# Homework Help: Reflect Impedance

1. Apr 23, 2010

### Lunat1c

For the circuit attached, I'm first asked to find $$I_1, I_2, V_1, V_2$$ by using the appropriate loop equations. I got these answers (which I checked and are correct):

$$\overline{I_1} = 1.556A\angle61.4^\circ$$
$$\overline{I_2} = 1.111A\angle-100.25^\circ$$
$$\overline{V_1} = 3.92V\angle139.8^\circ$$
$$\overline{V_2} = 1.57V\angle-55.25^\circ$$

My questions are these:

1. I got the value of V2 by multiplying I2 with (1 - j2 + j3). Why on earth don't I get the same answer if I use the fact that $$V_2 = I_{2}j2 + I_{1}X_{m}$$

Where Xm is the impedance as a result of the mutual inductance which I calculated using $$X_m = k\sqrt{j4 * j2}$$.

Also, shouldn't V2 be a stepped up/down version of V1 but phase shifted by 180 degrees? (from the dot convention)

2. How can I obtain the same results using the reflected impedance concept? Without knowing the turns ratio I can't do this and I can't seem to be able to find a way to find the turns ratio by using the coupling coefficient 'k' or any other of the known data. A hint would be much appreciated.

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Last edited: Apr 23, 2010
2. Apr 25, 2010

### Lunat1c

No replies? How come? Is my question that stupid? :/

3. Apr 26, 2010

### The Electrician

I do get the same result using V2 = I2*j2 + I1*jXm.

Don't forget to add that j in front of Xm.

If you'll show your arithmetic in detail, I may be able to see where you've gone wrong (assuming you can't find it yourself).

The usual conventions involving transformers, such as the assumption that the voltages on the windings are related by the turns ratio, and that impedances are transformed from one winding to the other by the square of the turns ratio, are only valid if the coefficient of coupling (K) is very close to 1 (it may be as high as .98 in a typical power transformer). Also, it is assumed that the self inductance (impedance, in other words) of the windings is very high compared to the other impedances in the circuit.

You can use the reflected impedance. For example, you can find V1 by calculating V1 = j8 - 4*I1. But, you could equally calculate it as V1 = I1*Zr1, where Zr1 is the reflected impedance at the left hand winding.

The turns ratio of an ideal transformer is the square root of the ratio of the inductances if the windings are closely coupled on the same core. The turns ratio of your problem transformer would ideally be 1.414:1, but because K is so low, the reflected impedance seen at the left hand inductor won't be just the square of the turns ratio (2) times the load impedance. That is, you might expect Zr1 to be 2 * (1+j1) = 2+j2, but it won't be that; I calculate Zr1 = .512 + j2.464.

4. Apr 26, 2010

### Lunat1c

First of all, thanks a lot for replying. I really appreciate you taking the time to do this.

Now,

I worked it again but I still don't get the same answer.

$$X_m = k\sqrt{L_1 * L_2} = 0.8\sqrt{4 * 2} = 2.26, \therefore jX_m = j2.26$$

$$V_2 = I_2j2 + I_1*jX_m = (1.111\angle-55.25^\circ)*j2 + (1.556\angle61.4^\circ)*j2.26 = 1.57\angle124.97^\circ$$

The phases are not equal at all and I can't figure why :(

I've never been told this before. Is there some kind of mathematical proof to this which I can look up? Not sure what I should search for in the first place

Ok so the turns ratio $$\frac{N_1}{N_2} = \sqrt{\frac{L_1}{L_2}} = \sqrt{2} = 1.414$$. Ignoring the impedance of the coil I would get $$Z_1 = 2*(1+j1) = 2+j2$$ just like you said at first.

What I don't get is how you got Z = 0.512 + j2.464. I know that probably considered the impedance of the coil but I'm not getting the same value as you

Last edited: Apr 26, 2010
5. Apr 26, 2010

### The Electrician

I'm not going to bother with tex. The problem is that you have I2 wrong; you're using I2 = 1.111<-55.25, but it should be I2 = 1.111<-100.25

You should be able to find a proof of this in a textbook dealing with transformers. Check your library. The classic book, "Magnetic Circuits and Transformers", MIT Staff, 1943, has a proof.

You can derive the impedance seen at the left hand winding be eliminating the 4 ohm resistor, changing the voltage source to 1 volt, and solving for I1 under those conditions. Then the impedance seen there will be equal to 1/I1 ohms.

6. Apr 27, 2010

### Lunat1c

Apparently that was a mistake when I was typing, I used the -100.25deg phase to get that answer. But now I realised that the drop on the coil, is in the same direction as the drop on the load of the second coil, but if the second coil is acting as a source, it's polarity should be reversed and hence a negative must be introduced. So If I just put a negative sign infront of it, i get exactly the answer I need.

7. May 26, 2010

### Lunat1c

My apologies. Seems like I have misunderstood what The Electrician said the first time round.

Last edited: May 26, 2010