1. ## More Complex Numbers

I am still working on these complex numbers. This one is giving me trouble:

Given the following:

$z_{1}=R_{1}+R+jwL$

$z_{2}=R_{2}$

$z_{3}=\frac{1}{jwC_{3}}$

$z_{4}=R_{4}+\frac{1}{jwC_{4}}$

Express R and L in terms of the constants R1, R2, C3 and C4.

We are also told that $z_{1}z_{3}=z_{2}z_{4}$

So this last bit is obviously the clue to solving the whole thing. I worked it out:

$\frac{R_{1}+R+jwL}{jwC_{3}}=R_{2}(R_{4}+\frac{1}{j wC_{4}})$

I did lots of algebraic manipulation after that, but nothing came of it. Would somebody please give me a hand?

(I may be crap at maths, but I'm getting good at LaTeX!)

Regards,

Evanator

2. Originally Posted by evanator

$\frac{R_{1}+R+jwL}{jwC_{3}}=R_{2}(R_{4}+\frac{1}{j wC_{4}})$
Given this is correct, I haven't actually checked!

To solve for $R$:

1. Multiply both sides by $jwC_3$

2. Take $jwL$ from both sides

3. Take $R_1$ from both sides

To solve for $L$:

1. Multiply both sides by $jwC_3$

2. Take $R_1$ from both sides

3. Take $R$ from both sides

d. Divide both sides by $jw$

Post what you get...

3. Thanks, man. It doesn't look right, but here is what I got:

$R_{1}+R+jwL=jwC_{3}(R_{2}R_{4}+\frac{R_{2}}{jwC_{4 }})$

$=jwC_{3}R_{2}R_{4}+\frac{jwC_{3}R_{2}}{jwC_{4}}$

$=jwC_{3}R_{2}R_{4}+\frac{C_{3}R_{2}}{C_{4}}$

$R=jwC_{3}R_{2}R_{4}+\frac{C_{3}R_{2}}{C_{4}}-jwL-R_{1}$

And now L:

$R_{1}+R+jwL=jwC_{3}R_{2}R_{4}+\frac{C_{3}R_{2}}{C_ {4}}$

$jwL=jwC_{3}R_{2}R_{4}+\frac{C_{3}R_{2}}{C_{4}}-R_{1}-R$

$L=C_{3}R_{2}R_{4}+\frac{C_{3}R_{2}}{jwC_{4}}+\frac {R_{1}}{jw}+\frac{R}{jw}$

It doesn't feel like that's what I'm supposed to do. I still have the imaginary parts hanging around for one thing.

4. Originally Posted by evanator
I still have the imaginary parts hanging around for one thing.
Sometimes these guys will cancel out, have you tried this?

It is also preferred that you have your answers in the form $z_i = a+bj$

Therefore you may have to pull those imaginary numbers from the demoninator by multiplying some complex conjugates.

5. Set the imaginary and real parts of the equation equal to each other and see what you get. What kind of variable is w? So take your resulting equation from $Z_1Z_3=Z_2Z_4$ multiply both sides by $jwC_3C_4$ and equate coefficients of j and then equate coefficients of the real part.

6. I have to admit to being stumped by this. I don't normally like doing this, but I looked at the answers in the back of the book and we are tantalizingly close.

It says that

$R=\frac{R_{2}C_{3}-R_{1}C_{4}}{C_{4}}$

and

$L=R_{2}R_{4}C_{3}$

So the real parts are right (pretty much).

7. So j and w are imaginary.

8. Nearly there. Just typing it up.

9. $jwC_{3}C_{4}(\frac{R_{1}+R+jwL}{jwC_{3}})=jwC_{3}C _{4}(R_{2}R_{4}+\frac{R_{2}}{jwC_{4}})$

$RC_{4}+R_{1}C_{4}+jw(LC_{4})=R_{2}C_{3}+jw(C_{3}C_ {4}R_{2}R_{4})$

Taking the real parts:

$RC_{4}+R_{1}C_{4}=R_{2}C_{3}$

$RC_{4}=R_{2}C_{3}-R_{1}C_{4}$

$R=\frac{R_{2}C_{3}-R_{1}C_{4}}{C_{4}}$

and the imaginary parts (coefficients of jw):

$LC_{4}=C_{3}C_{4}R_{2}R_{4}$

$L=C_{3}R_{2}R_{4}$

It felt good to get it in the end, even though I had lots of help. Thanks a lot, Krahl and Pickslides.