# Homework Help: Induced voltages and step-up transformers

1. Jul 21, 2008

### kitkat2950

I have been trying, but I still have 3 problems I can't get.

1)A simple generator has a 600 loop square coil 18.0 cm. on a side. How fast (in rev/s.) must it turn in a 0.730 T. field to produce a 120-V. peak output?

I think I need to use the equation Emax=NBAw. N=600 B=.730 A=.0324 Emax=120
I found w to be 8.456 but I didn't know if it was in rev/min or rev/s. So I tried that answer and then tried dividing it by 60 for seconds, but neither answer worked. Any help?

2)A step-up transformer increases 50 V. to 110 V. What is the current in amperes in the secondary as compared to the primary? Assume 100 percent efficiency.

I know the primary voltage is 50V and the secondary voltage is 110V. And I think Vp/Vs=Ip/Is, but I'm not sure how to find the answer w/o one of the currents.

3)A transformer has 32 turns in the primary and 181 turns in the secondary. Assuming 100 percent efficiency, by what factor does it change the voltage?

I think this one is basically the same as number 2, and I just don't understand what I am supposed to do. I think I'm supposed to divide one by the other, but when i put those answers in, they're not right.

2. Jul 21, 2008

### alphysicist

Hi kitkat2950,

The $\omega$ is angular frequency, which is not in rev/min or rev/s; what are the units of angular frequency?

Once you find $\omega$, you'll have to convert it to rev/s.

It's strange that the ask for the current in the secondary in amperes. When they say "as compared to the primary" that suggests that what they are looking for is the ratio of the currrents. Was there a diagram or anything else in the problem?

Your equation is not correct. If the transformer is 100% efficient, what is not lost by the transformer?

What is the relationship between voltage and turns in a transformer? (Without knowing what numbers you used and got, it's difficult to determine what you might have done wrong.)

Last edited: Jul 21, 2008
3. Nov 7, 2008

### annythewitch

2. the ratio you want is:

Vs/Vv=Ns/Np: Is/Ip=Np/Ns >>> Vs/Vp=Ip/Is >>> Is/Ip=Vp/Vs = 50/100= .5

4. Nov 7, 2008

### alphysicist

No, I don't believe that's the answer. You seem to have put in a wrong number.

Also, you have put in a few extra steps. There is 100% efficiency, so the power is the same on both sides, so you can immediately write down:

$$I_p V_p = I_s V_s$$

and get the ratio in the next step.

5. Nov 8, 2008

### annythewitch

you are right. I mistyped the fraction. it should have been 50/110, and yeah, your way is easier.

Last edited: Nov 8, 2008