MASSACHUSETTS INSTINUTE OF TECHNOLOGY ESG Physics 8.02 with Kai Spring 2003 Problem Set 5 Solution Problem 1: 27.8 An electric current is given by the expression I ( t ) = 100sin (120π t ) (1.1) where I is in amperes and t is in seconds. What is the total charge carried by the current 1 s? from t = 0 to t = 240 Solution: Since I = dq , dt 1 240 0 q = ∫ I dt = ∫ (100 ) sin (120π t ) dt = −100 π cos − cos 0 120π 2 (1.2) which gives q = 0.265 C 2003 Spring 8.02 with Kai Problem set 5 Solution (1.3) 1 Problem 2: 27.24 The rod in Figure P27.24 (not drawn to scale) is made of two materials. Both have a square cross section of 3.00 mm on a side. The first material has a resistivity of 40.0 × 10−3 Ω ⋅ m and is 25.0 cm long, while the second material has a resistivity of 6.00 ×10−3 Ω ⋅ m and is 40.0 cm long. What is the resistance between the ends of the rod? Solution: Applying the formula, R= ρ1l1 A1 + ρ 2l2 A2 = ρ1l1 + ρ 2l2 d2 (2.1) Substituting, we have R= ( 4.00 ×10 −3 Ω ⋅ m ) ( 0.250 m ) + ( 6.00 × 10−3 Ω ⋅ m ) ( 0.400 m ) ( 3.00 ×10−3 m ) 2 (2.2) which gives R = 378 Ω 2003 Spring 8.02 with Kai Problem set 5 Solution (2.3) 2 Problem 3: 27.53 A high-voltage transmission line with a diameter of 2.00 cm and a length of 200 km carries a steady current of 1000 A. If the conductor is copper wire with a free charge density of 8.00 × 1028 electrons/m3 , how long does it take one electron to travel the full length of the cable? Solution: We find the drift velocity from I = nqvd A = nqvd π r 2 (3.1) Therefore, vd = I nqπ r 2 = (8.00 ×10 1000 A 28 m -3 )(1.60 ×10−19 C ) π (10−2 m ) 2 (3.2) which gives vd = 2.49 × 10−4 ms -1 (3.3) For the time that one electron takes to travel the full length of the cable, we have t= l 200 × 103 m = = 8.04×108 s v 2.49 × 10−4 ms -1 (3.4) which is equavalent to 25.5 years. 2003 Spring 8.02 with Kai Problem set 5 Solution 3 Problem 4: 27.66 A resistor is constructed by shaping a material of resistivity ρ into a hollow cylinder of length L and with inner and outer radii ra and rb , respectively (Fig. P27.66). In use, the application of a potential difference between the ends of the cylinder produces a current parallel to the axis. (a) Find a general expression for the resistance of such a device in terms of L, ρ , ra and rb . (b) Obtain a numeraical value for R when L = 4.00 cm, ra = 0.500 cm , rb = 1.20 cm , and ρ = 3.50 × 105 Ω ⋅ m (c) Now suppose that the potential difference is applied between the inner and outer surfaces so that the resulting current flows radially outward. Find a general expression for the resistance of the device in terms of L, ρ , ra and rb . (d) Calculate the value of R, using the parameter values given in part (b). Solution: (a) R= ρl A = ρL (4.1) π ( rb2 − ra2 ) (b) Substituting the values given, we have ( 3.50 ×10 Ω ⋅ m ) ( 0.0400 m ) 5 R= 2 2 π ( 0.0120 m ) − ( 0.00500 m ) (4.2) which gives R = 37.4 MΩ (c) Since (4.3) dR ρ = , therefore dl A 2003 Spring 8.02 with Kai Problem set 5 Solution 4 dR = ρ dl A = Therefore R= ρ dr ρ dr = ( 2π r ) L 2π L r (4.4) ρ r dr 2π L ∫r r (4.5) b a which gives r ρ ln b 2π L ra (4.6) (d) Substituting that the value given, we have ( 3.50 ×10 Ω ⋅ m ) ln 1.20 R= 5 2π ( 0.0400 m ) which gives R = 1.22 MΩ 2003 Spring 8.02 with Kai Problem set 5 Solution 0.500 (4.7) (4.8) 5 Problem 5: 27.70 A material of resistivity ρ is formed into the shape of a truncated cone of altitude h, as shown in Figure P27.70. The bottom end has a radius b, and the top end has a radius a. Assuming that the current is distributed uniformly over any particular cross-section