Physics exercises_solution: Chapter 25

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Physics exercises_solution: Chapter 25

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 25

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Nội dung Text: Physics exercises_solution: Chapter 25

  1. 25.1: Q  It  (3.6 A)(3)(3600 s)  3.89  10 4 C. 25.2: a) Current is given by I  Q t  420 C 80 ( 60 s )  8.75  10 2 A. b) I  nqvd A I 8.75  102 A  vd   nqA (5.8  1028 )(1.6  1019 C)( π(1.3  10 3 m) 2 ) =1.78  106 m s . I 4.85 A 25.3: a) vd   nqA (8.5  10 )(1.6  10 C)( π 4)(2.05  103 m) 2 ) 28 19  1.08  104 m s  travel time  d vd  1.08 0.10 m m s  6574 s  110 min  71 4 b) If the diameter is now 4.12 mm, the time can be calculated using the formula above or comparing the ratio of the areas, and yields a time of 26542 s =442 min. c) The drift velocity depends on the diameter of the wire as an inverse square relationship. 25.4: The cross-sectional area of the wire is A  πr 2  π (2.06  103 m) 2  1.333  105 m 2 . The current density is I 8.00A J   6.00  10 5 A m 2 A 1.333  10 5 m 2 We have vd  J ne ; Therefore J 6.00  105 A m 2 electrons n  5 19  6.94  1028 vd e (5.40  10 m s)(1.60  10 C electron) m3 25.5: J  n q vd , so J vd is constant. J1 vd 1  J 2 vd 2 , vd 2  vd 1 ( J 2 J1 )  vd 1 ( I 2 I1 )  (1.20  104 m s)(6.00 1.20)  6.00  104 m s
  2. 25.6: The atomic weight of copper is 63.55 g mole, and its density is 8.96 g cm3 . The number of copper atoms in 1.00 m 3 is thus (8.96 g cm3 )(1.00  106 cm3 m 3 )(6.023  1023 atoms mole) 63.55 g mole  8.49  1028 atoms m 3 Since there are the same number of free electrons m 3 as there are atoms of copper m3 (see Ex. 25.1), The number of free electrons per copper atom is one. 25.7: Consider 1 m 3 of silver. density  10.5  10 3 kg m 3 , so m  10.5  10 3 kg M  107.868  103 kg mol , so n  m M  9.734  104 mol and N  nN A  5.86  1028 atoms m 3 If there is one free electron per m 3 , there are 5.86  1028 free electrons m 3 . This agrees with the value given in Exercise 25.2. 25.8: a) Qtotal  (nCl  n Na )e  (3.92  1016  2.68  1016 )(1.60  10 19 C)  0.0106 C Q 0.0106 C  I  total   0.0106 A  10.6 mA. t 1.00 s b) Current flows, by convention, in the direction of positive charge. Thus, current flows with Na  toward the negative electrode. 8 8 0.65 3 8 8 25.9: a) Q   I dt   (55  0.65 t 2 ) dt  55t |  t |  329 C. 0 0 0 3 0 b) The same charge would flow in 10 seconds if there was a constant current of: I  Q t  (329 C) (8 s)  41.1 A. 3.6 A 25.10: a) J  I A  ( 2.3  10  3 m ) 2  6.81  10 5 A/m 2 . b) E  ρJ  (1.72  10 8   m)(6.81  105 A/m 2 )  0.012 V m . c) Time to travel the wire’s length: l l nqA (4.0 m)(8.5  1028 m 3 )(1.6  1019 C)(2.3  103 m) 2 t    8.0  104 s vd I 3.6 A  1333 min  22 hrs! ρL (1.72  108   m)(24.0 m) 25.11: R    0.125 . A (π 4)(2.05  10 3 m) 2
  3. ρL RA (1.00 )(π 4)(0.462  103 m) 2 25.12: R  L   9.75 m. A ρ 1.72  108   m 25.13: a) tungsten: ρI (5.25  108  m 3 )(0.820 A) E  ρJ   3  5.16  10 3 V m . A (π 4)(3.26  10 m) 2 b) aluminum: ρI (2.75  108  m3 )(0.820 A) E  ρJ    2.70  103 V m . A (π 4)(3.26  10 3 m) 2 2 2 ρ Al L ρCu L πd A l πd Cu ρCu 25.14: R Al  RCu      d Cu  d Al AAl ACu 4 ρ Al 4 ρCu ρ Al 1.72  108   m  d Al  (3.26 mm)  2.6 mm. 2.75  108   m 25.15: Find the volume of one of the wires: ρL ρL R so A  and A R ρL2 1.72  108 Ohm  m)(3.50m) 2 volume  AL    1.686  10 6 mcb R 0.125Ohm m  (density)V  (8.9  10 kg m 3 )(1.686  10 6 m 3 )  15 g 3 25.16: 3.5 cm r1   1.75 cm 2 3.25 mm r2   1.625 mm 2
