Electric Currents, Solution Manual

Electric currents Synchrotron current The number of round trips that each electron makes per second is (3·108 m/s)/(240 m) = 1.25 · 106 s−1 . The charge per second passing any given point is therefore (1.25 · 106 s−1)N e = (1.25 · 106 s−1)(1011)(1.6 · 10 −19 C) = 0.02 A. (330) Combining the current densities The current density is given by J = qN u. So the magnitudes of the two J’s are Jion Je = = (2e)Nion uion = (2 · 1.6 · 10−19 C)(5 · 1016 m−3 )(105 m/s) = 1.6 · 103 eNe ue = (1.6 · 10−19 C)(1017 m−3 )(106 m/s) = 1.6 · 104 A . m2 A , m2 (331) Jion points west, and Je points southwest due to the negative charge of the electron; see Fig. 85. The components of the total current density J are therefore ( ) 1.6 · 104 A Je A √ Jwest = Jion + √ = 1.6 · 103 + = 1.29 · 104 2 , m m2 2 2 Je A (332) Jsouth = √ = 1.13 · 104 2 . m 2 = 41.2◦ , we see that J points in a direction 41.2◦ south of west. Since tan−1 (1.13/1.29) √ The magnitude of J is 1.292 + 1.132 · 104 A/m2 = 1.71 · 104 A/m2 . Current pulse from an alpha particle (a) Let x be the distance from the left plate, and let Qr be the charge on the right plate. From Exercise 3.37 we have Qr = −(2e)x/ℓ, where ℓ = 2 mm is the distance between the plates. The current flowing out of the right plate is therefore I=− dQr 2(1.6 · 10−19 C)(106 m/s) 2ev 2e dx = = 1.6 · 10−10 A. (333) = = ℓ 0.002 m ℓ dt dt N Jion W E 41.2 Jtotal Je Jion Figure 85 S I (10-10 A) 2 This current lasts for a time t = ℓ/v = (0.002 m)/(106 m/s) = 2 · 10−9 s, which is 2 nanoseconds. The current is constant during this time, so we have the bold line shown in Fig. 86. The total charge that flows during this time is It, which equals 2e as expected. ◦ If the path slopes upward at 45◦ , then dx/dt = v cos 45 √ . From above, the current pulse is therefore reduced in amplitude by a factor 1/ 2 and stretched out in time √ by a factor 2; see the dotted line in Fig. 86. Again the total charge transferred is It = 2e. 1 1 2 3 t (10-9 s) Figure 86 (b) Following the strategy of the solution to Exercise 3.37, we know that if Q1 and Q2 are the charges on the inner and outer electrodes (with radii a and b, respectively), then Q1 + Q2 = −2e. How is the charge of −2e distributed between Q1 and Q2 when the alpha particle is at radius r? As in Exercise 3.37, the key points are that (1) we can smear out the alpha particle into a cylinder of charge, and (2) the potentials of the two electrodes are the same, which means that the line integrals of the electric field from radius r to the two electrodes must be equal. The field inside radius r is proportional to Q1 /r (this points inward since Q1 is negative), and the field outside radius r is proportional to (2e + Q1 )/r = −Q2 /r (this points outward since Q2 is negative). Equating the two line integrals gives (note that both sides of the following equation are positive since dr is negative in the left integral) ∫ a r Q1 dr = r ∫ r b −Q2 dr r =⇒ Q1 ln(a/r) = −Q2 ln(b/r) =⇒ Q1 ln(r/a) = Q2 ln(b/r). (334) Combining this equation with Q1 + Q2 = −2e and solving for Q1 and Q2 gives Q1 = −(2e) ln(b/r) ln(b/a) and Q2 = −(2e) ln(r/a) . ln(b/a) (335) The current flowing out of the outer cylinder is then I I=− t Figure 87 dQ2 2e d(ln r) 2e 1 dr 2ev 1 = = = , ln(b/a) dt ln(b/a) r dt ln(b/a) a + vt dt (336) where we have used r = a + vt. We see that I(t) is not constant. A plot of the general shape of I(t) is shown in Fig. 87 (with b chosen to equal 4a). For a given value of b, if a is very small then the current starts out very large, because at t = 0 the smallness of a in the denominator in Eq. (336) wins out over the largeness of ln(b/a). In the case of a 45◦ angle of the path, the same modifications that applied in part horizontally by a factor √ (a) also apply here. That is, the curve is stretched √ of 2, and squashed vertically by a factor of 1/ 2. Transatlantic cable (a) The resistance of the seven wires together is R= (3 · 10−8 Ω m)(3 · 106 m) ρL = = 3.1 · 104 Ω. A 7 · π(3.65 · 10−4 m)2 Adding seven resistors in parallel would give the same answer. (337) (b) Our goal is to obtain a rough upper bound on the resistance of the ocean path, and then see if this is smaller than the answer to part (a). As stated, we’ll take the electrodes to be spheres with a radius of 0.1 m. (Any factors of order 1 will be irrelevant.) Let’s imagine the path of the current to be roughly a hemisphere expanding out from one sphere, then a tube with a large cross section, and then a hemisphere tapering down to the other sphere. (Even if the end parts are more conical than hemispherical, we’ll still get the same order of magnitude.) For the tubular middle part, it doesn’t matter much what cross section we pick, if our only goal is to obtain an upper bound on the resistance. Even if the cross section is just a kilometer in diameter, the resistance is (using a length of 3000 km) R= ρL (0.25 Ω m)(3 · 106 m) ≈ 1 Ω, = π(103 m)2 /4 A (338) which is negligible compared with the resistance of the cable. What is the resistance of the hemispheres? If d ≈ 0.1 m is their radius, then from dimensional analysis we expect the resistance to be roughly ρ/d, in order of magnitude. And indeed, from Problem 4.4 the resistance is proportional to ρ/d, assuming that the other radius involved is large. In the present case, ρ/d equals (0.25 Ω m)/(0.1 m) = 2.5 Ω. This is negligible compared with the cable’s resistance. Even if the electrode’s size was on the order of a millimeter, the hemisphere’s resistance would still only be 250 ohms, which is again negligible compared