no, doing so would mean R2 and Xc are in series which they are notOriginally posted by davidand@Jun 25 2005, 10:22 AM
I'm trying to solve this for Zt and It
do I add R1 and R2 for Rt
do I subtract Xc-XL =x
then add Rt squared + X squared
then take sqt root of result to equal Z
Should I solve for R1 and XL then R2 and XcOriginally posted by kinyo@Jun 25 2005, 01:24 PM
no, doing so would mean R2 and Xc are in series which they are not
[post=8713]Quoted post[/post]
Originally posted by davidand@Jun 25 2005, 04:33 PM
Should I solve for R1 and XL then R2 and Xc
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To correct myself......Originally posted by Semyazza@Jun 25 2005, 11:11 PM
3@0 + 6@90 + [(2@0 * 8@-90) / (2@0 + 8@-90)] = Ztotal = 7.376@48.56
7.376@48.56 / 30@0 = Itotal = 245.88mA @48.556
OR............
[(2 * -j8)/ (2 + -j8)] + j6 + 3 = Ztotal
Ztotal / 30 = Itotal
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What chapter of circuit analysis should I reference to learn how to solve thisOriginally posted by Semyazza@Jun 26 2005, 12:31 AM
To correct myself......
3@0 + 6@90 + [(2@0 * 8@-90) / (2@0 + 8@-90)] = Ztotal = 7.376@48.56
30@0 / 7.376@48.56 = Itotal = 4.07 @ -48.56
OR............
[(2 * -j8)/ (2 + -j8)] + j6 + 3 = Ztotal
30 / Ztotal = Itotal
[post=8730]Quoted post[/post]
here is a problem I'm working on now with the answer.Originally posted by Semyazza@Jun 26 2005, 12:31 AM
To correct myself......
3@0 + 6@90 + [(2@0 * 8@-90) / (2@0 + 8@-90)] = Ztotal = 7.376@48.56
30@0 / 7.376@48.56 = Itotal = 4.07 @ -48.56
OR............
[(2 * -j8)/ (2 + -j8)] + j6 + 3 = Ztotal
30 / Ztotal = Itotal
[post=8730]Quoted post[/post]
Originally posted by davidand@Jun 26 2005, 02:49 PM
here is a problem I'm working on now with the answer.
I dont understand how you got Z t = 7.376<48.56
[post=8747]Quoted post[/post]
If I plug my numbers in above equation I get 10.6 not 7.376Originally posted by Semyazza@Jun 26 2005, 05:54 PM
Celect: I just looked at the book and I would read Volume II Chapters 2,3,4,5 you can skip chapter 2 if you already are familiar with complex numbers (it dosn't hurt to review though). Chapter 5 puts everything all together so it can be skipped but its nice to know (and important as far as AC circuits go).
Davidand:
A resistor is always at 0˚ an inductor at 90˚(or j) and a capacitor at -90˚(or -j). R2 and Xc are in parallel because two different currents pass through them (Components are in series if the same current passes through them otherwise they are in parallel). R1 and XL are in series with the parallel combination so they are added.
(R2 * Xc) / (R2 + Xc) + R1 + XL = Ztotal
If you have any other questions just ask.
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When dealing with AC circuits, inductors, and capacitors in terms of impedance you are adding vectors. When adding vectors you can not simply add, multiply, and subtract the way you are doing, you must add, multiply, subtract vectors. http://en.wikipedia.org/wiki/Phasor_%28electronics%29 is a good starting place.Originally posted by davidand@Jun 26 2005, 06:53 PM
If I plug my numbers in above equation I get 10.6 not 7.376
(2*8) / (2+8) + 3 + 6 =
[post=8757]Quoted post[/post]
by Jake Hertz
by Jake Hertz