i don't know very well how to use kvl

 

I never called RRITESH stupid or dumb and I never thought such a thing. See my reply on the other thread.

What may appear to you as though I'm teaching to a 6th grader is my attempt to pierce the language barrier. If RRITESH feels offended he can certainly say so, but it seems to be working.

Mod edit: deleted quote of off topic material.

 

(180-30j ) // (0+30j)
0+5400j+0-900j²
5400j+900/180
30j+5

then 5+30j+5-30j
10ohm

Yes, so what is the current in Z, and after that, what is the voltage across Z?

When you have that, you can calculate the voltage across R2 because Xc2 and R2 form a voltage divider.

 

OK, 10V/10= 1amps.
10*30j+5/30j+5+5-30j
10*(30j+5)/ 10
5+30j can voltage be in complex form?

 

OK, 10V/10= 1amps.
10*30j+5/30j+5+5-30j
10*(30j+5)/ 10
5+30j can voltage be in complex form?

Yes, it's important to keep the voltage in complex form. Now work out the voltage divider consisting of Xc2 and R2.

 

5+30j*180/ 180-30j
900+5400j/ 180-30j
90+540j/18-3j
30+180j/ 6-j Volt

 

Apparently you think lack of aptitude is the same as dumb. I don't, and I hope other people see the distinction.

I, for example, lack aptitude at carpentry, but I don't think that means I'm dumb.

I'm trying my best to not convey any insult to RRITESH. I'm trying to calibrate my instruction to be in appropriate sized steps given the language barrier. It seemed to me over at the other thread that often RRITESH didn't understand what he was told because it was too much at once.

As I said, if RRITESH feels insulted, he can say so, and I'll try another tack.

Mod edit: deleted off topic quote.

 

5+30j*180/ 180-30j
900+5400j/ 180-30j
90+540j/18-3j
30+180j/ 6-j Volt

OK, but one thing you should be careful to do is include parentheses in the numerator and denominator, otherwise the order of operations on a calculator, for example, might not give the correct result. Do it like this:

(30+180j)/( 6-j)

Can you finish the calculation and end up with an expression in the form A + j B? You have to rationalize the denominator and divide into the numerator.

 

(30+180j)*(6+j) /( 6-j)*(6+j)
180+30j+1080j-180 / 36+6j-6j+1
1110j / 37
30j Volt

 

Electrician -- I say this then I'll stop cuz I'm interrupting lessons


Saying someone lacks talent or aptitude is to say they're congenitally defective! NOT COOL!!!

I disagree.

The word "congenitally" means something one is born with. Aptitude for some ability is a congenital quality for sure, but to universally call it a defect is being unnecessarily pejorative.

If a person has brown eyes, which is a congenital thing, in a culture where blue eyes are preferred, are the brown eyes a congenital defect?

Didn't you have a high school counselor that would give aptitude tests? If a person lacked aptitude for auto repair, but was great at journalism, would you say they had a congenital defect for auto repair?

As I said, I lack aptitude for carpentry. If you want to call it a defect, that's your privilege.

 

(30+180j)*(6+j) /( 6-j)*(6+j)
180+30j+1080j-180 / 36+6j-6j+1
1110j / 37
30j Volt

Absolutely correct!

So we see that a source that is supplying only 10 volts, is able to drive the 180 ohm resistor to 30 volts. This is due to the ability of resonance to cause a rise in voltage.

Are you satisfied that you understand how this circuit works?

This circuit started out with simple series resonance, but we didn't design a circuit to match to a predefined load. We just added on to the the series resonant circuit and accepted whatever load came out of it.

We can now go on to designing a circuit to match a predetermined source impedance and predetermined load.

It must be about time for bed in India. We can continue tomorrow if you like, or we can do some more now. What's your choice?

 

now.

i used to study whole night in exam.
So we see that a source that is supplying only 10 volts, is able to drive the 180 ohm resistor to 30 volts.
What does 30j mean?

 

If just for language barrier then I understand is ok but content is too basic I mean resonant networks without talking about Q?

If you think it would be better to discuss Q with RRITESH, then by all means start a new thread. In this thread Q comes later.

I thought I saw a better way to help him so I started a new thread. You could do the same.

 

now.

i used to study whole night in exam.
So we see that a source that is supplying only 10 volts, is able to drive the 180 ohm resistor to 30 volts.
What does 30j mean?

The reason j is there is because the voltage across R2 is shifted in phase, 90 degrees, with respect to Vs.