  4. 25.17: a) From Example 25.1, an 18-gauge wire has A  8.17  10 3 cm 2 I  JA  (1.0  105 A/cm2 )(8.17  103 cm 2 )  820 A b) A  I J  (1000 A) (1.0  106 A cm 2 )  1.0  103 cm 2 A  πr 2 so r  A π  (1.0  103 cm 2 π  0.0178 cm d  2r  0.36 mm 25.18: Assuming linear variation of the resistivity with temperature: ρ  ρ0 [1   (T  T0 )]  ρ0 [1  (4.5  10 3 C)(320  20)C]  2.35 ρ0 Since   E J , the electric field required to maintain a given current density is proportional to the resistivity. Thus E  (2.35)(0.0560 V m)  0.132 V m ρL ρL ρ 2.75  108   m 25.19: R   2    1.53  108  A L L 1.80m 25.20: The ratio of the current at 20C to that at the higher temperature is (0.860 A) (0.220 A)  3.909. Since the current density for a given field is inversely proportional to ρ( ρ  E J ), The resistivity must be a factor of 3.909 higher at the higher temperature. ρ  1   (T  T0 ) ρ0 ρ ρ0 1 3.909  1 T  T0   20C   666C  4.5  10 3 C V ρL ρL IL (6.00 A)(2.75  10 8   m)(1.20 m) R   r  25.21: I A πr 2 πV π (1.50 V)  2.05  10 4 m. RA VA (4.50 V)π (6.54  10 4 m) 2 25.22: ρ     1.37  10 7   m. L IL (17.6 A)(2.50 m)
  5. EA (0.49 V m)(π 4 (0.84  10 3 m) 2 ) 25.23: a) I  JA    11.1 A. ρ (2.44  10 8   m) IL (11.1 A)(2.44  10 8   m)(6.4 m) b) V  IR    3.13 V. A (π 4)(0.84  10 3 m) 2 V 3.13 V c) R    0.28 Ω. I 11.1A 25.24: Because the density does not change, volume stays the same, so LA  (2 L)( A 2) and the area is halved. So the resistance becomes: ρ ( 2 L) ρL R 4  4 R0 . A2 A That is, four times the original resistance. RAJ RI V 0.938 V 25.25: a) E  ρJ      1.25 V m . L L L 0.75 m RA V 0.938 V b) ρ     2.84  108   m. L JL (4.40  10 A m )(0.75 m) 7 2 R  R0 25.26:   (T f  Ti ) R0 R  R0 1.512   1.484        1.35  10 3 C 1 . (T f  Ti ) R0 (34.0 C  20.0 C)(1.484 ) 25.27:a) R f  Ri  Ri (T f  Ti )  R f  100   100 (0.0004C 1 )(11.5C)  99.54 . b) R f  Ri  Ri (T f  Ti )  R f  0.0160   0.0160 (0.0005C 1 )(25.8C)  0.0158 . R f  Ri R f  Ri 25.28: T f  Ti  ; T f  Ti  Ri Ri 215.8   217.3   1  4 o C  17.8o C. (0.0005 C )(217.3 ) o
  6. 25.29: a) If 120 strands of wire are placed side by side, we are effectively increasing the area of the current carrier by 120. So the resistance is smaller by that factor: R  5.60  106  120  4.67  108 . b) If 120 strands of wire are placed end to end, we are effectively increasing the length of the wire by 120, and so R  (5.60  10 6 Ω)120  6.72  10 4 . 25.30: With the 4.0  load, where r = internal resistance 12.6 V  (r  4.0 ) I Change in terminal voltage: VT  rI  12.6 V  10.4 V  2.2 V 2.2 V I r  2.2 V  Substitute for I: 12.6 V  (r  4.0 )   r  Solve for r: r  0.846  L 1.72  10 8 m)(100  103 m) 25.31: a) R    0.219 A π (0.050m) 2 V  IR  (125A)(0.219)  27.4V b) P  VI  (27.4 V)(125 A)  3422 W  3422 J/s Energy  Pt  (3422 J/s)(3600 s)  1.23  10 7 J 25.32: a) Vr  ε  Vab  24.0 V  21.2 V  2.8 V  r  2.8 V 4.00 A  0.700  . b) VR  21.2 V  R  21.2 V 4.00 A  5.30 . 25.33: a) An ideal voltmeter has infinite resistance, so there would be NO current through the 2.0  resistor. b) Vab  ε  5.0 V; since there is no current there is no voltage lost over the internal resistance. c) The voltmeter reading is therefore 5.0 V since with no current flowing, it measures the terminal voltage of the battery.