with the cable’s resistance. Intervals between independent events (a) If we divide the time t into a very large number, N , of equal small intervals, then the length of each interval is dt = t/N . The probability that an event happens in a particular one of these small intervals is therefore p dt = p(t/N ). So the probability that an event happens in none of these N intervals is (1 − pt/N )N . Multiplying this by the probability p dt that an event does happen in the next dt interval tells us that the probability of the next event happening between t and t + dt is (in the N → ∞ limit) (1 − pt/N )N p dt = e−pt p dt, (339) as desired. The integral of this probability must be 1, because the next event must happen at some time. And indeed, ∫ ∞ ∞  e−pt p dt = −e−pt = 1. (340) 0 0 We are free to pick the t = 0 point as the time of an event, so the probability in Eq. (339) is also the probability that an interval between events has length between t and t + dt. That is, if we look at a million successive intervals, then approximately (106 )(e−ptp dt) of them will have length between t and t + dt. (b) To find the average value (or expectation value) of a quantity, we must multiply the probability of a value occurring times the value itself, and then integrate over all the values. Equivalently, we can look at a million waiting times and calculate their average by adding up all the times and dividing by a million. So the average waiting time (starting at any given time, not necessarily the time of an event) until the next event is ∫ ∞ ∞  twait = t · e−pt p dt = −e−pt(t + 1/p) = 1/p. (341) 0 0 (You should check this integral by differentiating it.) It makes sense that this time decreases with p; if p is large, then the events happen more frequently, so the waiting time is shorter. Since this 1/p result holds for any arbitrary starting time, we are free to choose the starting time as the time of an event. A special case of this result is therefore the statement that the average waiting time between events is 1/p. This is consistent with the fact that pt is the average number of events that occur during a (not necessarily infinitesimal) time t. (c) If we pick a random point in time, then the average waiting time until the next event is 1/p, from part (b). And the average time since the previous event is also 1/p, because we can use the same reasoning that we used in part (a), going backward in time, to calculate the probability that the most recent event occurred at a time between t and t + dt earlier. The direction of time is irrelevant; the process is completely described by saying that p dt is the probability of an event happening in an infinitesimal time dt, and this makes no reference to a direction of time. The average length of the interval surrounding a randomly chosen point in time is therefore 1/p + 1/p = 2/p. (d) From part (a), an event-to-event interval with length between t and t + dt occurs with probability e−ptp dt (in the sense that out of a million successive intervals, (106 )(e−pt p dt) of them will have this length). But if you pick a random point in time, e−pt p dt is not the probability that you will end up in an interval with length between t and t + dt, because you are more likely to end up in an interval that is longer. Consider the simple case where there are only two possible lengths of intervals, 1 and 100, and these occur with equal probabilities of 1/2. If you look at 1000 successive intervals, then about 500 will have length 100. But if you pick a random point in time, you are of course 100 times more likely to end up in one of the large intervals. The probability of landing in each type of interval is not 1/2. The probability of landing in an interval of a given length (1/101 and 100/101 in the present example) does not equal the probability of that given length occurring in a list of the lengths (1/2 and 1/2 here). In this example, the average time between events is 50.5, while the average time surrounding a randomly chosen point in time is, as you can show, 99.02. (These results don’t have anything to do with the above results involving p, because the present example isn’t a random process described by a given probability per unit time. But it illustrates the basic point.) In short, the probability of landing in an interval with length between t and t + dt is proportional both to e−pt p dt (because the more intervals there are of a certain length, the more likely you are to land in one of them), and to the length t of the intervals (because the longer they are, the more likely you are to land in one of them). (e) Consider a large number N of intervals. The number of intervals with length between t and t + dt is N (e−pt p dt). The total length of these intervals with length between t and t + dt is therefore N (e−pt p dt)t. The total length of all of the N intervals is the integral of this, which you can quickly show equals N/p, as it should. The probability of picking a point in time that lands in one of the intervals with length between t and t + dt equals the total length associated with these intervals, divided by the total length of all of the N intervals, which gives (N e−ptpt dt)/(N/p) = e−pt p2 t dt. As mentioned in part (d), this probability is proportional to both e−pt p dt and t. The expectation value of the length of the interval that the given point lands in is obtained by multiplying this probability by the interval length t and integrating. This gives ∫ ∞ )∞ 2 e−pt ( e−pt p2 t2 dt = − 2 + 2pt + p2 t2  = , (342) 0 p p 0 as desired. (Again, you