Physics exercises_solution: Chapter 31

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Physics exercises_solution: Chapter 31

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 31

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Nội dung Text: Physics exercises_solution: Chapter 31

  1. V 45.0 V 31.1: a) Vrms    31.8 V. 2 2 b) Since the voltage is sinusoidal, the average is zero. 31.2: a) I  2 I rms  2 (2.10 A )  2.97 A. 2 2 b) I rav  I  (2.97 A)  1.89 A.   c) The root-mean-square voltage is always greater than the rectified average, because squaring the current before averaging, then square-rooting to get the root-mean-square value will always give a larger value than just averaging. V 60.0 V 31.3: a) V  IX L  Iω L  I    0.120 A. ωL (100 rad s) (5.00 H) V 60.0 V b) I    0.0120 A. ωL (1000 rad s) (5.00 H) V 60.0 V c) I    0.00120 A. ωL (10,000 rad s) (5.00 H)
  2. I 31.4: a) V  IX C   I  VωC  (60.0 V) (100 rad s) (2.20  10 6 F)  0.0132 A. ωC b) I  VωC  (60.0 V) (10000 rad s) (2.20  10 6 F)  0.132 A. c) I  VωC  (60.0 V) (10,000 rad s) (2.20  10 6 F)  1.32 A. d) 31.5: a) X L  ωL  2πfL  2π (80 Hz) (3.00 H)  1508 . X 120  b) X L  ωL  2πfL  L  L   0.239 H. 2πf 2π (80 Hz) 1 1 1 c) X C     497 . C 2fC 2 (80 Hz) (4.0  10 6 F) 1 1 1 d) X C  C    1.66  10 5 F. 2fC 2fX C 2 (80 Hz) (120 ) 31.6: a) X L  ωL  2πfL  2π (60 Hz)(0.450 H)  170. If f  600 Hz, X L  1700. 1 1 1 b) X C     1061 . If f  600 Hz, X C  ωC 2πfC 2 (60 Hz) (2.50  10 6 F) 106.1 . 1 1 1 c) X C  X L   ωL  ω    943 rad/s, ωC LC (0.450 H ) (2.50  10 6 Hz) so f  150 Hz. I I (0.850 A) 31.7: VC  C    1.32  10 5 F. ωC ωVC 2π (60 Hz) (170 V)
  3. VL (12.0 V) 31.8: VL  Iω L  f   3 4  1.63  10 6 Hz. 2πIL 2π (2.60  10 A) (4.50  10 H) v (3.80 V) cos ((720 rad s)t ) 31.9: a) i    (0.0253 A) cos((720 rad s)t ). R 150  b) X L  ωL  (720 rad s) (0.250 H )  180 . di c) vL  L   (ωL) (0.0253 A) sin ((720 rad s)t )   (4.55 V) sin((720 rad s)t ). dt 1 1 31.10: a) X C    1736 . ωC (120 rad s) (4.80  10 6 F) b) To find the voltage across the resistor we need to know the current, which can be found from the capacitor (remembering that it is out of phase by 90o from the capacitor’s voltage). v v cos(ωt ) (7.60 V) cos((120 rad s)t ) i C    (4.38  10 3 A) cos((120 rad s)t ) XC XC 1736   v R  iR  (4.38  10 3 A) (250 ) cos((120 rad s)t )  (1.10 V) cos((120 rad s)t ). 1 1 L 1 31.11: a) If ω  ω0   X  ωL  X    0. LC ωC LC C LC b) When ω  ω0  X  0. c) When ω  ω0  X  0. d) The graph of X against ω is on the following page.
