# Current and Impedance in a Transformer

1. Apr 3, 2013

### pious&peevish

My textbook doesn't cover transformers in very much detail at all, so I only have a very vague idea as to how they work...

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

The secondary voltage of an ignition transformer in a furnace is 13.9 kV. When the primary operates at an rms voltage of 120 V, the primary impedance is 22.6 Ω and the transformer is 91.1% efficient.
a) What turns ratio is required?
b) What is the current in the secondary?
c) What is the impedance in the secondary?

2. Relevant equations

V2/V1 = N2/N1 (where N2/N1 is the turns ratio)
V1 * I1 = V2 * I2
V1/I1 = R/(N2/N1)^2

3. The attempt at a solution

I already solved a) correctly, which was just 13.9 kV/120 V (i.e. N2/N1). But besides that, I honestly don't know what else to do. I guess I could start by plugging in 22.6 ohms into the equation for impedance [i.e. I(rms) = V(rms)/Z], but I'm getting completely mixed up. I'm also not sure how efficiency affects the final result...

2. Apr 3, 2013

### Sunil Simha

Efficiency refers to the ratio of V*I power of secondary and primary circuits. So, if you find Iprimary you can find Isecondary and hence the impedance of the secondary circuit.

3. Apr 3, 2013

### pious&peevish

Thanks! It doesn't say whether or not the secondary voltage (13.9 kV) is an rms voltage... should I assume that it is?

4. Apr 3, 2013

### Sunil Simha

Yes I think it should be assumed so.

5. Apr 3, 2013

### SammyS

Staff Emeritus
That's a reasonable assumption, since it says that the primary voltage is rms .