should check this integral by differentiating it.) To sum up, there are two different probabilities in this problem: (1) the probability that a randomly chosen interval has length between t and t + dt (this equals e−pt p dt), and (2) the probability that a randomly chosen point in time falls in an interval with length between t and t + dt (this equals e−pt p2 t dt). In the first case, by “randomly” we mean that we label each interval with a number and then pick a random number. The length of each interval is irrelevant in this case, whereas it is quite relevant in the second case. Mean free time in water From Eq. (4.23), the conductivity is σ = 2Ne2 τ /m, where the factor of 2 comes from the fact that in pure water there are both positive and negative ions. The mass of the OH− and OH3+ ions is essentially the mass of 17 or 19 nucleons, which is about 3 · 10−26 kg. So we have ( ) (3 · 10−26 kg) 4 · 10−6 (Ω m)−1 mσ τ= ≈ 4 · 10−14 s. (343) = 19 −3 −19 2 2 2(6 · 10 m )(1.6 · 10 C) 2N e The distance traveled in this time is vτ = (500 m/s)(4 · 10−14 s) = 2 · 10−11 m. As expected, this is smaller than the size of a water molecule, which is on the order of a couple angstroms (10−10 m). Drift velocity in seawater We know that the current density is J = N ev =⇒ v = J/N e. So our goal is to determine J. It is given by 1V . (344) J = σE = ρL Alternatively, we can find J from J= I V V /R V 1 = = = . A ρL/A ρL A A (345) The drift velocity is therefore (the factor of 2 in the N here comes from the fact that there are both Na+ and Cl− ions) v = = J V 12 V = = Ne ρLN e (0.25 Ω m)(2 m)(2 · 3 · 1026 m−3 )(1.6 · 10−19 C) 2.5 · 10−7 m/s. (346) Silicon junction diode We know that the densities of the slabs are equal and opposite, because ϕ is constant is each region outside the junction. It turns out that the slab in the n-type material is actually the positive slab (which is the opposite of what you might think), due to the fact that the slabs are brought about by the diffusion of holes from the p-type to n-type material, and the diffusion of electrons from the n-type to p-type material. But for the purposes of this exercise, it isn’t critical which is which. Poisson’s equation, ∇2 ϕ = −ρ/ϵ0 , tells us that ρ d2 ϕ =− dx2 ϵ0 =⇒ ϕ = A + Bx − ρ 2 x . 2ϵ0 (347) This can also be written in the alternative general form, ϕ = −(ρ/2ϵ0 )(x + C)2 + D. We see that ϕ varies quadratically with x if ρ is constant. The curvature is positive in the region where ρ is negative, and negative in the region where ρ is positive. We are told that the slope of ϕ is zero outside the charge layers. It must therefore also be zero just inside the boundaries. If there were a discontinuity in the slope of ϕ, then the second derivative would be infinite. Poisson’s equation would then imply an infinite ρ, such as that arising from a surface charge. But there are no surface charges in this setup. The zero slope at the boundaries implies that ϕ must look like the curve shown in Fig. 88. Let us define ϕ = 0 at the left boundary of the left slab, and let x = 0 be the location of the midplane. Let ℓ ≡ 10−4 m. The density is −ρ in the left (p-type) region, and +ρ in the right (n-type) region. In the left region we have half of a rightside-up parabola centered at x = −ℓ, and in the right region we have half of an upside-down parabola centered at x = ℓ. So if ∓ρ are the charge densities in the two regions, we quickly see that the potential ϕ in Eq. (347) (or rather the alternate form listed right after) must take the forms, ϕp = (ρ/2ϵ0 )(x + ℓ)2 , ϕn = (0.3 V) − (ρ/2ϵ0 )(x − ℓ)2 . (348) φ 0.3 V ρ -l l x -ρ (p-type material) (n-type material) Figure 88 These two forms must agree at x = 0, so we have ρℓ2 2ϵ0 = (0.3 V) − =⇒ ρ = ) s2 C2 8.85 · 10−12 kg C ϵ0 (0.3 V) m3 (0.3 V) = = 2.7 · 10−4 3 . ℓ2 m (10−4 m)2 ρℓ2 2ϵ0 ( (349) The electric field at x = 0 is obtained via Ex = −dϕ/dx. We can use either of the forms of ϕ in Eq. (348) for this, and we obtain Ex = −ρℓ/ϵ0 . Rather that plugging in the various numbers, we can be a little more economical by writing ( ) ϵ0 (0.3 V)/ℓ2 ℓ 0.3 V V ρℓ 0.3 V = − −4 = −3000 . Ex = − = − =− m ℓ 10 m ϵ0 ϵ0 (350) Note that this is twice the average field in the slabs, which is −(0.3 V)/(2ℓ). Unbalanced current The capacitance of the isolated box must be roughly that of a sphere intermediate between the inscribed and circumscribed spheres. ( So let’s use a sphere )with radius 6 cm. The capacitance is then C = 4πϵ0 r = 4π 8.85 · 10−12 s2 C2 /kg m3 (0.06 m) ≈ 7 · 10−12 F. The box is acquiring charge at a rate of 10−6 A, so the charge after time t is Q(t) = (10−6 C/s)t. The potential is then V (t) = Q(t)/C. Setting this equal to 1000 volts gives (10−6 C/s)t = 1000 V =⇒ t = 7 · 10−3 s. (351) 7 · 10−12 F 7 milliseconds is rather quick on an everyday timescale. Parallel resistors The two loop equations are E − (I1 − I2 )R1 −(I2 − I1 )R1 − I2 R2 = 0, = 0. (352) Adding these two equations quickly gives I2 = E/R2 (which corresponds to the loop around the whole circuit). Either equation then gives I1 = E E(R1 + R2 ) ≡ , R 1 R2 Reff where Reff ≡ R1 R 2 . R1 + R2 (353) The current through the battery is I1 , so E = I1 Reff tells us that the effective resistance is Reff . Keeping the same resistance We have an R1 resistor in series with the parallel combination of R1 and (R1 + R0 ). So we want R1 + R1 (R1 + R0 ) = R0 R1 + (R1 + R0 ) =⇒ (2R21 + R1 R0 ) + (R12 + R1 R0 ) = 2R1 R0 + R20 =⇒ R0 3R12 = R02 =⇒ R1 = √ . 