Resultant magnetic flux in a 3-phase induction motor

[rishi kesh]

Homework Statement

I have some attachements below from my textbook. According to it the value of resultant flux is constant in magnitude and 1.5 times the maximum value of flux.
But i have a doubt regarding how it is proved.
The magnetic flux waveforms are 120° apart from each other and at time 0, the instantaneous value of flux waveform of 1st phase is zero. Whereas 2nd phase has magnitude -√3/2.Φm and the third phase √3/2.Φm.The textbook says that:
Φr = 2.(√3/2)Φm.cos60/2
Here is my doubt
If the fluxes are 120° apart, shouldn't the equation be: Φr = 2.(√3/2)Φm.cos120/2 ?
Please explain me in the simple way how this works.i appreciate the help:) [/B]

Homework Equations

I know that when forces have same magnitudes then for resultant force we use the formula:
R=2.F.cosΘ/2[/B]

The Attempt at a Solution

I have been trying to think about it and draw a waveform to figure it out but couldn't understand it.

 

[cnh1995]

Looks like you're studying from B.L.Theraja.

You can do it yourself. Just draw two phasors of magnitudes √3/2 and -√3/2 , and 120° apart. Use the parallelogram rule of vector addition and you'll get Φ=1.5Φ_m.

 

[rishi kesh]

Looks like you're studying from B.L.Theraja.

You can do it yourself. Just draw two phasors of magnitudes √3/2 and -√3/2 , and 120° apart. Use the parallelogram rule of vector addition and you'll get Φ=1.5Φ_m.

Hi! Can you please solve this and send a picture right here? I will appreciate that. And also explain this little bit more.

 

[cnh1995]

Hi! Can you please solve this and send a picture right here? I will appreciate that. And also explain this little bit more.

That's one of the things you shouldn't ask for here. It is against the forum rules.
https://www.physicsforums.com/threads/guidelines-for-students-and-helpers.686788/

And also explain this little bit more.

Which part is not clear?

 

[rishi kesh]

That's one of the things you shouldn't ask for here. It is against the forum rules.
https://www.physicsforums.com/threads/guidelines-for-students-and-helpers.686788/ Which part is not clear?

Please explain me how to derive it to get resultant flux. I really need help.

 

[cnh1995]

Please explain me how to derive it to get resultant flux. I really need help.

Do you know the parallelogram law of vector addition?

Just draw two phasors of magnitudes √3/2 and -√3/2 , and 120° apart. Use the parallelogram rule

That's how you can derive it. I can't make it any simpler.

 

[rishi kesh]

Do you know the parallelogram law of vector addition?

That's how you can derive it. I can't make it any simpler.

This is what I am getting as answer

 

[rishi kesh]

This is what I am getting as answer

But the same one when i solve using cos60/2 :

 

[cnh1995]

The formula you used in #7 works for two 'equal' vectors. The fluxes here are √3/2 and
-√3/2, which are not equal. You can rotate the second vector by 180° and its magnitude will be √3/2. Now the angle between the two phasors is 60° and the formula works as it did in #8.

 

[rishi kesh]

The formula you used in #7 works for two 'equal' vectors. The fluxes here are √3/2 and
-√3/2, which are not equal. You can rotate the second vector by 180° and its magnitude will be √3/2. Now the angle between the two phasors is 60° and the formula works as it did in #8.

Oh i got it! Here is one more attachment. This is how i need to think about it?

 

[cnh1995]

Oh i got it! Here is one more attachment. This is how i need to think about it?

That's right.

 

[rishi kesh]

The formula you used in #7 works for two 'equal' vectors. The fluxes here are √3/2 and
-√3/2, which are not equal. You can rotate the second vector by 180° and its magnitude will be √3/2. Now the angle between the two phasors is 60° and the formula works as it did in #8.

Thanks so much i really got this concept clear today. I realized that what i didn't know is that vectors can't have negative magnitude now if i keep this in mind my concept about resultant flux is clearly understood. :)
I am so thankful to you :)

 

[cnh1995]

Thanks so much i really got this concept clear today. I realized that what i didn't know is that vectors can't have negative magnitude now if i keep this in mind my concept about resultant flux is clearly understood. :)
I am so thankful to you :)

You're welcome!

 

[rishi kesh]

You're welcome!

Hi !
Sorry, but one more question is overwhelming me right now.
Does the magnitude of resultant flux not get affected when we move -√3/2 by 180°.
Is it okay to do so ?

 

[cnh1995]

Hi !
Sorry, but one more question is overwhelming me right now.
Does the magnitude of resultant flux not get affected when we move -√3/2 by 180°.
Is it okay to do so ?

Yes.
Rotating a vector by 180° and changing its sign gives the same vector. It is like multplying a number by -1 twice.

 

[rishi kesh]

Yes.
Rotating a vector by 180° and changing its sign gives the same vector. It is like multplying a number by -1 twice.

Okay. So what does negative sign mean ? Does it mean opposite direction? Like if right side is positive then left is negative?

 

[cnh1995]

Okay. So what does negative sign mean ? Does it mean opposite direction? Like if right side is positive then left is negative?

Kind of like that, yes.