[SOLVED] Calculate values in a transformer

[Hatmpatn]

Hello Edaboarders! After a very successful thread I created on calculating the Thevenin Equivalent from a circuit I have now moved on to a more complex problem.
I have been given a problem which has much more than just voltage- and current sources and resistors in it.

My knowledge in this area is very restricted and I don't know much about capacitors and inductors and I hope that you guys know more about them and how to solve my problem.



I'm going to:

a) Calculate the voltage u(t)

b) Calculate the "active effect"(P=U*I*cosφ=RI²) and the "reactive effect" (Q=U*I*sinφ=XI²) that is obtained in the load R2-L2.

Note that the words active effect and reactive effect is just directly translated from my language and I don't know the corresponding term for this in english. I have searched using the formulas without results.

c) An assumption should be made that R1 is C2 is variables, and set the R1 and C2 so that the effect in A-B is maximized.

Can anyone point me inte right direction and tell me what to start with. I would greatly appreciate this!

 

[The Electrician]

In your first thread you said you were taking a basic course in electronics. This problem is not what I would call basic!

You need to know about AC circuits and complex arithmetic to solve this problem. Have you studied complex variable theory?

Do you perhaps have a degree in some other discipline, and now you're going back to learn electronics?

Hello Edaboarders! After a very successful thread I created on calculating the Thevenin Equivalent from a circuit I have now moved on to a more complex problem.
I have been given a problem which has much more than just voltage- and current sources and resistors in it.

My knowledge in this area is very restricted and I don't know much about capacitors and inductors and I hope that you guys know more about them and how to solve my problem.

View attachment 109611

I'm going to:

a) Calculate the voltage u(t)

b) Calculate the "active effect"(P=U*I*cosφ=RI²) and the "reactive effect" (Q=U*I*sinφ=XI²) that is obtained in the load R2-L2.

Note that the words active effect and reactive effect is just directly translated from my language and I don't know the corresponding term for this in english. I have searched using the formulas without results.

c) An assumption should be made that R1 is C2 is variables, and set the R1 and C2 so that the effect (power in English) in A-B is maximized.

Can anyone point me inte right direction and tell me what to start with. I would greatly appreciate this!

Have a look at https://en.wikipedia.org/wiki/Power_factor. The English terms are "active power" and "reactive power".

You might also have a look at this: https://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

You're going to have to find values for R1 and C2 so that the impedance looking into the terminals at A-B is the complex conjugate of the impedance of the source, assuming that varying only those two components will achieve the desired result.

Your problem gives the internal impedance of the source in an unusual format. Do you know how to show it in rectangular coordinates?

Do you know how to calculate the phasor impedance of a network?

 

[Hatmpatn]

This course sure escalated quickly. I've tried to read up on stuff and the links you showed me, but can't say I understand everything.

To convert the values in the internal impedance to cartesian coordinates should be like this:

Zi=100e^jπ/4 => 100/√2+j100/√2

I really don't know how to calculate the phasor impedance of a system, but i'll try to read up on that.

 

[Hatmpatn]

Alright. Apparently, you're supposed to convert the components so that they fit the domain.

I have done so in the following schematics.

 

[The Electrician]

The inductor on the far right is designated L2 on one of your schematics, and as L4 on the other--which is correct?

Now you need to begin combining series and parallel components until you have only one impedance at A-B.

 

[Hatmpatn]

L2 is correct. I need to double-check my drawings. Don't want it to go as the last time!
Here is the updated schematics:



I really have no clue on how to
combine series and parallel components to get only one impedance. Kinda lost here.

Too bad we're in different time zones, you and me, cause once I see your answer, it's bed time. Bet it's the other way around for you aswell

 

[The Electrician]

If you don't have a text book on the topic, search for "AC circuit analysis" or "AC steady state analysis":

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/accircon.html#c1

http://www.phy-astr.gsu.edu/nelson/ac.circuits.pdf

http://www.labsanywhere.net/circuit/lectures/lect9/lecture9.php

http://www.cedengineering.com/upload/Elec Eng Fundamentals - AC Power Circuits.pdf

http://www.g3ynh.info/zdocs/AC_theory/AC_theory.pdf

**broken link removed**

 

[Hatmpatn]

To be able to calculate the equivalent impedance of the system I need to know ω. How do I do to get that?

 

[The Electrician]

Look at the image in post #1. The source is labeled IO = 10 sin(1000t) (ma)

So ω = 1000

 

[Hatmpatn]

How stupid of me, ofcourse it is. Since the most tutorials had an i(t) with a cos in it, I converted the i(t)=10sin(1000t) to i(t)=10cos(1000t+Π/2)

=> j10 in the complex plane

Should be correct, right?

Here's the latest circuit drawing. Now.. how do I go about things to solve for u(t)??