3 (354) Automobile battery If the voltage drop across the 0.5 Ω resistor is 9.8 V , then the current in the circuit is I = V /R = (9.8 V)/(0.5 Ω) = 19.6 A. The voltage drop across the internal resistor is then Ri (19.6 A). But we know that this voltage drop is 12.3 V − 9.8 V = 2.5 V. Therefore, 2.5 V = Ri (19. ⇒ i 6 A) = R = 0.128 Ω. Equivalent boxes The resistances in the first box are simply Rab = 10 + 20 = 30, Rac = 10 + 50 = 60, and Rbc = 20 + 50 = 70, as desired. In the second box, in each case we have one resistor in parallel with the series combination of the other two resistors. So the the resistances in the second box are 34(170 + 85) = 30, Rab = 289 85(34 + 170) = 60, Rac = 289 170(85 + 34) = 70, (355) Rbc = 289 as desired. b For the two configurations pictured, the above resistances are the only possibilities that lead to the given resistances between the terminals. (Each configuration yields three equations in the three resistances, so they are uniquely determined.) However, there are other configurations that also work, for example, the one pictured in Fig. 89 (as you can check). 10 30 30 0 40 30 a c Figure 89 (b-a)(x/l) (b-a) x 2a dx 2b l Figure 90 The potential assumed by a free terminal when the potentials at the other two terminals are fixed is the same for the two boxes. For example, if the potentials at b and c in the first box are fixed at ϕb and ϕc , then the potential at a divides the difference ϕb − ϕc in the ratio of 20 to 50. And in the second box the ratio is 34 to 85, which is the same. The two boxes are therefore indistinguishable by external measurements (using direct currents). You can show that the configuration in Fig.89 also yields the same ratio of 2 to 5. Tapered rod Let the length of the rods be ℓ. Then the resistance of the cylindrical rod is ρL/A = ρℓ/πa2 . Now consider the tapered cone. The resistance of a cross-sectional disk is ρL/A = ρ dx/πr2 , where r = a + (b − a)(x/ℓ) from similar triangles in Fig. 90. So dx = dr ℓ/(b − a), and we have ) ( ∫ ℓ ∫ b 1 ρ ℓ ρ ℓ ρℓ 1 dr ρ dx R= − = = . (356) = 2 2 π π πab a r πr b b b − a − a a 0 This is a/b times the resistance of the cylindrical rod, as desired. If b < a (or b > a) then this fraction is larger (or smaller) than 1, which makes sense. If b → 0 then R → ∞. Note √ that the conical rod has the same resistance as a cylindrical rod with radius ab. See Problem 4.6 for a discussion of the approximate nature of this solution. As a check on this result, consider three objects, all with the same length: a cylinder with radius a, a tapered cone with radii a and b, and another cylinder with radius b. For concreteness, assume a < b, as in Fig. 90. Then Eq. (356) tells us that the resistance of the second of these objects is a/b times that of the first, and also that the resistance of the third is a/b times that of the second. So the first and third resistances are in the ratio of b2 to a2 , or equivalently 1/a2 to 1/b2 . This is correct, because the resistances of the two cylinders are inversely proportional to their cross-sectional areas, which are proportional to length squared. Laminated conductor extremum In the solution to Problem 4.5, if we replace the number 7.2 with n, and the fraction 1/3 with f , then we can repeat the same reasoning and derive the general result, σ⊥ 1 ][ ]. = [ σ∥ f + n(1 − f ) f + (1/n)(1 − f ) (357) (Or you can just look at Eqs. (12.204) and (12.206) and make the above replacements.) When the denominator is multiplied out, it equals −f 2 (n + 1/n − 2) + f (n + 1/n − 2) + 1. (358) Alternatively, if we don’t expand things as far, we can write the denominator as f 2 + (1 − f )2 + f (1 − f )(n + 1/n). Since n + 1/n ≥ 2 (by the arithmetic-geometricmean inequality, or by taking a derivative), the denominator is greater than or equal )2 ( to f 2 + (1 − f )2 + 2f (1 − f ) = f + (1 − f ) = 1. This means that σ⊥ is always less than or equal to σ∥ , for any values of f (between 0 and 1) and n. (a) For a given n, taking the derivative of the denominator given in Eq. (358), with respect to f , shows that it achieves a maximum when f = 1/2, which means that σ⊥ /σ∥ achieves a minimum when f = 1/2, that is, when the two thicknesses are equal. This result is independent of n. So no matter what the ratio of conductivities is, σ⊥ /σ∥ is minimum when the materials have the same thickness. The minimum value of σ⊥ /σ∥ is quickly found to be 4/(2 + n + 1/n). It makes sense that there should be an extremum of σ⊥ /σ∥ for an intermediate value of f (between 0 and 1), because if f = 0 or f = 1, the material consists of only one substance, so σ⊥ /σ∥ = 1. Therefore, unless σ⊥ /σ∥ is a flat curve (which it undoubtedly isn’t), it must reach a maximum or minimum for some f between 0 and 1. (b) For a given f (between 0 and 1), taking the derivative of the denominator given in Eq. (358), with respect to n, shows that it achieves a minimum when n = 1, which means that σ⊥ /σ∥ achieves a maximum when n = 1. This result is independent of f . So no matter what the ratio of thicknesses is, σ⊥ /σ∥ is maximum when the materials have the same σ. And the maximum value of σ⊥ /σ∥ is 1, of course, because the two materials are the same. It makes sense that there should be an extremum of σ⊥ /σ∥ for an intermediate value of n (between 0 and ∞), because setting one of the σ’s equal to zero makes σ⊥ (but not σ∥ ) equal to zero; and setting one of the σ’s equal to infinity makes σ∥ (but not σ⊥ ) equal to infinity. So σ⊥ /σ∥ is zero at both extremes. Therefore, unless σ⊥ /σ∥ is a flat curve (which, again, it undoubtedly isn’t), it must reach a maximum or minimum for some n between 0 and ∞. Effective resistances in lattices As in Problem 4.8, we