  7. 25.34: a) A voltmeter placed over the battery terminals reads the emf: ε  24.0 V. b) There is no current flowing, so Vr  0. c) The voltage reading over the switch is that over the battery: Vs  24.0 V. d) Having closed the switch: I  24.0 V 5.88   4.08 A  Vab  24.0 V  (4.08 A)(0.28 )  22.9 V. Vr  IR  (4.08 A)(5.60 )  22.9 V. Vs  0, since all the voltage has been “used up” in the circuit. The resistance of the switch is zero so Vs  IR  0. 25.35: a) When there is no current flowing, the voltmeter reading is simply the emf of the battery: ε  3.08 V. b) The voltage over the internal resistance is: V 0.11 V Vr  3.08 V  2.97 V  0.11 V  r    0.067 . I 1.65 A c) VR  2.97 V  (1.65 A) R 2.97 V R  1.8  1.65 A 25.36: a) The current is counterclockwise, because the 16 V battery determines the direction of current flow. Its magnitude is given by: ε 16.0 V  8.0 V I   0.47 A.  R 1.6   5.0   1.4   9.0  b) Vab  16.0 V  (1.6 )(0.47 A)  15.2 V. c) Vac  (5.0 )(0.47 A)  (1.4 )(0.47 A)  8.0 V  11.0 V. d)
  8. 25.37: a) Now the current flows clockwise since both batteries point in that direction: ε 16.0 V  8.0 V I   1.41 A.  R 1.6   5.0   1.4   9.0  b) Vab  16.0 V  (1.6 )(1.41 A)  13.7 V. c) Vac  (5.0 )(1.41 A)  (1.4 )(1.41 A)  8.0 V  1.0 V. d) 25.38: a) Vbc  1.9 V  I  Vbc Rbc  1.9 V 9.0   0.21 A. 5.48 b)  ε   IR  8.0 V  ((1.6  9.0  1.4  R ))(0.21 A)  R   26.1 . 0.21 c) 25.39: a) Nichrome wire: b) The Nichrome wire does obey Ohm’s Law since it is a straight line. c) The resistance is the voltage divided by current which is 3.88 .
  9. 25.40: a) Thyrite resistor: b) The Thyrite is non-Ohmic since the plot is curved. c) Calculating the resistance at each point by voltage divided by current: 25.41: a) r  ε I  1.50 V 14.8 A  0.101 . b) r  ε I  1.50 V 6.8 A  0.22 . c) r  ε I  12.6 V 1000 A  0.0126  . 25.42: a) P  V 2 R  R  V 2 P  (15 V) 2 327 W  0.688 . V 15 V b) V  IR  I    21.8 A. R 0.688  25.43: P  VI  (650 V)(0.80 A)  520 W. 25.44: W  Pt  IVt  (0.13 A)(9 V)(1.5)(3600 s)  6318 J.
