capa 9 - Phys 1120 CAPA#9 solutions 1 The trick here is(as...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Phys 1120, CAPA #9 solutions 1) The trick here is (as the problem hint suggested) to think of this as a stack of little pancakes, each of thickness "dx". (See the picture) They are all in a row, in series, so the total resistance is the sum of the little resistances, R = Integral(dR). So what is the resistance "dR" of a little pancake? Well, a pancake is simply like a tiny cylinder, and we know R of that, it's just R = rho * L / Area, or in this case dR = rho * dx / Area. They GAVE us rho, and we just need to stare at the picture and think about the area of the little pancakes. That would be "pi r^2", but r DEPENDS on x! So now it's a geometry puzzle, how does r vary as x moves from 0 (at the left) to h (at the right)? r=a when x=0, and r=b when x=h, and it varies linearly, so the formula must be r = a + (b-a) x/h (Do you see that? It's the formula for a straight line with intercept a and slope (b-a)/h. If you didn't come up with it on your own, *think about it* till you see how you could have gotten it yourself!) So we're all set up, R = rho*integral(from 0 to h) of (dx / [pi * (a + (b-a)x/h)^2] (Do you see that? it's just adding up rho dx / pi r^2. ..) The integral may look nasty, but it's not so bad, basically like integral(1/x^2) = 1/x, except here we have integral (1/(c + d x)^2) = (1/d) (1/[c+dx]) Do you see why? Once again, don't take my word for it, work out that integral! How do you normally do that - go back to your Calc 1 book, or just think about the rules of integration, but we do expect you do be able to do integrals of this difficulty level . .. If you can't, talk to someone for a little help or review, like one of the professors or your TA, or. ..) So I get R = (rho/pi) * (-h/(b-a)) / (a + (b-a) x/h), all evaluated from 0 to h. This gives me a brief mess, and then I factored and simplified, and got (rho/pi) * h/(ab). Wow, very simple! It almost makes me think there must be a neat trick that I missed, but anyway, that's my result. At this point, after doing algebra and integrals like this, I *really* want to stop and look at the result -does it make sense? It says R = rho h / (pi a b). If h gets longer, it gets bigger - definitely reasonable! Units are right: rho * (distance)/(distance^2), that's good. As the faces get bigger, (either a OR b), the resistance goes down. Also makes sense! So nothing about it seems crazy to me. In fact, given that R = rho *Length/Area, if you had to make a GUESS for a formula, this seems extremely logical. Length is just. .. length (h). And area is, well, some sort of "average"area - it's not pi a^2, it's not pi b^2, it's inbetween, pi a b! Kind of cool,.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2) Doesn't look at first like it should be hard, but this one does involve a little algebra. Let me define currents I1, I2, and I3 all to be going DOWN the page. Remember, that's arbitrary, doesn't matter what I pick! (Do you see this? If I guess wrong, they'll just come out negative, no big deal). They are the 3 unknowns. Loop 1, clockwise through the "left half":
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/28/2008 for the course PHYS 1120 taught by Professor Rogers during the Fall '08 term at Colorado.

Page1 / 5

capa 9 - Phys 1120 CAPA#9 solutions 1 The trick here is(as...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online