Series Parallel RLC example in Chap 5

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medimager

Joined Jan 3, 2005
1
To get Z of R//(L--C2), I used the

1
-------------------------------
1 1
--------- + -----------
ZR Z(L--C2)


where ZR = 1.523K <-90 degrees
Z(L--C2) = 470<0 degrees

I come up with an answer of 359<-90 degrees or 370- j14.073

What am I doing wrong? I can't figure out how you got your answer for this in Chap 5 Series Parallel RLC.
 
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