(4 - Xc ) Xl2 + (Xc² - 8Xc +52)=0

Xl2=- (Xc² - 8Xc +52) / (4 - Xc )

Yes!!!

Now we have to get our second equation by setting the real part (blue part) of our impedance expression in post #75 equal to 50 ohms.

Then we substitute the solution for Xl2 that you got above into the second equation. I'm going to save us both some pain by showing the result here. I think you deserve a rest.



You see how ugly and complicated it becomes when we make the substitution? The important thing to notice here is that this last equation only involves one variable, namely Xc now that we've made the substitution. I suppose I could ask you to simplify that equation, but I think I will just ask if you understand what we've done to get it?

 

no understand little bit

 

no understand little bit

Do you understand why we need two equations? It's because we have two unknowns, C1 and L2. We got one equation by setting the imaginary part of the impedance equal to zero. You solved that equation for Xl2 in post #80.

Then we got a second equation by setting the real part of the impedance to 50. Now we have to substitute the solution for Xl2 back into the second equation. This eliminated Xl2 from the second equation. Now the second equation only has Xc as an unknown. Thus we have one equation in one unknown, which is easy to solve. Our next step will be to actually solve the second equation.

Does this help?

 

6( 52 - 8Xc + Xc²)²/ (Xc-4)²(36 + (4-Xc + (52-8Xc+Xc²)/Xc-4)² =50

How you are writing it?

 

6( 52 - 8Xc + Xc²)²/ (Xc-4)²(36 + (4-Xc + (52-8Xc+Xc²)/Xc-4)² =50

How you are writing it?

I'm using Mathematica to write it and to actually do the algebra.

Are you ok with understanding how we got this equation? Do you realize that it got complicated because of the substitution we made. I won't ask you to simplify it--that would be a lot of algebra which wouldn't include complex arithmetic so it would be time consuming, but not impossible

 

then how to solve

 

then how to solve

OK. Now we just need to simplify that last equation so we can solve it. I'm going to let Mathematica simplify it and show you the result. Then you can solve it using the quadratic formula.

I hope you realize that we are at the end of this problem. Once you solve for Xc, then solving for Xl2 is just a matter of substituting the value for Xc back into the solution for Xl2.

Here's the simplification of the second equation:



You see that the last equation above only involves Xc, and it's a quadratic so it can be solved by using the quadratic formula.

 

8.66+Xc²*0.167 - 1.34Xc=50
Xc²*0.167 - 1.34Xc +8.66 =50
Xc²3.34x10^-3 - 0.0268Xc + 0.1732=0
Xc= -b+- (b²-4ac)^0.5/2a
Xc= 0.0238+-(7.1*10^-4 - 2.313*10^-3)/ 6.68*10^-3

Xc= 0.0238+-(7.1*10^-4 - 2.313*10^-3)/ 6.68*10^-3
Xc= 3.5 +- ..............


What is this Mathematica?

 

8.66+Xc²*0.167 - 1.34Xc=50
Xc²*0.167 - 1.34Xc +8.66 =50
Xc²3.34x10^-3 - 0.0268Xc + 0.1732=0
Xc= -b+- (b²-4ac)^0.5/2a
Xc= 0.0238+-(7.1*10^-4 - 2.313*10^-3)/ 6.68*10^-3

Xc= 0.0238+-(7.1*10^-4 - 2.313*10^-3)/ 6.68*10^-3
Xc= 3.5 +- ..............


What is this Mathematica?

Mathematica is a software for doing mathematics: http://www.wolfram.com/mathematica/?source=nav


8.66+Xc²*0.167 - 1.34Xc=50
Xc²*0.167 - 1.34Xc +8.66 =50

OK. You've made a mistake right here. If you divide by 50 then the right side can't be zero; it will be 1

Xc²3.34x10^-3 - 0.0268Xc + 0.1732=1

Then you have to subtract 1 from both sides like this:

Xc²3.34x10^-3 - 0.0268Xc + 0.1732 - 1=1 - 1

Xc²3.34x10^-3 - 0.0268Xc - .82666=0

You should probably carry a few more digits in your calculations.

 

Xc²3.34x10^-3 - 0.0268Xc - .82666=0

is there any tool to solve this?

 

Is it open source?

 

Xc²3.34x10^-3 - 0.0268Xc - .82666=0

is there any tool to solve this?

Yes. The quadratic formula which you were already using: Xc= -b+- (b²-4ac)^0.5/2a

 

Is it open source?

No. But there are free applications such as Macsyma that have similar capabilities.

 

OK.
Xc= -b+- (b²-4ac)^0.5/2a

Xc²3.34x10^-3 - 0.0268Xc - .82666=0

Xc= 0.0268 +- (7.1824*10^-4 + 0.011044)^0.5/ 6.68*10^-3

 

Now i don't go for free application.

 

OK.
Xc= -b+- (b²-4ac)^0.5/2a

Xc²3.34x10^-3 - 0.0268Xc - .82666=0

Xc= 0.0268 +- (7.1824*10^-4 + 0.011044)^0.5/ 6.68*10^-3

Can't you work it out on your calculator?

 

Xc= 0.0268 +- (7.1824*10^-4 + 0.011044)^0.5/ 6.68*10^-3

Xc = 0.0268 +- 0.10845/ 6.68*10^-3
Xc = 0.0268/ 6.68*10^-3 +- 0.10845/ 6.68*10^-3
Xc= 4.0119 +- 0.01623*10^-3

 

Xc= 0.0268 +- (7.1824*10^-4 + 0.011044)^0.5/ 6.68*10^-3

Xc = 0.0268 +- 0.10845/ 6.68*10^-3
Xc = 0.0268/ 6.68*10^-3 +- 0.10845/ 6.68*10^-3
Xc= 4.0119 +- 0.01623*10^-3

You've got an error in the very last calculation:

Xc= 4.0119 +- 16.235

Let's choose the positive root. So what do you get?

 

Xc=20.2654ohm

 

right??