Transformers and induced current

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Homework Help Overview

The discussion revolves around problems related to transformers, specifically a step-up transformer increasing mains voltage and a step-down transformer used in a soldering gun. Participants are exploring calculations involving turns ratio, current, and voltage relationships in transformers.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to calculate the turns ratio of a transformer and the primary current based on given secondary current. There are questions about the appropriate formulas to use for these calculations, particularly regarding the relationships between voltage, current, and turns.

Discussion Status

Some participants have provided partial answers and calculations, while others are seeking clarification on their approaches and results. There is a mix of correct answers and requests for further understanding, indicating an ongoing exploration of the concepts involved.

Contextual Notes

Participants are working within the constraints of homework rules, which may limit the information they can use or the methods they can apply. There are also discussions about the physical properties of materials used in the transformer setup, such as resistance considerations.

_Mayday_
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Hello.

3) A step up transformer increases the mains voltage of 230V to 5000V.

a) calculate the ratio Number of primary turns/Number of secondary Turns

I know the formula input voltage/output voltage = Primary turn/ Secondary Turns

So i did 230/5000 = 0.046. This as a fraction is 23/500. Is this correct?

b)Calculate the current in the primary when the secondary current is 0.5A.

I really am clueless here. I know of a formula Input voltage x input current = output voltage x output current. Is this the correct one to use

4) The diagram shows a solder gun. It is a step down transformer with a single turn winding. When the voltage across the primary coil is 240V, the primary current is 0.6A and the induced voltage in the secondary coil is 2.0V

a) Calculate the number of turns of wire in on the primary coil.

I don't know which formula to use here.

b) Calculate the current in the secondary winding.

This would have something to do with part a so if i find out which formula to use i should be able to do this.

c) Why is the secondary winding made from thick aluminium rod.

Is this to do with resistance?

d) Why is the nib of the solder gun thiner than the rest of the winding.

Is this also to do with resistance



Thanks for any help. I've tried my best to have a go. Thanks
 
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Your first answer is correct. To answer the second, you need to know that
I_sec = I_pri / R
where R is the ratio of secondary to primary turns (reciprocal of the ratio you found).

You could have found this from the two equations you listed!

4a) Use the equation you know for the voltages.
b) you have this equation, too.

Try these and get back to us if you need more help.
 
3b) input current x input voltage = output voltage x output current

240V x input current = 5000 x 0.5

input current = 10.42A

is this correct?

4a) i got 120 primary turns

b) I used the formula from 3b) and i got 72 in the end.

I would like some clarification here.

thanks a lot marcus
 
_Mayday_ said:
3b) input current x input voltage = output voltage x output current

240V x input current = 5000 x 0.5

input current = 10.42A

is this correct?
It would be if you used 230V as stated in the problem.

_Mayday_ said:
4a) i got 120 primary turns
b) I used the formula from 3b) and i got 72 in the end.
Both correct.

_Mayday_ said:
I would like some clarification here.

thanks a lot marcus
4c) Think about what kind of wire is needed to carry 72 amps!
4d) Now think about where you'd like this wire to get hot, and where you want it cool.
 

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