will superpose two setups, one with a current of 1 A entering at one node and heading out to infinity, and the other with a current of 1 A heading in from infinity and exiting at an adjacent node. The only modification we need to make to the solution to Problem 4.8 is the number of resistors connected to a given node. If there are n resistors connected to each node (n was 4 in Problem 4.8), then each of the above two setups has a current of 1/n A heading from one node to the other. (All of the setups have the necessary symmetry for the current to divide equally.) So the current between the two nodes is 2/n A when the setups are superposed. The voltage drop across the 1 Ω resistor is then 2/n V, so V = IReff gives 2/n V = (1 A)Reff . The effective resistance is therefore 2/n Ω . Hence: (a) The 3-D cubic lattice has n = 6, so Reff = 2/6 = 1/3 Ω. (b) The 2-D triangular lattice also has n = 6, so Reff = 1/3 Ω. (c) The 2-D hexagonal lattice has n = 3, so Reff = 2/3 Ω. b B a b a b A a 3 a's 3 b's A A B 1/3 1/6 1/3 B Figure 91 (d) The 1-D lattice has n = 2, so Reff = 2/2 = 1 Ω. This makes sense, because there is only one path between two adjacent nodes on a line, namely across the 1 Ω resistor connecting them. All the other resistors outside the two nodes are irrelevant. Resistances in a cube (a) In Fig. 91 the three vertices adjacent to A (which are labeled as “a”) are all at the same potential (by symmetry under rotations around the AB diagonal), so we can collapse them to one point. (Equivalently, if we connect them with resistance-less wires, no current will flow in these wires.) Likewise for the three vertices adjacent to B (which are labeled as “b”). So the circuit is equivalent to the second setup shown in Fig. 91 (the number of lines is still 12), which can be simplified as indicated. The equivalent resistance is therefore 5R/6. Alternatively, we can work in terms of currents. The input current I0 gets divided evenly, by symmetry, into three I0 /3 currents. It then divides into six I0 /6 currents, and then converges to three I0 /3 currents. The total potential drop across any of the possible paths from A to B is given by V = (I0 /3)R + (I0 /6)R + (I0 /3)R = (5/6)I0 R. The effective resistance is then V /I0 = 5R/6. (b) In Fig. 92 there are four vertices (labeled as “c”) that lie in the plane that is equidistant from A and B. These vertices are all at the same potential (halfway between VA and VB ), so we can collapse them to a point. (In the second setup shown, there are only 10 lines because 2 of the original 12 lines were collapsed). The circuit can then be simplified as shown, and the equivalent resistance is 3R/4. (c) From symmetry, the two points marked as a in Fig. 93 are at the same potential, so we can collapse them to a point. Likewise for the two b’s. The circuit can then be simplified as shown, and the equivalent resistance is 7R/12. As expected, this is smaller than the answer to part (b), which in turn is smaller than the answer to part (a). c B A 4 c's A A B c 1/2 1/2 1/2 1/2 1 c b a c A 1 b a 1/2 B 1/2 A 3/2 B 3/8 3/8 B 3/2 Figure 92 A A B a 2 a's 2 b's 1/2 1/2 d c A 1/2 1/2 b c B 1/2 b a 1 A B 1 1/2 B 1/2 1/2 2 Figure 93 A d 1 1/2 B 1/2 2/5 A 1 1 7/5 B A 7/12 B Note that the sum of the effective resistances across all 12 resistors is 12(7R/12) = 7R = (8 − 1)R, where the 8 here is the number of corners in the cube. This is a special case of the general result in Problem 4.9. R1 R2 R Attenuator chain If R is the effective resistance of the infinite chain, then the chain is equivalent to the circuit shown in Fig. 94. So we have Figure 94 R = R1 + R1 A' R1 I I1 R2 A R2 I2 B' R1 B Figure 95 R2 R R2 + R =⇒ R(R2 + R) = R1 (R2 + R) + R2 R =⇒ R 2 − R 1 R − R 1 R2 = 0 √ R1 + R12 + 4R1 R2 , R= 2 =⇒ (359) where we have chosen the positive root. To demonstrate the stated geometric-series result, consider four points A, A′ , B , B ′ that form a square somewhere within the circuit, as shown in Fig.95. Given the voltage V ′ between A′ and B ′ (so V ′ = V0 if A′ and B ′ are at the left end of the chain), what is the voltage V between A and B? Let the current flowing toward point A′ be I . This current splits into the currents I1 and I2 . The circuit to the right of A′ and B ′ is equivalent to a resistance R, so currents of I1 and I2 pass through resistances R and R2 , respectively. These currents are in the ratio of R2 /R; hence I1 = IR2 /(R2 + R). Now, the voltage V ′ between A′ and B ′ is proportional to the current I flowing into A′ (by dimensional analysis). Likewise, the voltage V between A and B is proportional to the current I1 flowing into A, with the same constant of proportionality (because we have the same infinite circuit, independent of where it starts). Therefore, the ratio of the voltage between A and B to the voltage between A′ and B ′ is V I1 R2 = = . R2 + R V′ I (360) This result is independent of where along the chain we pick the adjacent nodes, so the voltages across successive nodes decrease in a geometric series. If we want V /V ′ = 1/2, we must have R = R2 . Equation (359) then gives √ 2R2 = R1 + R12 + 4R1 R2 =⇒ (2R2 − R1 )2 = R12 + 4R1 R2 =⇒ 4R22 = 8R1 R2 =⇒ R2 = 2R1 . (361) ′ From Eqs. (359) and (360), we see that if we want V/V ≈ 1 (that is, the voltage hardly decreases), then we need R ≪ R2 , which implies R1 ≪ R2 . On the other hand, if we want V /V ′ ≪ 1 (that is, the voltage decreases quickly), then we need R ≫ R2 , which implies R1 ≫ R2 . These results make intuitive sense. To terminate the ladder after any section, without changing its resistance from that of the infinite chain, we can simply connect a single resistor R given by Eq. (359) in parallel with the last R2 , because this R mimics the rest of the infinite chain. For example, after one step of the ladder, we would have the setup in Fig. 94. (a) Some golden ratios (a) For simplicity, let’s set R = 1. Let the desired resistance be r. Following the strategy from Exercise 4.36, the circuit to the right of C and D in Fig. 96(a) is identical to the original infinite chain. So the infinite chain consists of a resistance of 1 connected in series with the parallel combination of 1 and r. The net result is r, so 1·r 1+ =r 1+r =⇒ =⇒ 1 + 2r = r =⇒ r2 − r − 1 = 0 1+r √ 1+ 5 = 1.618 ≡ r1 . r= 2 This number is the golden ratio. =⇒ =⇒ r2 + r − 1 = 0 √ −1 + 5 r= = 0.618 ≡ r2 . 