# RCL Series Circuit Question

Discussion in 'Homework Help' started by wildnixon, May 27, 2008.

1. ### wildnixon Thread Starter Member

May 1, 2008
12
0
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

2. ### mrmount Active Member

Dec 5, 2007
59
7
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.

3. ### wildnixon Thread Starter Member

May 1, 2008
12
0
I will proceed to use the 159 Hz...I see now where I botched it...it's always something simple...
Thanks!

4. ### silvrstring Active Member

Mar 27, 2008
159
0
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).
If you are keeping with peak voltage values, your pk V value is 10V at 0 degrees.
I = V / Z.
example (10V at 0 degrees/500 ohms at 60 degrees = 20 mA at -60 degrees). Not the answer by the way!

Hope this helps.

5. ### zamansabbir Member

May 27, 2008
15
0
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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