 

[The Electrician]

Start by converting L2 and R2 in to a single complex impedance on the N2 side of the transformer. Then find an explanation of how transformers change impedances (you might search for "transformer impedance ratio". This will give the value of the impedance appearing at the N1 side of the transformer. Then you just move to the left, combining series and parallel combinations of impedance.

 

[Hatmpatn]

Excuse me for taking so long, I was away for the weekend.
The single complex impedance on the N2 side should be 1/Z2=1/8+1/10j <=> (wolfram alpha) <=> Z2=4.88-3.90j

According to this pdf **broken link removed**, page 9,
the √(Z1/Z2)=N1/N2

So ill calculate for the N1 side aswell.

Input in wolfram alpha; 1/x(Z1)=1/(10i)+1/((100/sqrt(2))+(i(100/sqrt(2))))+1/(500i)+(1/(-1000i+4000i))+1/(-1000i)
Result: Z1: 0.5991-9.1856j

Using the √(Z1/Z2)=N1/N2 gives me the complex solution 0.1105-0.4902j, and that doesnt seem right.

 

[The Electrician]

Excuse me for taking so long, I was away for the weekend.
The single complex impedance on the N2 side should be 1/Z2=1/8+1/10j <=> (wolfram alpha) <=> Z2=4.88-3.90j

According to this pdf **broken link removed**, page 9,
the √(Z1/Z2)=N1/N2

So ill calculate for the N1 side aswell.

Input in wolfram alpha; 1/x(Z1)=1/(10i)+1/((100/sqrt(2))+(i(100/sqrt(2))))+1/(500i)+(1/(-1000i+4000i))+1/(-1000i)
Result: Z1: 0.5991-9.1856j

Using the √(Z1/Z2)=N1/N2 gives me the complex solution 0.1105-0.4902j, and that doesnt seem right.

I get different result for this: 1/Z2=1/8+1/10j <=> (wolfram alpha) <=> Z2=4.88-3.90j

I get 4.88+3.90j


Check your result again.

Also, consider that √(Z1/Z2)=N1/N2 is the same as Z1/Z2=(N1/N2)^2

The effect of the ideal transformer in this situation is to multiply the impedance on the N2 side to an impedance on the N1 side which is just 100 times larger. In other words, the impedance at the N1 side is 488+390j

At this point, you just move from N1 to the left, including the various parallel and series combinations in your impedance calculations, ultimately ending up with an impedance seen looking in at the A-B terminals.

At some point you will have to do the calculations involving R1 and C2 as symbolic variables for part C or your problem.

 

[Hatmpatn]

My input in wolfram alpha is: 1/x=1/8+1/10i and I still get 4.88-3.90j. Weird.

But looking at our history, I assume you are correct.

So I've got the 488+390j impedance on the N1 side, im not really sure what you are saying that i'm supposed to do next. Should I calculate for the impedance to the point that I get to the A-B terminal?

In that case, that should be: 1/(500i)+(1/(-1000i+4000i))+1/(-1000i)=(-1/750)j

 

[The Electrician]

My input in wolfram alpha is: 1/x=1/8+1/10i and I still get 4.88-3.90j. Weird.

But looking at our history, I assume you are correct.

You need to include some parentheses like this: 1/x=1/8+1/(10i).

Be very careful in your complex number calculations. When you have a denominator with more than a single number or variable in it, put the whole thing in parentheses.

So I've got the 488+390j impedance on the N1 side, im not really sure what you are saying that i'm supposed to do next.

Now you calculate the impedance of C2 in parallel with 488 + 390j (keep more than 3 digits in your computations); call the resulting impedance Za.

Be careful about the formulas for the impedance of an inductor and of a capacitor. The impedance of an inductor is (j ω L); the impedance of a capacitor is 1/(j ω C).

Now calculate the impedance of Za in parallel with the equivalent impedance of the series combination of C1 and L1; call that Zb. Next calculate the impedance of Zb in parallel with R1. That result will be the impedance seen at A-B.

 

[Hatmpatn]

1/Za=1/(-1000j)+1/(488+390j) <=> Za=799.680-0.400j

1/Zb=1/(-1000j)+1/(4000j)+1/(799.680-0.400j) <=> Zb=587.865-352.872j

1/Z(a-b)=1/500+1/(587.865-352.872j) <=> Z(a-b)=292.07-67.447j

So this is my impedance at A-B, interesting. How do I go about things to move on from here?

 

[The Electrician]

1/Za=1/(-1000j)+1/(488+390j) <=> Za=799.680-0.400j

You need to keep more digits in your intermediate calculations; keep at least 6 digits. Don't round to 3 digits until the end:

Furthermore, the formula for the impedance of a capacitor is 1/(j ω C), not 1/(-j ω C).