2 (363) This number is the inverse of the golden ratio. You should convince yourself why this setup is actually the same as the setup in Problem 4.7(b). As a double check, the only difference between the two given circuits is the extra vertical resistor connecting A to B in the second circuit. So r2 should equal the parallel combination of 1 and r1 . And indeed, 1 · r1 = r2 , 1 + r1 C B D (b) A C B D (362) (b) Again set R = 1, and let the desired resistance be r. The circuit to the right of C and D in Fig. 96(b) is identical to the original infinite chain. So the infinite chain consists of a resistance of 1 connected in parallel with the series combination of 1 and r. The net result is r, so 1 · (1 + r) =r 1 + (1 + r) A (364) as you can verify. This works out due to the various properties of the golden ratio. Two light bulbs (a) The power dissipated takes the form of V 2 /R. Both bulbs have the same voltage drop V , so if Bulb 1 is twice as bright as Bulb 2, it must have half the R. Bulb 2’s resistance is therefore larger by a factor of 2. (The larger resistor is dimmer.) (b) The power dissipated also takes the form of I 2 R. Both bulbs now have the same current I, so if Bulb 2 has twice the resistance, as we found in part (a), then it is twice as bright – the opposite of the case in part (a). (The larger resistor is brighter.) Note that in part (a) we used the expression P = V 2 /R because both bulbs (in parallel) had the same V , whereas now we are using the expression P = I 2 R because both bulbs (in series) have the same I . We can also compare the total power dissipated in each case. If the resistances are R and 2R, then in part (a) the total power dissipated is V 2 /R+V 2 /2R = 3V 2 /2R. In part (b) the total power is I 2 R + I 2 (2R) = 3I 2 R, where I = V /3R. So the power is V 2 /3R. This is 2/9 of the power in part (a). In units of V 2 /R, the powers in part (a) are 1 and 1/2, while in part (b) they are 1/9 and 2/9. Figure 96 Maximum power The Ri and R resistors are in series, so the current in the circuit is I = E/(R + Ri ). The power dissipated in the R resistor is therefore P = I 2 R = E 2 R/(R + Ri )2 . Taking the derivative with respect to R and setting the result equal to zero gives 0= (R + Ri )2 · 1 − R · 2(R + Ri ) Ri − R = (R + Ri )3 (R + Ri )4 =⇒ R = Ri . (365) This is indeed a maximum, because dP/dR > 0 for R < Ri , and dP/dR < 0 for R > Ri . Equivalently, the second derivative is negative at R = Ri , as you can check. It makes sense that a maximum exists for some finite value of R, because P = 0 both at R = 0 (because P = I 2 R, with I finite and R zero) and at R = ∞ (because P = V 2 /R, with V finite and R infinite). Consider a different question, “Given a fixed external resistance R, what value of the internal resistance Ri yields the maximum power delivered to the external resistor R?” In view of the above expression for the power, the answer is simply Ri = 0. This makes sense; we want the largest possible current passing through the given external resistor. Minimum power dissipation The power dissipated in the two resistors is P = I 21 R1 + I22 R2 = I12R1 + (I0 − I1 )2 R2 = I 21 (R1 + R2 ) − 2I0 R2 I1 + R2 I02. (366) Minimizing P by taking the derivative with respect to I1 gives 0= dP I 0 R2 = 2I1 (R1 + R2 ) − 2I0 R2 =⇒ I1 = , dI1 R1 + R2 (367) which is also what we obtain from Ohm’s law (that is, equating the voltage drops I1 R1 and I2 R2 , and using I1 + I2 = I0 ). Note that P is indeed a minimum, and not a maximum, because dP/dI1 is less than (or greater than) 0 if I1 is less than (or greater than) I0 R2 /(R1 + R2 ). Equivalently, the second derivative of P equals 2(R1 + R2 ), which is positive. Alternatively, we can set the differential of P = I12R1 + I22 R2 equal to zero, which gives dP = 2R1 I1 dI1 + 2R2 I2 dI2 = 0. But I1 + I2 = I0 tells us that dI2 = −dI1 , so we obtain R1 I1 = R2 I2 . This is simply the statement of equal voltage drops, as given by Ohm’s law. Combining this with I1 + I2 = I0 yields the above value of I1 . D-cell (a) The total charge produced by 0.1 A flowing for 30 hours is (0.1 C/s)(30 · 3600 s) = 10, 800 C. This charge passes through a potential difference of 1.5 V, so the total energy output is E = (10, 800 C)(1.5 J/C) = 16, 200 J. Since the mass of the battery is 0.09 kg, the energy storage in J/kg is (16, 200 J)/(0.09 kg) = 1.8 · 105 J/kg. In the example in Section 4.9, the 10 kg battery had an energy output of 8.6·105 J, which implies 8.6 · 104 J/kg. This is about half as much as the D cell. (b) Lifting a 70 kg person 1 m requires an energy of mgh = (70 kg)(9.8 m/s2 )(1 m) ≈ 700 J. The D cell with 50% efficiency can supply 8,100 J. This corresponds to about 11.5 m. Making an ohmmeter If adding 15 Ω to the circuit between the leads cuts the current through