help with simplification impedance

[mkforlife10]

Z1=1[(1/RA)+1/(-jXCA)]
XCA=-1/ω0CA
where RA=2.2kΩ, RB=2.2kΩ, Ca=0.022μF, Cb=0.022μF
how to simplify the impedance I am getting confuse with the imaginary
do I have to convert to polar and rectangular?

 

[MrAl]

Hi,

First thing you will have to do is learn to write out these equations properly. You dont seem to be paying attention to detail when you write them out.
This time you've got a '1' outside the brackets, which makes no sense. You've also included Cb and there's no Cb in the equation for Z.
You dont have to convert anything unless the result is required in a special form other than that given.

You have to be able to write these out carefully and accurately otherwise you'll never get the answer you really need.

 

[mkforlife10]

Hi,

First thing you will have to do is learn to write out these equations properly. You dont seem to be paying attention to detail when you write them out.
This time you've got a '1' outside the brackets, which makes no sense. You've also included Cb and there's no Cb in the equation for Z.
You dont have to convert anything unless the result is required in a special form other than that given.

You have to be able to write these out carefully and accurately otherwise you'll never get the answer you really need.

thats how the equation is written on my manual.
Z1==1/[(1/R_A )+1/(-jX_CA )]
Z2=Rb+1/jωCb
Sorry I forgot to include Z2
after finding Z1 and Z2 I have to find B=Z1/(Z1+Z2)

 

[WTP Pepper]

Which is not how originally posted. Z1 is now 1/ instead of 1[
What you asking in the second post is also different.

Try drawing us a circuit and let us work out what you are trying to achieve.

 

[MrAl]

thats how the equation is written on my manual.
Z1==1/[(1/R_A )+1/(-jX_CA )]
Z2=Rb+1/jωCb
Sorry I forgot to include Z2
after finding Z1 and Z2 I have to find B=Z1/(Z1+Z2)

Hi again,

Ok this time you've got two equal signs when you probably mean to have only one. Two equal signs are sometimes interpreted to mean a logical comparison so Z1==1 would be equal to 1 if Z1 was equal to 1 but would be 0 if Z1 was not equal to 1. So you see you have to be careful in writing these out.

I will assume you meant the following...

Declare D as:
D=1/Ra+1/(-j/(w*Ca))
So D is the denominator of Z1, which makes Z1 as:
Z1=1/D

Z2 is a little easier, with
Z2=Rb+1/(j*w*Cb)

Note i've made some assumptions in the above so you'll have to verify.

The first thing we can do is simplify D as:
D=1/Ra-(Ca*w)/j

and so Z1 is:
Z1=1/D=1/(1/Ra-(Ca*w)/j)

Now since we have j in the denominator of another denominator, we can simplify this by multiplying top and bottom by j. This gives us:
Z1=j/((j/Ra)-Ca*w)

and we can do the same with Ra, and so we get:
Z1=(j*Ra)/(j-Ca*Ra*w)

Now you can see we have reduced this to just two levels with just one numerator and one denominator.
To proceed from here you would multiply the top and bottom by the complex conjugate of the denominator.
Written slightly differently, we have:
Z1=(j*Ra)/(-Ca*Ra*w+j)

so the complex conjugate of the denominator is:
-Ca*Ra*w-j

We multiply top and bottom of Z1 by this, and we get:
Z1=(Ra-j*Ca*Ra^2*w)/(Ca^2*Ra^2*w^2+1)

Now you see we've changed the form of Z1 so that the denominator is completely real.

Next, we would do the same thing with Z2, then we find B by adding the resulting Z1 and Z2, taking the reciprocal, then simplify that using the same procedure, where the idea is to reduce to two levels with the denominator being purely real with no imaginary terms.

Does this make sense or are you still stuck?

I realize that everyone makes mistakes especially typos but the better you type out your questions the faster you'll get the answers, and those answers should be the right answers