of the cone so that the current density is not a function of radial position (although it does vary with position along the axis of the cone), show that the resistance between the two ends is given by the expression ρ h R= (5.1) π ab Solution: From the geometry of the longitudinal section of the resistor shown in the figure, we see that b−r b−a = y h (5.2) From this, the radius at a distance y from the base is r = (a − b) y +b h (5.3) For a disk-shaped element of volume, dR = ρ dy π r2 (5.4) Therefore R= ρ h π ∫0 dy y ( a − b ) h + b 2003 Spring 8.02 with Kai Problem set 5 Solution 2 (5.5) 6 Using the integral formula du ∫ ( au + b ) 2 = 1 a ( au + b ) (5.6) We have R= ρ h π ab 2003 Spring 8.02 with Kai Problem set 5 Solution (5.7) 7 Problem 6: 28.6 (a) Find the equivalent resistance between points a and b in Figure P28.6. (b) Calculate the current in each resistor if a potential difference of 34.0 V is applied between points a and b. Solution: (a) Rp = 1 1 1 + 7.00 10.0 = 4.12Ω Rs = R1 + R2 + R3 = 4.00 + 4.12 + 9.00 = 17.1 Ω (1.1) (1.2) (b) For the 4.00 Ω and 9.00 Ω resistors, since ∆V = IR , we have I= ∆V 34.0 = = 1.99 A R 17.1 (1.3) For the 7.00 Ω resistor, we have 8.18 = 1.17 A 7.00 (1.4) 8.18 = 0.818 A 10.0 (1.5) I= For the 10.0 Ω resistor, we have I= 2003 Spring 8.02 with Kai Problem set 5 Solution 8 Problem 2: 28.16. Two resistors connected in series have an equivalent resistance of 690 Ω . When they are connected in parallel, their equivalent resistance is 150 Ω . Find the resistance of each resistor. Solution: Denoting the two resistors as x and y, x + y = 690 (1.6) 1 1 1 = + 150 x y (1.7) and Substituting equation (2.1) to (2.2), we have ( 690 − x ) + x 1 1 1 = + = x ( 690 − x ) 150 x 690 − x (1.8) x 2 − 690 x + 103,500 = 0 (1.9) which gives By solving this quadratic equation, we have x = 470 Ω and y = 220 Ω 2003 Spring 8.02 with Kai Problem set 5 Solution (1.10) 9 Problem 3: 28.17 In Figures 28.4 and 28.5, let R1 = 11.0 Ω , let R2 = 22.0 Ω , and let the battery have a terminal voltage of 33.0 V. (a) in the parallel circuit shown in Figure 28.5, which resistor uses more power? (b) Verify that the sum of the power ( I 2 R ) used by each resistor equals the power supplied by the battery ( I ∆V ) . (c) In the series circuit, which resistor uses more power? (d) Verify that the sum of the power ( I 2 R ) used by each resistor equals the power supplied by the battery ( P = I ∆V ) . (e) Which circuit configuration uses more power? Solution: (a) We have ∆V = IR (1.11) 33.0 V=I1 (11.0 Ω ) and 33.0 V = I 2 ( 22.0 Ω ) (1.12) I1 = 3.00 A and I 2 = 1.50 A (1.13) Therefore Therefore, Since P = I 2 R , we have P1 = ( 3.00 A ) (11.0 Ω ) and P2 = (1.50 A ) ( 22.0 Ω ) 2 2 (1.14) which gives 2003 Spring 8.02 with Kai Problem set 5 Solution 10 P1 = 99.0 W and P2 = 49.5 W (1.15) Therefore, the 11.0- Ω resistor uses more power. (b) Since P1 + P2 = 148 W , we have P = I ( ∆V ) = ( 4.50 )( 33.0 ) = 148 W (1.16) Rs = R1 + R2 = 11.0 + 22.0 = 33.0 Ω (1.17) (c) Since ∆V = IR , I= V 33.0 Ω = = 1.00 A P 33.0 V (1.18) Since P = I 2 R , we have P1 = (1.00 A ) (11.0 Ω ) and P2 = (1.00 A ) ( 22.0 Ω ) 2 2 (1.19) and you will probably get P1 = 110 W and P2 = 22.0 W (1.20) Therefore the 22.0- Ω resistor uses more power. (d) P1 + P2 = I 2 ( R1 + R2 ) = (1.00 A ) ( 33.0 Ω ) = 33.0 W (1.21) P = I ( ∆V ) = (1.00 A )( 33.0 V ) = 33.0 W (1.22) 2 (e) The parallel configuration uses more power. 2003 Spring 8.02 with Kai Problem set 5 Solution 11 2003 Spring 8.02 with Kai Problem set 5 Solution 12 2003 Spring 8.02 with Kai Problem set 5 Solution 13