# Ideal Transformers & Power

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1. Mar 5, 2016

### Aristotle

1. The problem statement, all variables and given/known data

2. Relevant equations

Z_Impedance = Z_Load / n^2

3. The attempt at a solution
Hello everyone,
I understand that in order to provide maximum power to load, the load resistance (R_L) has to match the source resistance R_S. My main confusion is from the question itself on what my teacher meant by using "one additional component".
What I'm thinking is to change the load on the right circuit to frequency domain and add the two impedance. From there I would be able to add a impedance component on the left circuit to match the two, correct?
Can somebody possibly lead me to the right direction in this problem? Thank you :)

2. Mar 5, 2016

### Staff: Mentor

Your load has both resistance and reactance (an inductor) so the load is a complex impedance. The component you add to the primary will have to deal with matching the imaginary (reactive) part.

Take a look at the topic: "transformer impedance matching" or "transformer impedance reflection".

Hint: Start by choosing a transformer turns ratio to match the resistance in the source to that of the load.

3. Mar 5, 2016

### Aristotle

Okay so I did some reading on that topic of impedance reflection and from doing some math acknowledging that the Z_load = v2/i2 I came to realize that the impedance seen by the primary coil is equivalent to Z_L/(n^2). So what I did is I first found the impedance of the load by adding them (in series) & got Z_L= 100 + j2.05.
How would I choose a transformer turns ratio arbitrary if you don't mind me asking? Thank you for your reply!

4. Mar 5, 2016

### Staff: Mentor

Take a look at the hint I gave. The transformer ratio that you choose should match the resistances according to the maximum power transfer theorem.

5. Mar 5, 2016

### Aristotle

Hmm, well the transformer ratio (n) relies on the voltage across the secondary coil to voltage across primary..how would this correlate with the resistances?
Maximum power transfer theorem states that the load impedance is equivalent to the conjugate of thevenin impedance.

6. Mar 5, 2016

### Staff: Mentor

How is the load impedance reflected into the primary? What's the relationship?

7. Mar 5, 2016

### Aristotle

Z_Imp= Z_L/(n^2) soo Z_imp / Z_L = 1 / n^2
We know Z impedance and Z load...so we could potentially solve for n.

8. Mar 5, 2016

### Staff: Mentor

Yes. As I've mentioned, start by considering just the real part of the impedances in order to fix the ratio.

9. Mar 5, 2016

### Aristotle

So the component is 92 - j2.05 ohms then ?

10. Mar 5, 2016

### Staff: Mentor

I'm not sure what you're showing me there.

You want the reflected load resistance to match the source's resistance. What turns ratio will make the 100 Ohm load resistance "look like" 8 Ohms from the primary's viewpoint?

11. Mar 5, 2016

### Aristotle

If we only consider the real parts, then the turn ratio is just 1

12. Mar 5, 2016

### Staff: Mentor

No, then the load's 100 Ohms would look like 100 Ohms from the primary. That would not match the source's 8 Ohms.

13. Mar 5, 2016

### Aristotle

Solving for n I got an answer of 1.04.

Work:
Zimp/Zload = 1/(n^2). n= sqrt(100/92) = 1.04

So turn ratio is 1: 1.04

14. Mar 5, 2016

### Staff: Mentor

Where does the 92 come from? The source resistance is 8 Ohms, the load impedance is 100 Ohms. You want the load resistance to look like 8 Ohms when "viewed" from the primary. Then it will satisfy the maximum power transfer theorem.

15. Mar 5, 2016

### Aristotle

Ooh right. 1:3.53

16. Mar 5, 2016

### Staff: Mentor

That's better

Now, using that same ratio, what does the load's inductive impedance look like from the primary?

17. Mar 5, 2016

### Aristotle

Since Zimp/ZL = 1/3.53, solving for Zimp, we get:
Zimp = 100-j2.05 / 3.53
= 28.33 - j .5807

This is what the load impedance look like from primary

18. Mar 5, 2016

### Staff: Mentor

Okay, a couple of things. First, note that your 100 Ohms is now 28.3 rather than 8. This is because you forgot to square the turns ratio.
Second, the impedance of an inductor is a positive imaginary value: ZL = jωL. It's a capacitor that ends up with a negative imaginary value.

So, fix the sign of the inductor impedance and square the turns ratio and give it another go.

19. Mar 5, 2016

### Aristotle

Woops pretend you did not see that
8.025 - j .1654

Oh that makes sense.. Capacitor impedance is -j*1/(wc)..
So if we equate -j.1654 to -j*1/(wc) we can find the capacitor value!

20. Mar 5, 2016

### Staff: Mentor

Yes!

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