# Solve for Z

#### davidand

Joined Jun 2, 2005
43
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

#### kinyo

Joined Jun 6, 2005
13
Originally 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
no, doing so would mean R2 and Xc are in series which they are not

#### davidand

Joined Jun 2, 2005
43
Originally 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]​
Should I solve for R1 and XL then R2 and Xc

#### celect

Joined May 31, 2005
18
Originally posted by davidand@Jun 25 2005, 04:33 PM
Should I solve for R1 and XL then R2 and Xc
[post=8717]Quoted post[/post]​

I believe Z t = 3ohms + 6ohms + (2ohms) (8ohms)/2ohms+J8

#### Semyazza

Joined Jun 25, 2005
12
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

#### Semyazza

Joined Jun 25, 2005
12
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
[post=8728]Quoted post[/post]​
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

#### celect

Joined May 31, 2005
18
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]​
What chapter of circuit analysis should I reference to learn how to solve this

#### davidand

Joined Jun 2, 2005
43
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]​
here is a problem I'm working on now with the answer.

I dont understand how you got Z t = 7.376<48.56

#### Semyazza

Joined Jun 25, 2005
12
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]​

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.

#### davidand

Joined Jun 2, 2005
43
Originally 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.
[post=8749]Quoted post[/post]​
If I plug my numbers in above equation I get 10.6 not 7.376

(2*8) / (2+8) + 3 + 6 =

#### Semyazza

Joined Jun 25, 2005
12
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]​
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.