  4. 31.12: a) Z  R 2  (ωL) 2  (200 ) 2  ((250 rad/s) (0.400 H)) 2  224 . V 30.0 V b) I    0.134 A Z 224  c) VR  IR  (0.134 A) (200 )  26.8 V; VL  Iω L  (0.134 A) (250 rad/s) (0.400 H)  VL  13.4 V. v   13.4 V  d)   arctan  L   arctan  v    26.8 V   26.6 , and the voltage leads the current.   R   e) 31.13: a) Z  R 2  (1 / ωC ) 2  (200 ) 2  1 /((250 rad/s) (6.00  10 6 F)) 2  696 . V 30.0 V b) I    0.0431 A. Z 696  VR  IR  (0.0431 A) (200 )  8.62 V; c) I (0.0431 A) VC    28.7 V. ωC (250 rad/s) (6.00  10 6 F) V   28.7 V  d)   arctan  C   arctan  V   8.62 V    73.3, and the voltage lags the current.   R  
  5. 31.14: a) 1 Z  (ωL  1 / ωC )  (250 rad/s) (0.400 H)   567 . (250 ad/s) (6.00  10 6 F) V 30.0 V b) I    0.0529 A. Z 567  c) VC  Iω L  (0.0529) (250 rad/s) (0.400 H)  5.29 V I (0.0529 A) VC    35.3 V. ωC (250 rad/s) (6.00  10 -6 F)  V  VC  d)   arctan  L  V   arctan ()   90.0, and the voltage lags the current.   R  e) 31.15: a) b) The different voltages are: v  (30.0 V) cos(250t  26.6), vR  (26.8 V) cos(250t ), vL  (13.4 V) cos(250t  90 At t  20 ms : v  20.5 V, vR  7.60 V, vL  12.85 V. Note vR  vL  v. c) At t  40 ms : v   15.2 V, v   22.49 V, v L  7.29 V. Note v R  v L  v. Be careful with radians vs. degrees in above expressions!
  6. 31.16: a) b) The different voltage are: v  (30.0 V) cos(250t  73.3), v R  (8.62 V) cos(250t ), vC  (28.7 V) cos(250t  90) At t  20 ms : v   25.1 V, vR  2.45 V, vC   27.5 V. Note v R  vC  v. c) At t  40 ms : v   22.9 V, v R  7.23 V, vC  15.6 V. Note v R  vC  v. Careful with radians vs. degrees! 31.17: a) Z  R 2  (ωL  1 / ωC ) 2  Z  (200 ) 2  ((250 rad/s) (0.0400 H)  1 / ((250 rad/s) (6.00  10 6 F))) 2  601 . V 30 V b) I    0.0499 A. Z 601   ωL  1 / ωC   100   667   c)   arctan   arctan      70.6  , and the voltage lags   R   200   the current. d) VR  IR  (0.0499 A) (200 )  9.98 V; VL  Iω L  (0.0499 A) (250 rad s)(0.400 H)  4.99 V; I (0.0499 A) VC    33.3 V. ωC (250 rad/s) (6.00  10 6 F) e) Because of the charge-storing nature of the capacitor, its voltage will tag the source voltage. That is, the capacitor’s voltage will peak after the source voltage.
  7. 31.18: a) The different voltages plotted above are: v  (30 V) cos(250t  70.6), v R  (9.98 V) cos(250t ), vL  (4.99 V) cos(250t  90)vC  (33.3 V) cos(250t  90). b) At t  20 ms : v   24.3 V, v R  2.83 V, v L  4.79 V, vC   31.9 V. c) At t  40 ms : v   23.8 V, v R   8.37 V, v L  2.71 V, vC   18.1 V. In both parts (b) and (c), note that the voltage equals the sum of the other voltages at the given instant. Be careful with degrees vs. radians! 31.19: a) Current largest at the resonance frequency 1 f0   113 Hz. At resonance, X L  X C and Z  R. I  V / R  15.0 mA 2π LC b) X C  1 / ωC  500 ; X L  ωL  160  Z  R 2  ( X L  X C ) 2  (200 ) 2  (160   500 ) 2  394.5  I  V / Z  7.61 mA X C  X L so source voltage lags the current. 2  1   ωL  1 /(ωC )  31.20: Using Z  R   ωL   and   arctan  2 , along with the  ωC   R  6 values R  200 , L  0.400 H, and C  6.00  10 F : a) ω  1000 rad/s : Z  307 ,   49.4; ω  600 rad/s : Z  204 ,    10.7; ω  200 rad/s : Z  779 ,    75.1. V b) The current increases at first, then decreases again since I  . Z c) The phase angle was calculated in part (a) for all frequencies.
