Xc=20.2654ohm

That's pretty good. I get 20.2481

Now substitute that value for Xc in the expression you derived in post #80 for Xl2. What do you get for Xl2?

Don't forget that Xc will need to be a negative value in the circuit because Xc is the reactance of a capacitor.

 

Xl2=- (Xc² - 8Xc +52) / (4 - Xc )
Xl2= - (409.9855- 162.1232+52)/ -16.2654
Xl2= 18.4355

 

Xl2=- (Xc² - 8Xc +52) / (4 - Xc )
Xl2= - (409.9855- 162.1232+52)/ -16.2654
Xl2= 18.4355

Yes. I get 18.4637, but I am carrying 12 digits for the calculation.

Congratulations!!!

So Xc = -20.2481
Xl2 = 18.4637

Here's the result from Mathematica:



After all this work I hope you understand why we did each step of the calculation. This was the brute force way to solve the problem. "Brute force" is an English phrase that means "the most difficult way".

You can imagine that somebody that designs a lot of impedance matching networks would want a simpler way to make the calculations.

If we were to leave the source and load resistances in symbolic form, we could derive a formula for each case that would be much easier to use.

The conventional formulas are derived in these two resources:

http://www.ece.ucsb.edu/~long/ece145a/Notes5_Matching_networks.pdf
http://rfic.eecs.berkeley.edu/142/pdf/book_chap7.pdf

To show you how easy it is to use the formulas, let's do the calculation for this network.

First, the system Q is given by Q = SQRT(tratio-1), where tratio = transformation ratio = 50/6

So we get Q = SQRT(50/6-1) = SQRT(66)/3

The source resistance Rs = 6, and the load resistance Rl = 50

The definition of Q for a series branch is Q = Xc/Rs, so Xc = Q*Rs =-SQRT(66)/3 * 6 = -2*SQRT(66) = -16.2481

But, remember there is a small inductive reactance of 4 ohms in series with Rs, so we need to subtract 4 to compensate our calculated value. Thus Xc = -2*SQRT(66) - 4 = -20.2481

The definition of Q for a parallel branch is given by Q = Rl/Xl2 so Xl2 = Rl/Q = 50/(SQRT(66)/3) = 18.4637

This is much easier than all that work we did for the hard way, isn't it?

But, by having done it the hard way, you have a permanent appreciation for the standard formulas.

Congratulations for hanging in there, RRITESH. You earned it.

 

First, the system Q is given by Q = SQRT(tratio-1), where tratio = transformation ratio = 50/6

So we get Q = SQRT(50/6-1) = SQRT(66)/3
50/6-1 =7.333 not 66/3
The definition of Q for a series branch
what is this?

 

First, the system Q is given by Q = SQRT(tratio-1), where tratio = transformation ratio = 50/6

So we get Q = SQRT(50/6-1) = SQRT(66)/3
50/6-1 =7.333 not 66/3
The definition of Q for a series branch
what is this?

Be careful. There's a difference between SQRT(66)/3 and SQRT(66/3)

Have a look here: https://en.wikipedia.org/wiki/Inductor (near the bottom of the page)

Also here: https://en.wikipedia.org/wiki/Q_factor (near the bottom of the page)

The Q of an inductor or a capacitor is a number that gives us an idea of how lossy the inductor or capacitor is.

In case of a series R and L, the Q = ω L/R. For a parallel R and L, Q = R/(ω L)

 

lossy having or involving the dissipation of electrical or electromagnetic energy

 

next what to do?

 

next what to do?

We're done with this problem. I hope you enjoyed the process. I'm going to bed. I'm sure there are more problems you can work on, but I'm going to have to take a rest for a while.

 

Absolutely fantastic work!
@RRITESH KAKKAR You should refer back to this entire thread several times to further firm up the process so that you can apply it yourself to solve problems.
@The Electrician I am locking this thread since the problem is complete. If you decide to fire up another problem you can start another thread OR if it suits your purposes, PM a mod and we will reopen this one for you.