Impedance matching transformer question

[ch5497]

1. It is required to pass a current of between 6.0A and 6.5A through an impedance of (1.6 + 1.2i) ohms. With only a 230V supply available, it is proposed to use a transformer for impedance matching. A transformer is available with a turns ratio of 16:1. A laboratory test showed that with the secondary winding short-circuited, the input to the primary winding was 1A and 102.4W for an excitation voltage of 128V.

Determine whether the transformer would be suitable for this application



2. None specified



3. So I've got the following so far;

P = I²R
102.4W = 1²R
R = 102.4 ohms


V = IZ
128V = 1 . Z
Z = 128 ohms


X²= Z²-R²
X²= 128²-102.4²
X = 384/5 = 76.8 ohms

Now, do I need to refer this impedance over to the secondary by multiplying by the turns ratio squared, and then add on the impedance given in the question? I could then find the current being drawn from the 230V and see if it's in the range required? I've done this, and gotten an answer, but without knowing if my method is correct, I don't know whether my decision on whether the transformer is suitable or not is accurate.

Thanks in advance.

 

[skeptic2]

1. It is required to pass a current of between 6.0A and 6.5A through an impedance of (1.6 + 1.2i) ohms. With only a 230V supply available, it is proposed to use a transformer for impedance matching. A transformer is available with a turns ratio of 16:1. A laboratory test showed that with the secondary winding short-circuited, the input to the primary winding was 1A and 102.4W for an excitation voltage of 128V.

Determine whether the transformer would be suitable for this application
2. None specified
3. So I've got the following so far;

P = I²R
102.4W = 1²R
R = 102.4 ohmsV = IZ
128V = 1 . Z
Z = 128 ohmsX²= Z²-R²
X²= 128²-102.4²
X = 384/5 = 76.8 ohms

Now, do I need to refer this impedance over to the secondary by multiplying by the turns ratio squared, and then add on the impedance given in the question?

No, you already know the impedance at the secondary, it's a short, remember? You should take the short off the secondary and put the 1.6 + 1.2i ohm impedance on the secondary instead. What impedance would you then see at the primary? Don't forget that with a short on the secondary you still have some impedance at the primary. That has to be added in.

I could then find the current being drawn from the 230V and see if it's in the range required?

There are two things you need to find.
1. You have to be sure the transformer can supply all the current that load requires and
2. if it can, is the load current in the range of 6 to 6.5 amps?

To find the first you have to transform the load impedance back to the primary to see how much current the primary will draw and then convert that into current available at the secondary.

To find the second calculate how much voltage there is at the secondary if the primary is connected to 230 V and calculate the current flowing through the load. If it is less than the available current, is it in the range of 6 to 6.5 amps?