In the RLC series circuit shown, the amplitude of the source is Em = 200 V opera
ID: 1399968 • Letter: I
Question
In the RLC series circuit shown, the amplitude of the source is Em = 200 V operating at a driving angular frequency, Wd = 8000 rad/s.
L = 0.25 H
R1 = 2000 (omega)
R2 = 500 (omega)
C1 = C2 = 0.5 uF
A) what is the current amplitude, Im (the correct answer is 68.6 i just need to see the work of how it is done)
B) what is the amplitude Vr1, m of the voltage across the resistor, R1? (the correct answer is 137 V i need to see the work of how it is done)
C) What is the average power delivered by the emf? (the correct answer is 5.88 W i need to see the work though)
2 2Explanation / Answer
Cureent amplitute I=V/Z
C1 and C2 in series so
!/Ceq=1/0.5+1/0.5
Ceq=0.25uF
R1 and R2 are in series so
Req=R1+R2
Req=2500ohm
XL=wL=8000*0.25=2000
XC=1/wC=1/8000*0.25*10^-6
XC=500
so Z=undrt(R^2+(XL-Xc)^2)
=undrt(2500^2+(2000-500)^2
Z=2915.45
I=V/Z=200/2915.45
I=0.0686
I=68.6mA--answer to a
b)V=IR
V=0.686*2000
V=137.1V--answer to b
c)P=I^2 R
=(Vrms/ Z)^2R
=(V/undrt(2)/Z)^2 R
=V^2R/2Z^2
=200^2*2500/2*(2915.45)^2
=5.88watt--answer to c