the ammeter in half (compared with the R = 0 case where the leads are connected together), then the current I in the external (to the ammeter) circuit must be in order of magnitude (1.5 V)/(15 Ω) = 0.1 A. This is much larger than the maximum current 50 µA = 5 · 10−5 A in the ammeter. This implies that R1 must be much smaller than the resistance Ra = 20 Ω of the ammeter’s coil, so that nearly all of the current I is shunted through R1 . Said in a different way, assume that R1 is not much smaller than Ra . Then with the leads connected together, the current through R1 will be no larger (in order of magnitude) than the current through Ra (that is, 5 · 10−5 A), because R1 and Ra are in parallel. But by looking at the bottom loop in the circuit, this means that R2 must be very large, roughly (1.5 V)/(5 · 10−5 A) = 3 · 104 Ω in order of magnitude. This implies that the insertion of an R = 15 Ω resistor in series with R2 will hardly change the current in the circuit. In particular, there is no way it can cut the current in half. Our initial assumption (that R1 is not much smaller than Ra ) must therefore have been incorrect. Having shown that R1 ≪ Ra , we see that the resistance of the whole circuit is essentially equal to R2 + R. Therefore, if I is to be half as large for R = 15 Ω as for R = 0, then we must have R2 = 15 Ω. This R2 = 15 Ω result then implies that the current I shown in Fig. 4.53 is 0.1 A when R = 0. Now, the fraction of I that goes through the ammeter’s coil is given by Ia /I = R1 /(R1 + Ra ) ≈ R1 /Ra . The stated conditions tell us that I = 0.1 A (which corresponds to R = 0) must yield Ia = 5 · 10−5 A. So Ia /I = R1 /Ra gives (5 · 10−5 A)/(0.1 A) = R1 /(20 Ω) =⇒ R1 = 0.01 Ω. If R = 5 Ω, all currents are reduced by a factor of essentially 15/(15 + 5) = 3/4, compared with the R = 0 case. So we have 3/4 of full deflection, which corresponds to the 37.5 µA mark. If R = 50 Ω, we have 15/(15 + 50) = 23% of full deflection, which corresponds to the 11.5 µA mark. We made two approximations above, namely R1 ≪ R2 and R1 ≪ Ra . If you want to solve the problem exactly, you can show that R2 should be 14.99 Ω instead of 15 Ω. And R1 should be (0.1 Ω)/(1 − 0.0005) ≈ 0.010005 Ω instead of 0.01 Ω. Neither of these refinements would ordinarily be necessary. R3 Using symmetry The symmetry is evident in Fig. 97. The first missing term in the numerator must be R1 R3 R4 (the mirror-image of the R1 R2 R3 term). The second missing term in the numerator must be R1 R4 (the mirror-image of the R2 R3 term). The missing term in the denominator must be R2 R3 (the mirror-image of the R1 R4 term). So we have Req = R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4 + R5 (R1 R3 + R2 R3 + R1 R4 + R2 R4 ) . R1 R2 + R1 R4 + R2 R3 + R3 R4 + R5 (R1 + R2 + R3 + R4 ) (368) (a) If R5 = 0 we equivalently have the circuit shown in Fig. 98(a), where we have collapsed R5 to a point since it is short circuited. The parallel combination of R1 and R3 is in series with the parallel combination of R2 and R4 . So the resistance is R2 R 4 R1 R 2 R3 + R1 R 2 R4 + R1 R3 R 4 + R 2 R 3 R 4 R1 R 3 + = , R1 R 2 + R 1 R 4 + R 2 R 3 + R 3 R 4 R1 + R3 R 2 + R 4 (369) R1 R5 R4 Figure 97 R2 (a) (b) (c) R3 R1 R3 R1 R4 R2 R4 R2 (d) R4 R2 R R R R Figure 98 which agrees with the formula in Eq. (368) when R5 = 0. (b) If R5 = ∞ we equivalently have the circuit shown in Fig. 98(b). The series combination of R1 and R2 is in parallel with the series combination of R3 and R4 . So the resistance is (R1 + R2 )(R3 + R4 ) , (370) R1 + R 2 + R 3 + R 4 which agrees with the formula when R5 = ∞. (c) If R1 = R3 = 0 we equivalently have the circuit shown in Fig. 98(c). The R5 resistor doesn’t matter, since it is short circuited via R1 and R3 . So we just have R2 and R4 in parallel, and the resistance is R2 R4 /(R2 + R4 ). This agrees with the formula when all terms containing R1 or R3 are dropped. (d) If R1 through R4 have the common value R, then by symmetry no current flows through R5 , so we can collapse that resistor to a point. We then have the circuit shown in Fig. 98(d). R1 and R3 are in parallel, so the effective resistance across them is R/2. And likewise for R2 and R4 . So the total resistance is R. This agrees with the formula; the numerator is 4R3 + 4R5 R2 , and the denominator is 4R2 + 4R5 R, so the whole fraction equals R. More generally, if R1 = R3 ≡ a and R2 = R4 ≡ b, then no current flows through R5 , so we again have two sets of parallel resistors. You should check that the formula agrees with what you calculate directly. Likewise for R1 = R2 ≡ a and R3 = R4 ≡ b. Using the loop equations The three loop equations are E − (I3 − I1 )R3 − (I3 − I2 )R4 = 0, = 0, −I2 R2 − (I2 − I3 )R4 − (I2 − I1 )R5 = 0. −I1 R1 − (I1 − I2 )R5 − (I1 − I3 )R3 (371) Solving these equations via Mathematica (for the unknowns I1 , I2 , I3 ) yields I3 = E/Req , where Req is given in Eq. (4.48). The equivalent resistance, E/I3 , is then Req, as desired. Battery/resistor loop If the circuit is open, we can ignore the leads to A and B, so we just have the given loop. The current around this loop is (3 · 1.5 V)/(5 · 100 Ω) = 9 · 10−3 A. The opencircuit voltage is then (using the upper part of the loop, with two cells and three resistors) (372) VB − VA = 2(1.5 V) − 3(9 · 10−3 A)(100 Ω) = 0.3 V. A 1.5 V 100 Ω −3 The lower part of the loop gives the same result: VA −VB = 1.5 V−2(9·10 A)(100 Ω) = −0.3 V. To find the short-circuit current, we can draw the circuit in the manner shown in Fig. 99. We quickly find that the two loop currents are I1 = (3 V)/(300 Ω) = 