  8. 31.21: V 2  V R2  (V L  VC ) 2 V  (30.0 V) 2  (50.0 V  90.0 V) 2  50.0 V 31.22: a) First, let us find the phase angle between the voltage and the current 1 ωL  2 (1.25  10 3 Hz) (20.0  10 3 H)  2 (1.25  103 Hz ) (140  109 C ) 1 tan( )  ωC      65 R 350  The impedance of the circuit is 1 2 Z  R 2  (ωL  )  (350 ) 2   (  752 ) 2  830 . ωC The average power provided by the supply is then V2 (120 V) 2 P  Vrms I rms cos( )  rms cos( )  cos(65.1)  7.32 W Z 830  b) The average power dissipated by the resistor is PR  I rms R  2   120 V 2 830  (350 )  7.32 W 31.23: a) Using the phasor diagram at right we can see: IR R cos    . I R2  X  X 2 2 Z L C 1V2 V2 b) Pav  cos   rms cos  2 Z Z 2 V R  Pav  rms  I rms R. 2 Z Z
  9. 2 2 Vrms Vrms R 31.24: Pav  cos   Z Z Z 2 V (80.0 V) 2  rms R  (75.0 )  43.5 W. Z2 (105 ) 2 R R 31.25: a) cos    Z  1  2 R   ωL  2   ωC  240   2  1  (240 )   2π (400 Hz) (0.120 H)  2    2π (400 Hz) (7.30  10 F)  6  240    0.698 344     cos 1 (0.698)  45.8. b) From (a), Z  344 . c) Vrms  I rms Z  (0.450 A) (344 Ω)  155 V. d) Pav  Vrms I rms cos   (155 V) (0.450 A) (0.698)  48.7 W. e) PR  Pav  48.7 W. f) Zero. g) Zero. For pure capacitors and inductors there is no average energy flow. 31.26: a) The power factor equals: R R (360 ) cos      0.181. Z R  (ωL) 2 2 (360 )  (((2π )60 rad / s) (5.20 H)) 2 2 b) 1 V2 1 (240 V) 2 Pav  cos   (0.181)  2.62 W. 2 Z 2 (360 ) 2  (((2π )60 rad / s) (5.20 H)) 2
  10. 31.27: a) At the resonance frequency, Z  R. V  IZ  IR  (0.500 A) (300 Ω)  150 V b) VR  IR  150 V X L  ωL  L(1/ LC  L / C  2582 ; VL  IX L  1290 V X C  1/ (ωC )  L / C  2582 ; VC  IX C  1290 V c) Pav  1 V I cos   1 I 2 R, since V  IR and cos   1 at resonance. 2 2 Pav  1 (0.500 A) 2 (300 )  37.5 W 2 31.28: a) The amplitude of the current is given by V I R  (ωL  ωC ) 2 2 1 Thus, the current will have a maximum amplitude when ωL  1C  C   2 L  ( 50.0 rad/s1)2 ( 9.00 H )  44.4 F. 2 1 b) With the capacitance calculated above we find that Z  R , and the amplitude of the current is I  V  120 V  0.300 A. Thus, the amplitude of the voltage across the R 400  inductor is V  I (ωL)  (0.300 A) (50.0 rad / s) (9.00 H)  135 V. 31.29: a) At resonance, the power factor is equal to one, because the impedance of the R circuit is exactly equal to the resistance, so  1. Z V 2 rms 1 150 V  2 b) Average power: Pav    75 W . R 2 150  c) If the capacitor is changed, and then resonance is again attained, the power factor again equals one. The average power still has no dependence on the capacitor, so Pav  75 W again. 1 1 31.30: a)  0    15.4  10 3 rad s . LC 0.350 H  1.20  10 8 F  b) VC  I ωC    I  VC ωC  550 V  15.4  10 3 rad s 1.20  10 8 F  0.102 A   Vmax source  IR  0.102 A  400    40.8 V.
  11. 31.31: a) At resonance: 1 1 ω0   LC 0.400 H  6.00  10 6 F   0  645.5 rad s  103 Hz . b) V 30.0 V V V 21.2 V c) V1  Vrms  source     21.2 V, I rms  rms  rms  2 2 Z R 200   0.106 A V2  I rmsω0 L  0.106 A  645.5 rad s  0.400 H   27.4 V. V3  I rms  0.106 A   27.4 V  V2 ,  ω0C 645.5 rad s  6.00  10 6 F  V4  0 , since the capacitor and inductor’s voltages cancel each other. V 30 V V5  Vrms source   21.2 V . 2 2 d) If the resistance is changed, that has no affect upon the resonance frequency: ω0  645.5 rad s  103 Hz V V 21.2 V e) I rms  rms  rms   0.212 A . Z R 100  1 1 31.32: a) ω0    945 rad s . LC 0.280 H  4.00  10 6 F V 120 V b) I = 1.20 A at resonance, so: R  Z    70.6  I 1.70 A c) At resonance: Vpeak R   120 V, Vpeak L   Vpeak C   Iω L  1.70 A  945 rad s  0.280 H   450 V.
