RCL Series Circuit Question

Thread Starter

wildnixon

Joined May 1, 2008
12
I am having trouble figuring out the phasor voltage and phasor current of an RCL series circuit. I am given the following values: v(t)=10sin(wt) (that w is a small Omega) and that the value of w is 1000 rad/sec. The inductance is 120mH, the capacitance= 1 micro Farad, and the resistance =100 ohms. I have calculated the frequency given the omega value, but I end up with a really small number : 6.3x10-3, and then I end up with a really small XL number and a really huge XC!! I know the math once I get the value of the frequency right, but I am just beating my head against the wall here!!
Thanks for any help!
D
 

mrmount

Joined Dec 5, 2007
59
I am not sure how you came up with that value for w. w=2*Pi*f; f=w/2*pi. So you will be getting 159.23 Hz! The value you have found is 'T' or the time taken for one complete cycle of the sine wave.
 

silvrstring

Joined Mar 27, 2008
159
wildnixon,

omega = 2*pi*f. So f = omega/(2*pi) = 159.2Hz.
X(L) = 2*pi*f*L, and will be at 90 degrees (or * j).
X(c) = 1/(2*pi*f*C), and will be at -90 degrees (or * -j).
Z = R + jX(l) - jX(c) = sqrt(R^2 + X^2) at angle tan^(-1) (X/R).
10V is your peak Voltage.
If you are keeping with peak voltage values, your pk V value is 10V at 0 degrees.
I = V / Z.
Subtract your denominator angle from your numerator angle to obtain your current angle (phase).
example (10V at 0 degrees/500 ohms at 60 degrees = 20 mA at -60 degrees). Not the answer by the way!

Hope this helps.
 

zamansabbir

Joined May 27, 2008
15
XL=w*L (w=small omega)

Xc=1/(w*c)
R= 100 ohm
Hence Z=R+J(XL-Xc) ohm
in phasor form Z= sqrt(R^2+(XL-Xc)^2)< arc tangent of((XL-Xc)/R)
then use V= I*Z;
to get the current through every component and then use voltage divider rule or same ohms law to get the voltage phasor
Hope u understand
 
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