  10. P I 2 R J 2 A 2 R J 2 A( L A) 25.45: a) P  I R  p  2     J 2   p  JE since vol AL AL L E  J . b) From (a) p  J 2  . c) Since J  E ρ, (a) becomes p  E 2 ρ. 25.46: a) I   ε Rtotal  8.0 V 17   0.47 A  P5  I 2 R  (0.47 A) 2 (5.0 )  1.1 W and P9  I 2 R  (0.47 A) 2 (9.0 )  2.0 W. b) P V  εI  I 2 r  (16 V)(0.47 A)  (0.47 A) 2 (1.6 )  7.2 W. 16 c) P8V  εI  Ir 2  (8.0 V)(0.47 A)  (0.47 A) 2 (1.4)  4.1 W. d) (b)  (a )  (c) 25.47: a) W  Pt  IVt  (60 A)(12 V)(3600 s)  2.59  10 6 J. b) To release this much energy we need a volume of gasoline given by: 2.59  10 6 J m 0.056 kg m  56.0 g  vol    6.22  10 5 m 3  0.062 liters. 46,000 J g  900 kg m 3 c) To recharge the battery: t  (Wh ) P  (720 Wh ) (450 W )  1.6 h. 25.48: a) I  ε ( R  r )  12 V 10   1.2 A  P  εI  (12 V)(1.2 A)  14.4 W. This is less than the previous value of 24 W. b) The work dissipated in the battery is just: P  I 2 r  (1.2 A) 2 (2.0 )  2.9 W. This is less than 8 W, the amount found in Example (25.9). c) The net power output of the battery is 14.4 W  2.9 W  11.5 W. This is less than 16 W, the amount found in Example (25.9). 25.49: a) I  V R  12 V 6   2.0 A  P  εI  (12 V) (2.0 A)  24 W. b) The power dissipated in the battery is P  I 2 r  (2.0 A) 2 (1.0 )  4.0 W. c) The power delivered is then 24 W  4 W  20 W. 25.50: a) I   ε / R  3.0 V / 17   0.18 A  P  I 2 R  0.529 W. b) W  Pt  IVt  (0.18 A)(3.0 V)(5.0)(3600 s)  9530 J. c) Now if the power to the bulb is 0.27 W, 2  3.0 V  P  I R  0.27 W   2  17   R  (17 )  (17   R)  567   R  6.8 .  2 2  
  11. 25.51: a) P  V 2 R  R  V 2 P  (120 V) 2 / 540 W  26.7 . b) I  V R  120 V / 26.7   4.5 A. c) If the voltage is just 110 V, then I  4.13 A  P  VI  454 W. d) Greater. The resistance will be less so the current drawn will increase, increasing the power. m 25.52: From Eq. (25.24), ρ  . ne 2 τ m 9.11  10 31 kg τ 2  3 19  1.55  10 12 s. ne ρ (1.0  10 m ) (1.60  10 C) (2300   m) 16 2 b) The number of free electrons in copper (8.5  10 28 m 3 ) is much larger than in pure silicon (1.0  1016 m 3 ). RA (0.104 ) (π 4) (2.50  10 3 m) 2 25.53: a) ρ    3.65  10 8   m. L 14.0 m EA (1.28 V m) (π 4) (2.50  103 m)2 b) I  JA    172 A. ρ 3.65  108   m J E 1.28 V/m c) vd    nq ρnq (3.65  10   m) (8.5  1028 m  3 ) (1.6  1019 C) 8  2.58  10 3 m/s. 25.54: r = 2.00 cm T = 0.100 mm V V VA V (2πrT ) I    R ρl A ρl ρl (12 V) (2π )(2.00  10 2 m) (0.100  10 3 m)  (1.47  10 8   m) (25.0 m)  410 A
  12. 25.55: With the voltmeter connected across the terminals of the battery there is no current through the battery and the voltmeter reading is the battery emf; ε  12.6 V. With a wire of resistance R connected to the battery current I flows and ε  Ir  IR  0 Call the resistance of the 20.0-m piece R1 ; then the resistance of the 40.0-m piece is R2  2R1 . ε  I1r  I1R1  0; 12.6 V  (7.00 A)r  (7.00 A) R1  0 ε  I 2 r  I 2 (2 R1 )  0; 12.6 V  (4.20 A)r  (4.20 A)(2 R1 )  0 Solving these two equations in two unknowns gives R1  1.20. This is the resistance of 20.0 m, so the resistance of one meter is [1.20 /(20.0m)] (1.00m)  0.060 V V 25.56: a) I   R RCu  R Ag and ρCu LCu (1.72  10 8   m) (0.8 m) RCu    0.049 , ACu (π/4) (6.0  10  4 m) 2 and ρ Ag LAg (1.47  10 8   m) (1.2 m) RAg    0.062  AAg (π/4) (6.0  10  4 m) 2 5.0 V I  45 A. 0.049   0.062  So the current in the copper wire is 45 A. b) The current in the silver wire is 45 A, the same as that in the copper wire or else charge would build up at their interface. IR (45 A) (0.049 ) c) ECu  JρCu  Cu   2.76 V m . LCu 0.8 m IR (45 A) (0.062 ) d) E Ag  JρAg  Ag   2.33 V m . LAg 1.2 m e) V Ag  IR Ag  (45 A ) (0.062 )  2.79 V.
