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kwatson18

CIV1 #21 vs NCEES #116 - Three legged Magnetic Ckts

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Felt ok about these types of problems, after going through NCEES #116 which involved a picture of a 3-legged magnetic circuit with legs A-B-C from left to right.

Leg A is 3.0x10^-3 Wb, then they ask assuming linear magnetic properties, the approx flux (Wb) in Leg is mostly:

In the picture they give dimensions for the widths and lengths of the legs and the answer basically has you caclulate the path using like a perimeter/ratio calculation. Made sense....

Now in regards to Complex Imaginary Vol 1, #21 I get a little confused. When i first looked at it, i thought, very similar, should be cake.

Again, 3 legged magnetic core, with flux in leg A being 0.004 Wb. They ask "If the flux density in leg A is 1.6 Teslas (t), what is the dimesion of "x" (referring to the width of Leg A). Again they give you tons of dimensions to every part of the core.

I (flux) = B (teslas) x Area (m)

So my question is : HOW / WHY do they calculate the Area, A, of Leg A as " x^2" ?

Should the area be x * 15'' (which is the height of Leg A?) Is it because it has depth???

You kind of have to have the picture in front of you, but they have all these dimensions all over the diagram that are very similar to the NCEES#116. Just trying to make sure I really know whats going on so if there's a variation on Friday i know what's up

Thanks!!!

kW

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For the CI problem: The cross-sectional area at any point in the core is x^2. If you look at the top right of the core you'll see an <--x--> in white. It's hard to see on the grayscale shading. For CI the only dimension that matters is the area of the core A. 1.6T = 1.6Wb/m2. If 1.6Wb/m2 * x^2 = 0.004Wb, then x^2 = (0.05m)^2, so x = 5cm. Hope that clarifies!

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I have a similar question only I am confused with the NCEES method. Why wouldn't the path distance for leg B be 2d instead of d?


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