Electrical Engineering - Transformers

[YoshiMoshi]

Homework Statement

Please see the picture attached.

Homework Equations

The Attempt at a Solution

I was able to solve part A correctly, at least I believe I was able to do so. I'm struggling with part (b) however. I'm not sure were to start on this. My test that has this problem doesn't have anything on MLT. I pulled out my Power Electronics book from my undergraduate and found that

R = (rho * l_b)/A_w

R -> Resistance of the winding
rho -> resistivity of the conductor material
l_b is the length of the wire
A_w is the wire bare area

I believe I need to find l_b? I don't understand exactly how I can go about this. I believe that for A_w

A_w = (5 cm)(3 cm) = 15 cm^2
I used 3 cm because it's a 3 cm stack as specified in the problem

Am I just supposed to assume for rho that I'm using copper winding?

I don't understand the hint and how I'm supposed to solve the problem using Ohm's law. The wire has a resistance R, and I found the MMF in part A, but not the voltage. Hence I'm stuck because

V = IR

I don't know V or I. and I know that

F = IN

Where I solved for F in part (a) but don't know I or N

Any help would be greatly appreciated.

 

[rude man]

What is μ for the laminations? What does "S.F. = 0.90" mean?
You can get an approximate answer by neglecting reluctance everywhere except in the gap.

 

[cabraham]

What is μ for the laminations? What does "S.F. = 0.90" mean?
You can get an approximate answer by neglecting reluctance everywhere except in the gap.

I used to design xfmrs. "S.F. = 0.90" means that the "stacking factor" value is 0.90. "SF" is the percentage of the wound metal tape that consists of ferromagnetic material. The tape is coated with an insulation layer on both sides. When the tape is wound some of the core volume is insulation and this contributes little to flux path. At high frequencies thinner tape is needed to keep eddy current losses low. But insulation thickness remains about constant. So for a 12-mil tape thickness, and a 0.4-mil insulation thickness on each surface of the tape, stacking factor can be computed as follows. The total tape thickness = 12 + 0.4 + 0.4 mils = 12.8mils. Of the 12.8 mils, 12 mils is actual ferrous material. So SF = 12/12.8 = 0.938. If the tape thickness was only 4 mils, needed for higher frequency operation, with 0.4-mil insulation thickness on each side, the stacking factor is 4 / (4 + 0.4 + 0.4) = 0.833.

In general the thinner the tape, the lower the SF. Did I help? Regards.

Claude

 

[rude man]

Thank you Claude for the information but the OP seems to have evanesced as usual. He/she did not respond with your info nor with a permeability number.