  13. 25.57: a) The current must be the same in both sections of the wire, so the current in the thin end is 2.5 mA. ρI (1.72  108   m) (2.5  103 A) b) E1.6mm  ρJ   3  2.14  10 5 V/m. A (π 4) (1.6  10 A) 2 ρI (1.72  10 8   m) (2.5  10 3 A) c) E 0.8mm  ρJ   A (π 4) (0.80  10 3 A) 2 = 8.55  10 5 V/m ( 4 E1.6mm ). d) V  E1.6 mm L1.6 mm  E 0.8 mm L0.8 mm  V  (2.14  105 V/m) (1.20 m)  (8.55  105 V/m) (1.80 m)  1.80  104 V. K 1 2 25.58: a)  n  mvd  volume 2  K 1   (8.5  10 28 m  3 ) (9.11  10  31 kg) (1.5  10 4 m/s) 2 volume 2  8.7  1010 J / m 3. b) U  qV  ne( volume)V  (8.5  1028m 3 ) (1.6  1019C) (106 m 3 ) (1.0 V)  13600 J. And the kinetic energy in 1.0 cm 3 is K  (8.7  1010 J/m3 ) (106 m)  U 13600 J 8.7  10 16 J. So   1.6  1019. K 8.7  10 16 J
  14. 25.59: a) ρL ρdx r r  dR   2 where r  r1   1 2  x. A πr  h  h r2 ρ dx ρh du R   0   x  π r1  r1  r2 h 2 π(r1  r2 ) r1 u2 r2 ρh 1 ρh  1   R  . π (r1  r2 ) u r π  r1r2    1 ρh ρL b) When r1  r2  r , R  2  . πr A b b ρdr ρ dr ρ 1 ρ 1 1 25.60: a) dR  4πr 2 R 4π  r 2   4π r a  4π  a  b . a   V V 4πab I Vab 4πab Vab ab b) I  ab  ab J   . R ρ(b  a) A ρ(b  a )4πr 2 ρ(b  a)r 2 c) If the thickness of the shells is small, we have the resistance given by: ρ  1 1  ρ(b  a) ρL ρL R     2  , where L  b  a. 4π  a b  4πab 4πa A 25.61: E  ρJ and E  σ Kε 0  Q AKε0  ρJ  Q AKε0  AJ  I  Q Kε0 ρ  leakage current.
  15. 25.62: a) I  V  J  A  RA    LV A  A   L . So to make the current density a R I V / V maximum, we need the length between faces to be as small as possible, which means L  d . So the potential difference should be applied to those faces which are a distance d apart. This maximum current density is J MAX  ρd . V b) For a maximum current I  V  VA  JA must be a maximum. The maximum area R ρL is presented by the faces that are a distance d apart, and these two faces also have the greatest current density, so again, the potential should be placed over the faces a distance d apart. This maximum current is Vd I MAX  6 . ρ ρL (9.5  107   m) (0.12 m) 25.63: a) R    0.057 . A (π 4) (0.0016 m)2 b) ρ(T )  ρ0 (1  αT )  ρ(60 C)  (9.5  10 7   m) (1  (0.00088(C) 1 ) (40C)  ρ(60C)  9.83  107   m    3.34  108   m. c) V  βV0 T  AL  A ( βL0 T )  L  βL0 T  (18  10 5 (C) 1 )  (0.12 m) (40C)  L  8.64  10 4 m  0.86 mm. The volume of the fluid remains constant. As the fluid expands the container, outward expansion “becomes” upward expansion due to surface effects. ρL ρL ρL d) R   R   A A A (3.34  108   m) (0.12 m) (95  108   m) (0.86  103 m)  R   (π/4) (0.0016 m)2 (π/4) (0.0016 m)2  2.40  103 . e) From Equation (25.12), α  1 T  R R0  1  1 40 C  ( 0.057   2.40  103  ) 0.057   1  1.1  10 3 (C) 1. This value is greater than the temperature coefficient of resistivity and therefore is an important change caused by the length increase.
