Induced voltages and step-up transformers

In summary, for the three problems discussed, the first problem can be solved by finding the angular frequency and converting it to rev/s. The second problem involves finding the ratio of currents in the primary and secondary coils, and the third problem requires finding the ratio of turns in the primary and secondary coils to determine the voltage change factor.
  • #1
kitkat2950
7
0
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.
 
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  • #2
Hi kitkat2950,

kitkat2950 said:
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?

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

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

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.

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?

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.

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.)
 
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  • #3
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
annythewitch said:
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

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:

[tex]
I_p V_p = I_s V_s
[/tex]

and get the ratio in the next step.
 
  • #5
you are right. I mistyped the fraction. it should have been 50/110, and yeah, your way is easier.
 
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FAQ: Induced voltages and step-up transformers

What is an induced voltage?

An induced voltage is the voltage that is generated in a conductor when it is placed in a changing magnetic field. This voltage is caused by the movement of electrons within the conductor, which creates an electric current.

How does a step-up transformer increase voltage?

A step-up transformer works by increasing the number of turns in the secondary coil compared to the primary coil. This creates a greater magnetic flux, which in turn induces a higher voltage in the secondary coil.

What is the purpose of a step-up transformer?

A step-up transformer is used to increase the voltage of an alternating current (AC) power supply. This is necessary for efficient transmission of electricity over long distances and for use in high-voltage equipment.

Can a step-up transformer also decrease voltage?

Yes, a step-up transformer can be used in reverse to decrease voltage, as long as the secondary coil has fewer turns than the primary coil. However, the majority of step-up transformers are designed for increasing voltage.

What are some potential dangers of using a step-up transformer?

When working with high voltages, there is always a risk of electric shock. Additionally, step-up transformers can also generate heat and pose a fire hazard if not used properly. It is important to follow safety protocols and use caution when working with step-up transformers.

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