# AC Series Parallel ?

Discussion in 'General Electronics Chat' started by netwillie, Aug 14, 2010.

1. ### netwillie Thread Starter New Member

Aug 10, 2010
11
3
I'm working through the Vol. II - AC
On page 121 the answer to my calculation for Z(R//(l--C2) just do not match up to the value shown. has any one found this or am I missing something?
Any in put will be greatly appreciated.

Willie

2. ### debjit625 Well-Known Member

Apr 17, 2010
790
186
Yes that is ok, the equation is Z = C1--[R1//(L1--C2)].

Good Luck

3. ### netwillie Thread Starter New Member

Aug 10, 2010
11
3
I agree with Z = C1--[R1//(L1--C2)]
My Problem is with Z[R1//(L1--C2)] = 1/[(1/Zr) +(1/Z(L--c2))]
My answer for Z of this component of the circuit is 246.7/_-9.32degrees or 243.44-j39.95

4. ### debjit625 Well-Known Member

Apr 17, 2010
790
186
No ,
The values of (Using this "\" as angle symbol )
R1 = 470 \ 0 deg
L1 = 245.04 \ 90 deg
C2 = 1768 \ -90 deg
So
1/(1/R1 + 1/(L1+C2))
1\0 /(1\0 / 470\0) + (1\0 /(245.04 \ 90) + (1768 \ -90))
1\0 /(0.002128\0) + (1\0 /(245.04 \ 90) + (1768 \ -90))
1\0 /(0.002128\0) + (1\0 /(1522.96\-90))
1\0 /(0.002128\0) + (0.0006566\90)
1\0 /0.002227\17.14
449.04\-17.14

Hope this helps

5. ### netwillie Thread Starter New Member

Aug 10, 2010
11
3
My Mistake - I recorded the wrong R value,
Thank You for taking the time to look at this for me.

Have a Great Day

Willie