[URGENT] Doubts on coupled inductors, pls advise...

[powersys]

"By making... For example, a Coiltronix CTX10-4 is a 10uH inductor with two windings. With the windings in parallel, 10uH inductance is obtained with a current rating of 4A... Splitting the two windings creates two 10uH inductors with a current rating of 2A each."

Question:
Why is the inductance when the windings connected in parallel similar to the inductance of each splitting winding?

Thanks

 

[Kral]

powersys,
Consider the situation where you have two un-coupled inductors, with inductance = L, and inductive reactance = wL. If you connect them in parallel (with no coupling), the effective inductive reactance is wL/2. Therefore the effective inductance of the parrallel combination is L/2. However, if you couple the two inductors, the inductance goes to L because of the mutual inductance m.
Regards,
Kral

 

[zorro]

Kral is right.
You can consider also this situation:
You have a unique winding. Now, split the wire longitudinally along all the winding, getting two parallel windings of half section and completely coupled. As the two parallel windings are equipotential at every point on their length, the longitudinal cut had not effect.
So, it is the same to have only one winding or the two in parallel.
Regards

Z

 

[powersys]

Reply from Coiltronics Application Engineer:

The CTX10-4 consists of two windings of equal turns on the same core structure. The inductance is determined by the number of turns squared times the core material factor. Since the windings are on the same core structure, the only changing parameter is the turns. If both winding have the same number of turns the inductance will be the same on each winding. If the windings are connected in parallel, the turns are the same so the inductance does not change.

 

[roy2li]

hi

could i say that for coupled two inductors,
each self-inductance =L/2, and the mutual inductance m=L/2,
thus, the total inductance =L?

And actually, the mutual inductance would be changed in the range of [0,L/2].

Am I right?

Thanks for your clarify!
Roy

powersys,
Consider the situation where you have two un-coupled inductors, with inductance = L, and inductive reactance = wL. If you connect them in parallel (with no coupling), the effective inductive reactance is wL/2. Therefore the effective inductance of the parrallel combination is L/2. However, if you couple the two inductors, the inductance goes to L because of the mutual inductance m.
Regards,
Kral

 

[Kral]

roy2li,
Yes, your equations is correct for the case where the coupling is 100%. For a nice discussion of mututal inductance, see
https://en.wikipedia.org/wiki/Inductance
Regards,
Kral

 

[olu]

how to calculate no of mosef needed to build various watts of inverter

 

[olu]

Is there any formula for calculating the number of power moseft needed for buiilding an inverter

Added after 7 minutes:

the moderator
Is there any formula for calculating the number of power moseft needed for buiilding an inverter