1120, Phys 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,.
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": +I1*R1 + V1 - I2*R2 = 0. (See the concept tests or last week's homework for the explanation of signs Be sure you can get them all right by yourself without any help! The key is, it's a POSITIVE term if you're walking past a resistor opposite the way the current arrow points (walking "up the creek") and it's a NEGATIVE term if you're walking past an R in the same way as the current arrow, walking "down the hill")) Loop 2, clockwise through the "big circle": +I1*R1 + V1 + V2 - I3*R3 = 0. You might have chosen the other small loop, that's o.k. But, doing all three loops won't help any, the equation is equivalent to the sum of the two I just wrote down! The third equation is thus NOT a loop equation, it's the current conservation equation: I1+I2+I3=0. (Do you see why those are all + signs? It matters! Of course, if you picked some different directions for YOUR current arrows, your equations will look slightly different. But we'll have to agree in the end!) I think I'll use the last equation to eliminate I2, since I don't care about I2! I2 = -I3-I1. Now plugging this in for I2 in the loop 1 equation, that one now looks like I1*R1 + V1 + (I3+I1)*R2 = 0 And, The loop 2 equation still looks like I1*R1 + V1 + V2 - I3*R3 = 0 At this point, you can solve these two equations formally (they have 2 unknowns, I1 and I3). I'm going to simplify at this point with the givens: all three R's are the same (call them "R"), and V2=2V1. So I have I1*R + V1 + (I3+I1)*R = 0 I1*R + 3 V1 - I3*R = 0. If you add those equations, you get 2*I1*R + 4 V1 + I1*R =0, (Nice, I3 is gone!) So, 3 (I1)* R = -4 V1, ir I1 = -4 (V1)/3 R. (The minus sign tells me I1 was really UP the page, which actually makes sense if you look at how the two batteries are oriented, they clearly drive current overall clockwise around the whole circuit) CAPA wanted the magnitude, though, so use 4/3 (V1)/R. That's it - you could now go back and figure out I2 and I3 if you needed those...
3) BEFORE the switch is thrown, the battery has already charged up the capacitor. (Initially, current flows across the resistor to "charge up" the top plate of C, but after a while, when the top plate of C gets to a voltage Vc=Vb, there is no more voltage drop across the resistor, no current flows, and Vc=Vb. That's all the setup, it's all "before". Now we throw the switch. The battery is out of the picture! You have a capacitor and a resistor in a simple little RC circuit, just like on P. 985 of the text. The voltage on the right side of the resistor is still Vc (remember, charge on the capacitor cannot instantly change, and Q=CV means V across the capacitor cannot instantly change. So if you call the bottom of the picture V=0, the top of the capacitor is still at the same V as it was before, Vc) ,But the LEFT side of the resistor has been switched, now the left side is hooked to the V=0 wire! So the resistor DOES have a voltage drop across it now! Just BEFORE the switch is thrown, we just said there's no current flowing, Ohm's law says there is ZERO VOLTAGE drop across (and ZERO Current through) the resistor. Just AFTER the switch is thrown, the voltage across the resistor is now Vc, it is NOT zero. (That's what I just argued above) Ohm's law says that if there is a voltage drop across there resistor, there IS a current through it. It is NOT zero. So the current just "turns on", very fast! Just after the switch is thrown, the voltage across the the capacitor is also Vc, still it is NOT zero. Conservation of charge says the current is always the same throughout a loop, so there IS a current "through" the capacitor. (The capacitor is discharging, current is flowing OFF the top plate, through the resistor, onto the bottom plate) As always, the current in a loop is the same EVERYWHERE. 4) Eq 31-39 of your text says I = V0/R*exp(-t/RC). (Do you understand where it comes from? Don't just treat formulas as "black boxes", it's more important that you think about why it's true, and what it says! This is the classic deal with CAPA - you can get a "yes" by plugging and chugging into appropriate formulas, but our goal is to encourage you to think about what's going on, where the formulas COME from... Here, V0 is "Vb", the battery voltage is the SAME as the starting voltage across the capacitor. (That was described in the previous problem) I(given) = V0/R exp(-t/RC), which means I(given)*R/V0 = exp(-t/RC) Take the ln of both sides (ln is the inverse of exp, it's the natural log) to get ln(I(given)/(V0/R)) = -t/RC, and finally, t = -R*C*ln(I(given)*R/V0) Be careful you use ln, not log, on your calculator.