  16. 25.64: a) I   ε  8.0 V  4.0 V  0.167 A R 24.0   Vad  8.00 V  (0.167 A) (8.50 )  6.58 V. b) The terminal voltage is Vbc   4.00 V  (0.167 A) (0.50 )   4.08 V. c) Adding another battery at point d in the opposite sense to the 8.0 V battery: I  ε  10.3 V  8.0 V  4.0 V  0.257 A, and so R 24.5   Vbc  4.00 V  (0.257 A) (0.50 )  3.87 V. 25.65: a) Vab  ε  Ir  8.4 V  ε  (1.50 A) r and 9.4 V  ε  (3.50 A) r  9.4 V  (8.4 V  (1.50 A)r )  (3.50 A)r 9.4 V  8.4 V r  0.2 . 5.00 A b) ε  8.4 V  (1.50 A) (0.20 )  8.7 V. 25.66: a) I  V / R  14 kV / (10 k  2 k)  1.17 A. b) P  I 2 R  (1.17 A) 2 (10,000 )  13.7 kW. c) If we want the current to be 1.0 mA, then the internal resistance must be: 14,000 V Rr  1.4  107   R  14 M  10 k  14 M. 0.001 A ρL (5.0   m) (0.10 m) 25.67: a) R    1000 . A π (0.050 m)2 b) V  IR  (100  10 3 A) (1000 )  100 V. c) P  VI  (100V) (100  10 3 A)  10 W.
  17. 25.68: a) V  2.50 I  0.360 I 2  4.0 V. Solving the quadratic equation yields I  1.34 A or  8.29 A, so the appropriate current through the semiconductor is I  1.34 A. b) If the current I  2.68 A,  V  (2.50 V / A) (2.68 A)  (0.36 V / A 2 ) (2.68 A) 2  9.3 V. 25.69: V  IR  V ( I )  IR  αI  βI 2  (α  R) I  βI 2  βI 2  ( R  α ) I  V  0  (1.3) I 2  (3.8  3.2) I  12.6  0  I  1.42 A. 25.70: a) r  ε  7.86 V  0.85   I  ε  7.86 V  2.42 A. I 9.25 A Rr 0.85   2.4  b) βI 2  (α  r ) I  ε  0  0.36 I 2  (2.50  0.85) I  7.86  0  I  1.94 A c) The terminal voltage at this current is Vab  ε  Ir  7.86 V  (1.94 A) (0.85 )  6.21 V. 25.71: a) With an ammeter in the circuit: I ε  ε  I A (r  R  R A ). r  R  RA So with no ammeter: ε  r  R  RA   IA   RA  I rR  r  R   I A 1  r  R  .        b) We want: I  RA  RA  1     1.01    0.01  RA (0.01) (0.45   3.8 ) IA  r  R rR  0.0425 . c) This is a maximum value, since any larger resistance makes the current even less that it would be without it. That is, since the ammeter is in series, ANY resistance it has increases the circuit resistance and makes the reading less accurate.
