How do I relate flux equation to transformers?

[pghaffari]

http://imgur.com/VsDrQ,dJ2iv#1

So according to the picture, I found the transformer voltage equation and current equation to be the following:


Precondition:
i1 is current in left loop
i2 is current in 2nd loop

Transformer Voltage Eq:

V1/N1 = 4i2/N2


Current Eq:

Ni(i1) = N2(-i2)


But since we are given that flux(t) = flux_initial sin(wt) How do i use this and relate it to the problem..? I don't get it.

Thanks for any help.

 

[jim hardy]

seems to me they are asking you to find V1 as function of flux? After all, flux was described as "known".


Then you'd know V2, hence I2 and hence I1 all as functions of flux and N.
Does the source where you got your transformer equations give you a relationship between voltage and flux?

 

[pghaffari]

That's my question, how do i relate the voltage and flux? I am only given equation for flux in terms of t and then for transformers i only know the current equation and the voltage equation. What's the equation to relate voltage/flux or current/flux or something of that sort?

 

[jim hardy]

Have you really not studied induction?

Take a quick look at Faraday's law. I'd wager it is the whole point of that exercise.
here's some google results



http://www.google.com/url?sa=t&rct=...r4WPAg&usg=AFQjCNFEMtIAm0jJUOWdXYiV42uU5ROBpg

http://www.engineering.com/Library/.../articleId/207/Faradays-Law-of-Induction.aspx

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html

and if that doesn't clear it up, c'mon back...
probably you knew it already and just didnt make the connection.
That's what a lot of learning is - discovering what you already know.
hint - you're not going to get a solid number for answer, just an expression in terms of ω and [itex]\Phi[/itex]₀.


old jim

 

[pghaffari]

Actually I haven't even learned induction yet. I will learn it in another class in about 2 weeks from now. For some reason they didn't make that class (where you learn Faraday's law) a pre-requisite..

I'll check it out and get back to you.
Thanks.

 

[pghaffari]

I looked through those links you sent me and I can't find an equation that relates voltage and flux and I'm still not sure on how to proceed with this question.

 

[jim hardy]

"""I looked through those links you sent me and I can't find an equation that relates voltage and flux """



every one of them said e = -N * d[itex]\Phi[/itex]/dt ; [itex]\Phi[/itex] being flux

have you had first semester calculus where they introduce derivative?

Faraday Basics
Faraday's law of induction is one of the important concepts of electricity. It looks at the way changing magnetic fields can cause current to flow in wires. Basically, it is a formula/concept that describes how potential difference (voltage difference) is created and how much is created. It's a huge concept to understand that the changing of a magnetic field can create voltage.


Faraday's Work
Michael Faraday was an English physicist working in the early 1800's. He worked with another scientist named Sir Humphrey Davy. Faraday's big discovery happened in 1831 when he found that when you change a magnetic field, you can create an electric current. He did a lot of other work with electricity such as making generators and experimenting with electrochemistry and electrolysis.

Faraday's experiments started with magnetic fields that stayed the same. That setup did not induce current. It was only when he started to change the magnetic fields that the current and voltage were induced (created). He discovered that the changes in the magnetic field and the size of the field were related to the amount of current created. Scientists also use the term magnetic flux. Magnetic flux is a value that is the strength of the magnetic field multiplied by the surface area of the device.


Faraday's Law
You're going to have to review your Greek letters when you memorize the real formula. Here are the basics...
E=dB/dt


"E" is the value of voltage induced (the old name for voltage was "ElectroMotive Force", or EMF. That's the "E" in the equation). The change in time for the experiment is "dt". Time is measured in seconds. Last is "dB" which stands for the change in magnetic flux. The magnetic flux is the field lines of the magnetic field. The flux is equal to BA, where B is the magnetic field strength, and A is the area. This formula is a bit harder than those you may have seen before.

In English: the amount of voltage created is equal to the change in magnetic flux divided by the change in time. The bigger the change you have in the magnetic field, the greater amount of voltage.

that author used "B" for flux instead of [itex]\Phi[/itex].
http://www.physics4kids.com/files/elec_faraday.html

 

[pghaffari]

Voltage = d(flux)/dt for 1 turn in the coil (thats what my TA said today)

so if i have these 2 equations:

v(t) = 4 i2 * (n1/n2)

and v(t) = -4 i1 ( n1^2 / n2^2)

I set v(t) = flux initial omega cos(omega t) and then what.. ?

i need to solve for i1, and v1..

so is that my answer?

v(t) = the derivative of flux which is flux w cos(wt) ?

then plug that in the equation and solve for i 1 ?

 

[jim hardy]

You got it - good job!

glad you persevered.
Careful algebra will solve for all in terms of flux, cos(wt), and turns.
Now you have a headstart on induction for that next class.

v(t) = the derivative of flux which is flux w cos(wt) ?

don't forget N.

old jim

 

[pghaffari]

Wow that was simple enough.. Thanks a lot for your help!

So my final answer I get is:


v1(t) = -N1 flux w cos (wt)

i1(t) = (N2)^2 flux w cos (wt)

v1(t) / i1(t) = -N1 / (N2)^2


Is this correct?

 

[jim hardy]

Hi pghaff...

well let's see... i would have expected there'd be a 4 in there someplace because that was one of the few hard facts they gave us.

Glad you got the derivative concept. That's important, and gets real interesting when you have a non-sine wave...

Okay let's set that derivative w * flux * cos(wt) to DFLUX just to simplify typing.

V1 = - DFLUX * N1, as you wrote

I2 is (volts over there) / (resistance over there) = (-DFLUX * N2) / 4
and I1 = I2 * N2/N1

so I1 = (-DFLUX * N2/4 ) * N2/N1,
which = (-DFLUX * N2^2) / (4 * N1) did one of us mess up on N's?

and V1/I1 = (-DFLUX * N1) / (-DFLUX * N2^2) / (4 * N1)
which = (N1) / (N2^2) / (4 * N1)
which = (4*N1^2) / (N2^2)

which = 4 (N1/N2)^2

they're telling us ohms are transformed by square of turns ratio.

Check both our arithmetics and see what's right...

now i see why teachers make us show our work...

Thanks for your interest. Makes an old man feel less useless!

old jim

 

[pghaffari]

Yours is right I believe, but here:

and I1 = I2 * N2/N1I believe it should be:

I1 = I2 * -N2/N1

Because the current i2 is flowing OUT of the transformer at the dot, so shouldn't it be Negative value?

Thx again

 

[jim hardy]

hmmm i worked quickly and didnt draw an arrow for current over on secondary.
my bad - inattention to detail. I should've assigned a direction.

i think you're right, and most important you have got the point of the exercise !

I wish you success in your studies.

This stuff isn't so hard, what is hard is to apply the necessary rigor and discipline.

Form the habit of being meticulously neat in your algebra - align your equations, show every step and life will go easier. Better to work a problem once neat and slow than five times sloppy and fast producing four different answers.

Probably you're already neat and orderly , evidenced by your catching my sign mistake.

have fun !