181 W
resonance frequency, f = 1/2*pi*sqrt(LC) = 1/(2*3.14*sqrt(5*10^-3*5.066*10^-6))
= 1000.5 Hz
at resonance, Xl = 2*3.14*1000.5*5*10^-3 = 31.42 ohm
Xc = 31.42 ohm
Z = R = 100 ohm
Im = Vm/R = 200/100 = 2A
Vmr = 200 V
Vml = Im*Xl = 2*31.42 = 62.84 V
Vmc = -62.84 V
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