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 Z = 5.4 ohms E / Z = I = 5.57A
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
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
here is a problem I'm working on now with the answer. I dont understand how you got Z t = 7.376<48.56
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.
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.