  18. 25.72: a) With a voltmeter in the circuit: I ε   Vab  ε  Ir  ε 1  r  . r  RV  r  RV    b) We want: Vab  r  r  1    0.99   0.01 ε  r  RV    r  RV r  0.01r  RV   99r  99  045   44.6 . 0.01 c) This is the minimum resistance necessary—any greater resistance leads to less current flow and hence less potential loss over the battery’s internal resistance. 25.73: a) The line voltage, current to be drawn, and wire diameter are what must be considered in household wiring. P 4200 W b) P  VI  I    35 A, so the 8-gauge wire is necessary, since it can V 120 V carry up to 40 A. I 2 ρL (35 A) 2 (1.72  108   m) (42.0 m) c) P  I 2 R    106 W. A ( 4) (0.00326 m)2 d) If 6-gauge wire is used, I 2 ρL (35 A) 2 (1.72  10 8 Ω  m) (42 m) P   66 W A (π 4) ) (0.00412 m) 2  E  Pt  (40 W ) (365) (12 h )  175 kWh  Savings  (175 kWh ) ($0.11 kWh )  $19.25. 25.74: Initially: R0  V I 0  (120 V) (1.35 A)  88.9 . Finally: R f  V I f  (120 V) (1.23 A)  97.6 . Rf 1 R  1  97.6   And  1  α (T f  T0 )  (T f  T0 )   f  1  R  4.5  10 4C 1   88.9   1  R0 α 0     T f  T0  217C  T f  217C  20C  237C. b) (i) P0  VI 0  (120 V) (1.35 A)  162 W (ii) Pf  VI f  (120 V) (1.23 A)  148 W
  19.  ε 12.0 V  8.0 V 25.75: a) I    0.40 A. R 10.0  b) Ptotal  I 2 Rtotal  (0.40 A) 2 (10 )  1.6 W. c) Power generated in ε1 , P  ε1 I  (12.0 V) (0.40 A)  4.8 W. d) Rate of electrical energy transferred to chemical energy in ε2 P  ε2 I  (8.0 V)  (0.40 A)  3.2 W. e) Note (c)  (b)  (d), and so the rate of creation of electrical energy equals its rate of dissipation.  L (2.0  10 7   m) (2.0 m) 25.76: a) Rsteel    1.57  10 3  A ( 4) (0.018 m) 2 ρL (1.72  10 8   m) (35 m) RCu    0.012  A (π 4) (0.008 m) 2  V  IR  I ( Rsteel  RCu )  (15000 A) (1.57  10 3   0.012 )  204 V. b) E  Pt  I 2 Rt  (15000 A) 2 (0.0136 ) (65  10 6 s)  199 J. |q| a 25.77: a)  F  ma  | q | E   . m E | q | aL b) If the electric field is constant, Vbc  EL   . m Vbc c) The free charges are “left behind” so the left end of the rod is negatively charged, while the right end is positively charged. Thus the right end is at the higher potential. V | q | (1.0  103 V) (1.6  1019 C) d) a  bc   3.5  108 m/s 2 . mL (9.11  1031 kg) (0.50 m) e) Performing the experiment in a rotational way enables one to keep the experimental apparatus in a localized area—whereas an acceleration like that obtained in (d), if linear, would quickly have the apparatus moving at high speeds and large distances.
  20. 25.78: a) We need to heat the water in 6 minutes, so the heat and power required are: Q  mcv T  (0.250 kg) (4190 J/kgC) (80C)  83800 J Q 83800 J P   233 W. t 6(60 s) V2 V 2 (120 V) 2 But P  R   61.8 . R P 233 W  L  L2 R  vol (61.8 ) (2.5  10 5 m 3 ) b) R   L   39 m. A vol  1.00  10 6   m Now the radius of the wire can be calculated from the volume: vol 2.5  10 5 m 3 vol  L(πr 2 )  r    4.5  10 4 m. πL π (39 m) 25.79: a) Vab  ε  Ir  12.0 V  (10.0 A) (0.24 Ω)  14.4 V. b) E  Pt  IVt  (10 A) (14.4 V) (5) (3600 s)  2.59  10 6 J. c) E diss  Pdiss t  I 2 rt  (10 A) 2 (0.24 ) (5) (3600 s)  4.32  10 5 J. d) Discharged at 10 A: I ε  R  ε  Ir  12.0 V  (10 A) (0.24 )  0.96 . rR I 10 A e) E  Pt  IVt  (10 A) (9.6 V) (5) (3600 s)  1.73  106 J. f) Since the current through the internal resistance is the same as before, there is the same energy dissipated as in (c): E diss  4.32  10 5 J. g) The energy originally supplied went into the battery and some was also lost over the internal resistance. So the stored energy was less than was needed to charge it. Then when discharging, even more energy is lost over the internal resistance, and what is left is dissipated over the external resistor.
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