0.01 A counterclockwise, and I2 = (1.5 V)/(200 Ω) = 0.0075 A counterclockwise. The net short-circuit current from B to A is therefore Isc = 0.0025 A. (You should verify that you obtain the same result if you simply connect A and B in Fig. 4.57 and use the two resulting loops.) In the Thevenin equivalent circuit, Eeq is simply the open-circuit voltage, 0.3 V. And Req is given by 0.3 V Eeq = 120 Ω. (373) = Req = 0.0025 A Isc As a double check, we can calculate Req by setting all the emf’s equal to zero and finding the equivalent resistance of the resulting circuit. A and B are connected by the parallel combination of 200 Ω and 300 Ω, which quickly yields 120 Ω, as desired. Maximum power via Thevenin The open circuit (with the bottom resistor R absent) has current flowing only in the top loop. This current is (120 V)/(10 Ω + 10 Ω) = 6 A. The open-circuit voltage, which equals Eeq , is the same as the voltage across the lower of the 10 Ω resistors (because no current flows in the 15 Ω resistor when the circuit is open), so we have Eeq = (6 A)(10 Ω) = 60 V. The resistance Req can be found by ignoring (that is, shorting) the emf. As far as A and B are concerned, we then have a circuit with 15 Ω in series with the parallel combination of 10 Ω and 10 Ω. So Req = 15 Ω + 5 Ω = 20 Ω. Alternatively, we can find Req by finding the short-circuit current Isc, and then using Req = Eeq /Isc . To find Isc, note that the short-circuit setup consists of 10 Ω in series with the parallel combination of 10 Ω and 15 Ω (which is equivalent to 6 Ω). The whole circuit therefore has a resistance of 16 Ω. The total current is then (120 V)/(16 Ω) = 7.5 A. A fraction 10/(10 + 15) = 2/5 of this goes through the bottom branch of the circuit. So Isc = (2/5)(7.5 A) = 3 A. Hence Req = Eeq/Isc = (60 V)/(3 A) = 20 Ω, as above. You can also find Isc by using Kirchhoff’s rules with two loops. From Exercise 4.39, the power dissipated in R will be greatest when R = Req = 20 Ω. The current through R is then I = Eeq/(Req + R) = (60 V)/(40 Ω) = 1.5 A. So the power dissipated in R is P = I 2 R = (1.5 A)2 (20 Ω) = 45 J/s. Alternatively, we can solve the problem from scratch. A two-loop exercise (which is quicker than using the series/parallel rules) shows that for a general value of R, the current through R is 60/(20 + R) (ignoring the units). So we want to maximize I2R ∝ R/(20 + R)2. You can quickly show that this is maximized when R = 20. Discharging a capacitor From Eq. (4.44) the current is I(t) = (V0 /R)e−t/RC . The power dissipated in the resistor is P = I 2 R = (V02/R)e−2t/RC , so the total energy dissipated is ∞ ∫ ∞ ∫ ∞ 2 V0 −2t/RC V 02 RC −2t/RC  1 2 E= P dt = (374) e e dt = −  = 2 CV0 , R R 2 0 0 0 which is the initial energy in the capacitor, as desired. Suppose we have a 1 microfarad capacitor charged to 100 volts. The initial charge is Q = CV0 = (10−6 F)(100 V) = 10−4 C. From Eq. (4.43) the charge decreases I1 I2 B Figure 99 according to Q(t) = Q0 e−t/t0 , where t0 = RC is the time constant. Since the charge of an electron is 1.6·10−19 C, we will have roughly one electron left when 1.6·10−19 C = (10−4 C)e−t/t0 =⇒ t = −t0 ln(1.6 · 10−15 ) = 34t0 . So if the time constant were, say, 1 second (which would mean R = 106 Ω here), we would be down to roughly one electron in a little over half a minute. For a 1 kΩ resistor, the time would be 0.034 s. Charging a capacitor The total work done by the battery is Qf E, where Qf is the final charge on the capacitor. This is true because the battery transfers a charge Qf through a constant potential difference of E . The final energy of the capacitor is Qf ϕ/2 = Qf E/2, because the final potential ϕ across the capacitor equals the voltage E across the battery. (There is no current flowing after a long time, so there is no voltage drop across the resistor.) ∫ The energy dissipated in the resistor is the integral of the power, that is, RI 2 dt. −t/RC . Therefore, From the solution to Problem 4.17, we have I(t) = (E/R)e ∞ ∫ ∞ ∫ ∞ 2 2 2  CE 2 E E E Qf E RC RC = e−2t/RC  = RI 2 dt = R 2 e−2t/RC dt = − , = R 0 R 2 R 2 2 2 0 0 (375) where we have used Qf = CE. The conservation-of-energy statement is then Wbattery = Ucapacitor + Eresistor =⇒ Qf E = Qf E Qf E , + 2 2 (376) which is indeed true. Remark: It is also possible to use the general formulas for Q(t) and I(t) from Problem 4.17 to show that energy is conserved at all times (not just t → ∞), as we know it must be. But we can show this in a quicker manner by demonstrating that the conservation-of-energy statement is equivalent to the Kirchhoff loop equation, E − Q/C − RI = 0. We can do this either by differentiating the former to obtain the latter, or by integrating the latter to obtain the former. Let’s take the first of these routes. The conservation-of-energy statement at any intermediate time is ∫ t Q(t)2 (377) EQ(t) = RI 2 dt. + 2C 0 Differentiating with respect to t gives (using dQ/dt = I and canceling a factor of I ) E dQ Q dQ + RI 2 = C dt dt =⇒ E = Q + RI, C (378) which is the Kirchhoff loop equation, as desired. If you want to go in the reverse direction, just multiply by I and then integrate with respect to t (using the fact that Q = 0 at t = 0). Displacing the electron cloud The charge density on each of the sheets is σ = ϵ0 E. The effective number of electrons per unit area on the sheets is N d, where d is the displacement of the electron cloud. The field will be neutralized if (N d)e = σ. Hence, ( N d) e = ϵ 0 E =⇒ d = ) ( s2 C2 (3 · 104 V/m) 8.85 · 10−12 kg ϵ0 E m3 = 1.7 · 10−9 m. = 21 −3 (10 m )(1.6 · 10−19 C) Ne (379)