Start by considering the 'laws' regarding EMF and current distribution among components of RLC networks -- especially at resonance...

Impedance transformation is fundamental to electrical engineering (at least!)...

this

Ok...?

 

@RRITESH KAKKAR

My time has come to be gone --- Please think about the exercise and I'll see you tomorrow!

Best regards
HP

 

anyway how to do that?
what is answer

 

The answer you're looking for is figuratively 'staring you in the face' -- You need only stop 'looking' and begin seeing!


Best regards
HP

 

@RRITESH KAKKAR

It occurs to me that the 'block' you seem to be experiencing with the current exercise may owe to misapprehension as regards impedance transformation via networks (arrangements) of discrete reactors...
Here I present an example of transformation of 10v to over 60V via a series resonant RLC circuit --- While not directly applicable to the exercise, this is intended to set you upon the right path!
Note that, with some qualification, impedance may be thought of as EMF to current ratio!

Please note that following is not to be confused with other impedance transformation schemes/phenomena (i.e. mutual inductance, 'buck/boost', etc...)

I hope you find this helpful!!!
/////////////////////////////

Within the confines of this discussion, resonance is a condition wherein 'opposite' reactor types exhibit reactances of equal magnitude but opposite sign and, hence, 'cancel' -- Below is a representation of a series RLC circuit at resonance -- Please consider the following:

Because EMF lags current by 90° in a purely reactive, net capacitive circuit, capacitive reactance is said to be negative.
Because EMF leads current by 90° in a purely reactive, net inductive circuit, inductive reactance is said to be positive.

With reference to the below attached schematic:
Reactor 1 (The capacitor) exhibits a reactance of -2πΩ @ a frequency of 100MHz
Reactor 2 (The inductor) exhibits a reactance of 2πΩ @ a frequency of 100MHz
Hence the circuit is said to be resonant @ 100MHz

Owing to their series connection, the net reactance = the sum of the reactances = 0Ω. Thus the net impedance = the value of R = 1Ω
Thus it is that the current through the circuit is determined by the EMF of the power supply (1oV Peak) and the resistance of R (1Ω) → 10V (Peak)/1Ω= 10A (Peak)

Ohm's law tells us that the EMF across a pure reactance = I*X = 10A*2π ≈ 62.83V in the case of the inductor --- And 10A*2π ≈ -62.83V in the case of the capacitor (note that the signs indicate the respective phase shifts)
In theory any finite transformation ratio is possible via proper choice of the reactors and the resistor.

Illustration of a series RLC circuit at resonance -- Post continued below image...


As an aside please consider a parallel resonant circuit.

Via the technique of calculating the equivalent value of paralleled resistors or pure reactances:
1/(1/Xc+1/Xl)=Xnet but since Xc=-Xl at resonance: 1/(1/-x + 1/x) = 1/(0/x^2)=1/0 = ∞ (For these purposes)
Thus we see that a parallel LC circuit at resonance presents as an 'open circuit'

Again, hope to have been of help!

Best regards
HP

 

You need only stop 'looking' and begin seeing!

Mean?

 

I hope you find this helpful!!!

I have not finish the last exercise ...
nee more time to understand.

 

Mean?

I mean that if you focus on solving the problem instead of searching for the answer you'll find it easier than you think!

 

The question of topic maximum power transfer i think?

What to think in it?

 

I have not finish the last exercise ...
nee more time to understand.

Please know that post #145 is not an exercise -- but, rather, an illustration intended to assist you with the exercise of post #113 -- Know also that appropriate questions and requests for clarification are always welcome!

 

The question of topic maximum power transfer i think?

Indeed! That is the 'goal' of impedance matching!

 

an illustration intended to assist you with the exercise of post #113

That mean it is solution #145

 

That mean it is solution #145

No, not a solution -- Merely an example of the way in which reactors may transform impedance

 

As an aside please consider a parallel resonant circuit.

Via the technique of calculating the equivalent value of paralleled resistors or pure reactances:
1/(1/Xc+1/Xl)=Xnet but since Xc=-Xl at resonance: 1/(1/-x + 1/x) = 1/(0/x^2)=1/0 = ∞ (For these purposes)
Thus we see that a parallel LC circuit at resonance presents as an 'open circuit'

 

Did you have a question regarding the text copied to post #154?

 

i don't got the purpose and answer of #149
what is the frequency ??
for calculating Xl=2*3.14*F*L

 

i don't got the purpose and answer of #149
what is the frequency ??
for calculating Xl=2*3.14*F*L

Note that post #149 is excerpted from post #113
Inasmuch as the exercise deals exclusively with reactances and resistances - specification of frequency is unnecessary

 

Note that post #149 is excerpted from post #113
Inasmuch as the exercize deals exclusively with reactances specification of frequency is unecessary

Yes it is.

OK, it is already given the reactance then where to move i mean what to do?

if we will connect port a to b what will happen?

 

What are the reactances of the network reactors and how are they connected?

 

What are the reactances of the network reactors and how are they connected?

6+4j 50+0j