the flux will decrease if number of turns are increased

[PG1995]

the flux will decrease if number of turns is increased

Hi

Could you please help me with the queries included in the attachment? Thank you.

Regards
PG

PS: After looking at davenn's post below, I have corrected the mistake and have added some extra information.

 

[davenn]

why would you think the inductor is an open cct to DC ?
its going to be a short cct or at least low resistance depending on the number of turns

Dave

 

[MrAl]

Hi,

For simple inductor calculations the length of the magnetic path is of concern. That's the total length the flux would pass through. So for a core with square OR rectangular OR circular cross section (as long as it was uniform) and say it was a rod bent into a square shape with ends touching perfectly with no gap, then the magnetic path would be one side multiplied by 4 (because a square circumference is simply c=4*side). The actual length though is the *mean* length, so if the inside of the core on one side was 2 and the outside of the core on one side was 4, then the mean length of that one side would be 3 (just the average of the two). That would make the total magnetic mean path length equal to 12.

If the core was a circle and the inside diameter was 2 and the outside 4, then the mean diameter would be 3 and so the magnetic path length would be 3*pi.

So the length used for simple inductor calculations is the mean magnetic path length which is the mean magnetic path length of the entire core.

What this means is that if we had core material that was 1 inch square cross section for example and we made a core that had 4 inches mean mag path and one that had 8 inches mean mag path length, the smaller one would have a higher inductance than the second.

 

[RCinFLA]

Not sure if you got your answer on why current goes down.

For a given magnetic length, (not changing the size of the core), the flux density will be proportional to N*I. The current is a direct function of inductance (1/L) but the inductance is proportional to turns squared (N^2). For the same excitation voltage the inductance ratio squared wins, reducing the current, more then offseting the effect of adding more turns.

 

[Bob Scott]

Hi

Could you please help me with the queries included in the attachment?

In the first paragraph, the equation for inductance is given as:

L = mu * K * N^2 * A / l

The "l" is the effective core length le, not lx.

We can derive the above equation. The full equation for inductance is:

L = 0.4 * pi * N^2 * mu *Ae / Le * 10^-8.

If we create the constant K as being equal to "0.4 * pi * 10^-8" then:

L = mu * K * N^2 * Ae / Le