  12. N1 120 31.33: a)   10 . N2 12 V 12.0 V b) I rms  rms   2.40 A R 5.00 Ω c) Pav  I rmsVrms  2.40 A  12.0 V   28.8 W . V 2 rms 120 V  2 d) R    500  , and note that this is the same as P 28.8 W 2 2   5.00   N1   5.00    120   500  . N     2  12.0  N 2 13000 31.34: a)   108. N1 120 b) P  I 2V2  0.00850 A  13000 V   110.5 W . N2 c) I 1  I 2  0.00850 A  108  0.918 A . N1 2  N1  N1 R1 12.8  103  31.35: a) R1  R2 N   N  R    40.  2 2 2 8.00  N  1 b) V2  V1  2   60.0 V  N   1.50 V  1 40 31.36: a) Z tweeter  R 2  (1 ωC ) 2 b) Z woofer  R 2  ωL  2 c) If Z tweeter  Z woofer , then the current splits evenly through each branch. d) At the crossover point, where currents are equal:   R 2  1 ωC 2  R 2  ωL   ω  2 1 . LC
  13.  ωL  R R 31.37:   arctan   L  tan   tan   R  ω 2f  48.0    2π 80 Hz   tan 52.3   0.124 H.    31.38: a) If ω  200 rad s : Z  R 2  ωL  1 ωC  2 Z  200 2  200 rad s  0.400 H   1 200 rad s 6.00  10 6 F2  779 . V 30 V 1 I   0.0385 A  I rms   0.0272 A. Z 779  2 So, V1  I rms R  0.0272 A  200    5.44 V, V2  I rms X L  I rms ωL  0.0272 A  200 rad s  0.400 H   2.18 V, V3  I rms X C  I rms  0.0272 A   22.7 V,  ωC 200 rad s  6.00  10 6 F  V4  V3  V2  22.7 V  2.18 V  20.5 V, and V5  ε rms  30.0 2 V  21.2 V. b) If ω  1000 rad s, using the same steps as above in part (a): Z  307 , V1  13.8 V, V2  27.6 V, V3  11.5 V, V4  16.1 V, V5  21.2 V. π 3π π 31.39: a) I rav  0 when ωt  n  1 2π  t1  , t2   t 2  t1  . 2ω 2ω ω I t2 I 2I 2I b)  idt   I cosωt dt  sin ωt   sin 3π 2  sin π 2   t2 t2  , t1 t1 ω t1 ω ω ω since it is rectified. 2I ω 2I 2I c) So, I rav t 2  t1    I rav   . ω π ω  XL 250  31.40: a) X L  ωL  L    0.332  ω 2π 120 Hz  R b) Z  R 2  X L  2 400  2  250  2  472 , cos   . Z V 2 rms R P 800 W Pav   Vrms  Z av  472    668 V. Z Z R 400 
  14. 31.41: a) If the original voltage was lagging the circuit current, the addition of an inductor will help it “catch up,” since a pure LR circuit would have the voltage leading. This will increase the power factor, because it is largest when the current and voltage are in phase. b) Since the voltage is lagging, the impedance is dominated by a capacitive element so we need an inductor such that X L  X 0 , where X 0 is the original capacitively dominated reactance (this could include inductors, but the capacitors “win”). R  0.720 Z  0.72060.0    43.2   Z  R2  X C  X 0  Z 2  R2  2 60  2  43.2  2  41.6 . XC 41.6  X L  X C  41.6   ωL  L    0.132 H ω 2π 50 Hz   80.0   R 2  X C  R 2  50.0   . Thus, V rms 2 31.42: Z  I rms  240 V 3.00 A 2 R  80.0    50.0    62.4 . . The average power supplied to this circuit is 2 2 equal to the power dissipated by the resistor, which is P  I 2 rms R  3.00 A  62.4    562 W 2 31.43: a) ω0  1 LC  3162 rad s; ω  2ω0  6324 rad s X L  ωL  31.62  ; X C  1 ωC   7.906  Z  R 2   X L  X C   X L  X C  23.71  2   I  V Z  5.00  10 3 V 23.71    2.108  10 4 A VC  IX C  1.667  10 3 V; this is the maximum voltage across the capacitor.    Q  CVC  20.0  10 6 F 1.667  10 3 V  33.34 nC b) In part (a) we found I = 0.211 mA c) X L  X C and R = 0 gives that the source and inductor voltages are in phase; the voltage across the capacitor lags the source and inductor voltages by 180.  1   2  X L2 31.44: a) X L2  ω2 L  2ω1 L  2   ω C   2  ω C   4X C2  X  4 , and so the     1   2  C2 inductor’s reactance is greater than that of the capacitor. ωL  1   1  1 X L2 1 b) X L3  ω3 L  1   3  3ω C    9ω C   9 X C3  X  9 , and so the     1   3  C2 capacitor’s reactance is greater than that of the inductor. c) Since X L  X C at ω1 , that is the resonance frequency.