5) A hard way to do this is to integrate I(t)^2*R. After all, d(energy) = power*d(time), so total energy = Integral[Power*dt]. (It's a good exercise, and it will give you the right answer) The EASY way is to use conservation of energy! All the energy is initially stored in the capacitor, and it is ALL dissipated in the resistor. So, the answer is (1/2) C V^2, where C was given, and V=V0 = "Vb", the original voltage on the capacitor. You might still want to try the integral, and see that it gives you the same answer, if you like that sort of thing! 6) As discussed in the virtual office hours, this question only makes sense if you assume that they're asking all questions "after waiting a bit, for the circuit to settle down". If you look at the circuit as shown, it says "the capacitors are uncharged". Well, that must have been BEFORE we hooked up the voltage V there. Since they started off uncharged, and since the switch S is open, no charges can ever enter or exit the region BETWEEN the capacitors. So conservation of charge says Q1=Q2, whatever charge is on capacitor #1 must ALSO be on #2 (because they are hooked together in series) Of course, at first Q1=Q2=0, they start off uncharged, but once the battery is hooked up, the capacitors WILL pick up charges very fast! So the circuit is hooked up as shown, and sits for a second. Some charge will rush onto C1 (and the same number on C2) from the battery. Of course, current will ALSO be flowing down the line of resistors, since there's a Delta V across them! So current FLOWS down the resistors, and continues to flow. Current can NOT continue to flow through the capacitors, almost immediately the ONLY current flowing is straight down the resistors. So look down the resistors: there is a current I, which must be given by V/(R1+R2). (Since R1 and R2 are in series with each other!) And so, there is a voltage drop across R1, given by Delta V(1) = I R1 = V R1/(R1+R2) and a voltage drop across R2 given by Delta V(2) = I R2 = V R2 / (R1+R2) (I get that by using I = V/(R1+R2) Since Delta V(2) = V(a) -0, we immediately have the answer, V(a) = V R2/(R1+R2). The capacitors really had nothing to do with this, did they? 7) Meanwhile, to get V(b), we can similarly ignore those resistors! Just look down the "row" of capacitors. We have V = V(across C1) + V(across C2) (Do you see that? Pure Kirchhoff!) So V = Q(1)/C(1) + Q(2)/C(2), but earlier we argued Q(1)=Q(2). And in fact Q(1) is easy to find, because the two capacitors are just in series, and so they act like an effective single capacitor With C(eff) = 1/(1/C(1)+ 1/C(2)), since that's how capacitors in series add up. Thus, V = Q/C(eff), and so I can solve for Q. But that's the SAME as Q(1) and also Q(2). So now V(b)-0 = Delta V(across C2) = Q(2)/C(2) = C(eff)*V / C(2). That's it, V(b) = (V/C2)*C(eff).
8) When you CLOSE switch S, points a and b have to be at an equipotential. Charges can now freely flow from or to those capacitors. But in steady state, charge cannot CONTINUE to flow to or from them, so in the long run, current will again just flow straight down through the resistors and go nowhere else. This is exactly the same situation as back in question 6, and so the answer here is just exactly the answer for that one too! 9) Think of the solar cell as like a battery with an internal resistance r. Call the battery voltage "V". Then, you have a simple circuit: V, then internal resistance r in series, then external resistor R in series, and back to the battery. So V(ideal, cell) -ir - i R =0 for the circuit, or you could say V(ideal) = i(r+R). Now, when CAPA tells you the "potential difference", it must be telling you the potential difference across the external resistor. (What else COULD it mean? You can't measure anything else here! It can't mean V(ideal, cell) because there's no way to measure that directly. All you can measure it's the output voltage!) So we know V(given, across R) = iR, thus we solve: i = Given V / given external R. So for the FIRST case: V(ideal) = i1(r+R1) = V1(given) /(500 Ohms)(r+500) (since R1=500) For the SECOND case: V(ideal) = i2(r+R2) = V2(given)/(1000 Ohms) (r+1000) (since R2=1000) Thus, setting the two V(ideals) equal, and cross multiplying by 500 Ohms and 1000 Ohms, V1(given)(r+500)*1000 = V2(given)(r+1000)*500 This means 2V1(r+500) = V2(r+1000), or r(2V1-V2) = 1000*(V2-V1) That's all we need! 10) You should draw yourself a simple little circuit diagram. It has a battery (with V(ideal)=12 V), then a small r(internal), then an r(wire), finally an R(motor). That's it, close the circuit! It's just a simple series of three resistors and one battery. We are given current (same through all of them!) We are given V(effective, battery), which must be V(ideal)-i r(internal). Since we know V(ideal), and i, we can immediately solve for r(internal) and check to see if it's within specs. They also give us Delta V across the cable, so r(wire) = Delta V/i, both of which are given - so we check that too. Then we need Delta V across the motor, but by Kirchoff, that has to be the V(effective,battery)V(wire). (Think about why I subtracted that! It's the "leftover voltage"!) Here again, R = Delta V/i, so you can figure out R(motor), and see if it's the bad part. CAPA was designed so one (and only one) would come out over specs.