The problem now is to design a matching network to match predetermined source and load impedances.

Here's an incomplete circuit:


Inside the box we need to determine a network consisting of inductors and capacitors to transform the 50 ohm load to match the 6 + j4 ohm source impedance.

We know that we will need to use the power of resonant circuits to cause a voltage rise, because the 50 ohm load is greater than the 6 + j4 ohm source.

There are several topologies that will do the job, but let's use this one:



There are already worked formulas to determine the values of Xc and Xl.

The derivations can be found on the web: http://www.ece.ucsb.edu/~long/ece145a/Notes5_Matching_networks.pdf

But I think it would be instructive to derive the component values without recourse to those formulas.

What you need to do now is to derive an expression for the voltage across the 50 ohm resistor like you did for the previous circuit, but with Xc and Xl in symbolic form instead of numerical values.

So, please give it a go. If you run into trouble just post here.

We may both have to go to bed in a couple of hours.

 

The reason j is there is because the voltage across R2 is shifted in phase, 90 degrees, with respect to Vs.

The problem now is to design a matching network to match predetermined source and load impedances.

if there is j that mean 90 phase out?

 

I think as you said first step is to find impedance at load 50ohm by Thevenin.
(6+4j-Cj) // L1

 

if there is j that mean 90 phase out?

Exactly. In English we would say "90 degrees out of phase".

 

I think as you said first step is to find impedance at load 50ohm by Thevenin.
(6+4j-Cj) // L1

I've edited post #25 because I inadvertently labeled both inductors L1; now the second one is L2.

You need to treat all impedances as complex quantities. For example, we will take Xc to be the value of the capacitor's reactance, but when you use it in a complex impedance expression you have to use the correct sign, and include j where appropriate.

So the impedance at the load will be (6 + 4j - Xc*j) || (Xl2*j). I have written Xc*j and Xl2*j because if I wrote Xcj that would look like a new symbol.

Now you need to do a lot of algebra. You need to work out \(\frac{ (6 + j(4 - Xc))*(Xl2*j)}{(6 + j(4 - Xc))+(Xl2*j)}\)

This is going to be a lot more work (algebra) than just using the canned formulas in the link I gave previously. If you don't want to do it this way, just say so.

 

(6 + 4j - Xc*j) || (Xl2*j)

why we have not consider 50ohm?

(6 + 4j - Xc*j) || (Xl2*j)
(6 + 4j - Xc*j)*(Xl2*j) / (6 + 4j - Xc*j) + (Xl2*j)
6*Xl2*j + 4j*Xl2*j - Xc*j*Xl2*j / (6 + 4j - Xc*j) + (Xl2*j)
6*Xl2*j - 4Xl2 + - XcXl2 / (6 + 4j - Xc*j) + (Xl2*j)

 

(6 + 4j - Xc*j) || (Xl2*j)

why we have not consider 50ohm?

(6 + 4j - Xc*j) || (Xl2*j)
(6 + 4j - Xc*j)*(Xl2*j) / (6 + 4j - Xc*j) + (Xl2*j)
6*Xl2*j + 4j*Xl2*j - Xc*j*Xl2*j / (6 + 4j - Xc*j) + (Xl2*j)
6*Xl2*j - 4Xl2 + - XcXl2 / (6 + 4j - Xc*j) + (Xl2*j)

Hi, RRITESH

I just woke up and I see you have been at work.

We will consider the 50 ohm load later, but right now we want to calculate the impedance looking back into the circuit, when the circuit looks like this:



What you have to do now is some more algebra. You have: 6*Xl2*j - 4Xl2 + - XcXl2 / (6 + 4j - Xc*j) + (Xl2*j)

It looks like you have both a + and - sign where you should only have a + sign. Fixing that problem, you have this expression; I have separated the real and imaginary parts in both the numerator and denominator:

\(\frac{(-4 Xl2+Xc*Xl2)+j*(6*Xl2)}{(6)+j*(4+Xl2-Xc)}\)

Next you must rationalize the denominator like you did before, by multiplying both the numerator and denominator by the conjugate of the denominator:

\(\frac{(-4 Xl2+Xc*Xl2)+j*(6*Xl2)}{(6)+j*(4+Xl2-Xc)} *\frac{(6)-j*(4+Xl2-Xc)}{(6)-j*(4+Xl2-Xc)}\)

The reason we're doing this is to get rid of the imaginary part in the denominator.

This is a lot of algebra, and I can also make mistakes, so help me watch out for mistakes.