To calculate the equivalent impedance of two impedances in parallel you add the reciprocals of the impedances and then take the reciprocal of that. The impedance of a capacitor is 1/(j ω C), so the reciprocal of the impedance of a capacitor is (j ω C).

Your first calculation should be 1/Za=(+.001j)+1/(487.805+390.244j) <=> Za=800 -0.00016j

The part in red is .001j and not 1/(1000j) because the impedance of a capacitor is 1/(j ω C), and when you take the reciprocal of that you get 1/(1/(j ω C)), which is (j ω C); this is the admittance of a capacitor.

The reciprocal of impedance is called admittance. Another way of expressing the previous description is to say that the equivalent impedance of two impedances in parallel is calculated by adding the admittances of the two impedances and then taking the reciprocal. The sum of two admittances is still an admittance, and when you take the reciprocal you are converting back to an impedance.

The impedance of an inductor is j ω L; the admittance of an inductor is 1/(j ω L).
The impedance of a capacitor is 1/(j ω C); the admittance is (j ω C). Watch out for signs.

1/Zb=1/(-1000j)+1/(4000j)+1/(799.680-0.400j) <=> Zb=587.865-352.872j
1/Z(a-b)=1/500+1/(587.865-352.872j) <=> Z(a-b)=292.07-67.447j

So this is my impedance at A-B, interesting. How do I go about things to move on from here?

Now redo these calculations, fixing the error I pointed out in your calculation for Za.

You could take the admittance seen at N1 (the reciprocal of the impedance) and add the admittance of C2, leaving it as an admittance (don't take the reciprocal, in other words). Then add the admittance of the C1 L1 combination, again leaving the result as an admittance. Finally, add the admittance of R1; your result at this point will be an admittance. Taking the reciprocal will give the impedance at A-B.

What you are doing now is to take the reciprocal (convert back to an impedance) at every step of the way. Doing this is OK, but more calculations are involved; that's why you need to keep more digits for the intermediate calculations.

 

[Hatmpatn]

I must blame Wolfram Alpha for doing that error. Alright, so it should be like this ofcourse:



I'll make every calculation 3 digits more accurate;

Z2= 4.87805+3.90244j
Impedance at N1 side= 487.805+390.244j

Impedance at Za: 1/Za=(+.001j)+1/(487.805+390.244j) <=> Za=800 -0.00016j

Impedance at Zb: 1/Zb=(0.001j+0.00025j)+1/(800 -0.00016j) <=> Zb=/input in wolfram alpha: 1/x=((0.001i)+(0.00025i))+1/(800-(0.00016i))/ <=> Zb=400-400j

Impedance at Z(A-B): 1/Z(A-B)=(0.002j)+1/(400+400j) <=> Z(A-B)=103.093 -268.041j

Really hope this is correct. Thanks The Electrician!

 

[The Electrician]

I must blame Wolfram Alpha for doing that error. Alright, so it should be like this ofcourse:

View attachment 109914

I'll make every calculation 3 digits more accurate;

Z2= 4.87805+3.90244j
Impedance at N1 side= 487.805+390.244j

Impedance at Za: 1/Za=(+.001j)+1/(487.805+390.244j) <=> Za=800 -0.00016j

Impedance at Zb: 1/Zb=(0.001j+0.00025j)+1/(800 -0.00016j) <=> Zb=/input in wolfram alpha: 1/x=((0.001i)+(0.00025i))+1/(800-(0.00016i))/ <=> Zb=400-400j

Impedance at Z(A-B): 1/Z(A-B)=(0.002j)+1/(400+400j) <=> Z(A-B)=103.093 -268.041j

Really hope this is correct. Thanks The Electrician!

There's a problem with the part in red. That should be the admittance (the reciprocal of the impedance) of the series combination of C1 and L1. The way to calculate that is to first calculate the impedance of that series combination.

When you have components in series, the impedance of the series combination is just the sum of the individual impedances. When you have components in parallel, the admittance of the parallel combination is just the sum of the individual admittances.

First, calculate the impedance of the C1 L1 branch. The impedance of C1 is 1/(.001j) = -1000j. The impedance of L1 is 4000j. The impedance of the series combination is the sum of the two, 3000j, so your calculation should look like this:

Impedance at Zb: 1/Zb=1/(3000j)+1/(800 -0.00016j) <=> Zb=746.888 + 199.17j

Carry on from here and see what you get at A-B

 

[Hatmpatn]

Feels like I'm doing so much errors as it is possible to do here, thanks for sticking with me through this!

So the impedance at Zb is: 1/Zb=1/(3000j)+1/(800 -0.00016j) <=> Zb=746.888 + 199.17j

Assuming I did the calculation in Z(A-B) correctly, ill just replace the figures; Impedance at Z(A-B): 1/Z(A-B)=(0.002j)+1/(746.888+199.17j) <=> Z(A-B)=288 -384j