  15. Vs R 2  ωL  2 31.45: Vout  VR2  VL2  I R 2  (ω L ) 2  Z R 2  ωL  2 Vout   . Vs R  ωL  1 ωC  2 2 V R ωR It  is small: out    ωRC. Vs R 2  1 ωC  ω 2 R 2  1 C  2 2 V ωL 2 If  is large: out   1. Vs ωL 2 I V 1 31.46: Vout  VC   out  . ωC Vs ωC R 2  ωL  1 ωC  2 V 1 1 1 If ω is large: out    . Vs ωC R  ωL  1 ωC  2 2 ωC ωL  2 LC ω 2 V 1 ωC If  is small: out    1. Vs ωC 1 ωC  ωC 2
  16. V V 31.47: a) I   . Z R 2  ωL  1 ωC  2 2 1 2 1 V  V2 R 2 b) Pav  I R    R  2 . R  ωL  1 ωC  2 2 2 Z c) The average power and the current amplitude are both greatest when the 1 1 denominator is smallest, which occurs for ω0 L   ω0  . ω0C LC d) Pav  100 V 2 200  2 . 200 2  ω2.00 H   1 ω 5.00  10 6 F2 25ω 2  Pav  .  40,000ω 2  2ω 2  2,000,000  2 Note that as the angular frequency goes to zero, the power and current are zero, just as they are when the angular frequency goes to infinity. This graph exhibits the same strongly peaked nature as the light red curve in Fig. (31.15).
  17. Vω L Vω L 31.48: a) VL  Iωω   . Z R 2  ωL  1 ωC  2 I I 1 b) VC    . ωC ωCZ ωC R 2  ωL  1 ωC 2 c) d) When the angular frequency is zero, the inductor has zero voltage while the capacitor has voltage of 100 V (equal to the total source voltage). At very high frequencies, the capacitor voltage goes to zero, while the inductor’s voltage goes to 100 1 V. At resonance, ω0   1000 rad s , the two voltages are equal, and are a LC maximum, 1000 V.
  18. 2 1 1 1 2 1  1  1 2 31.49: a) U B  Li 2  U B  L i 2  LI rms  L    LI . 2 2 2 2  2 4 2 1 1 1 2 1 V  1 U E  Cv 2  U E  C v 2  CVrms  C    CV . 2 2 2 2 2  2 4 b) Using Problem (31.47a): 2 1 1  V2  LV 2 U B  LI 2  L    4 4  R 2  ωL  1 ωC 2      4 R 2  ωL  1 ωC  2 . Using Problem (31.47b): 1 2 1 V2 V2 U E  CVC  C 2 2 2  4  4 ω C R  ωL  1 ωC 2   4ω 2C R 2  ωL  1 ωC  2 . c) Below are the graphs of the magnetic and electric energies, the top two showing the general features, while the bottom two show the details close to angular frequency equal to zero. d) When the angular frequency is zero, the magnetic energy stored in the inductor is zero, while the electric energy in the capacitor is U E  CV 2 4 . As the frequency goes to infinity, the energy noted in both inductor and capacitor go to zero. The energies equal 1 LV 2 each other at the resonant frequency where ω0  and U B  U E  . LC 4 R 2
  19. 31.50: a) Since the voltage drop between any two points must always be equal, the parallel LRC circuit must have equal potential drops over the capacitor, inductor and resistor, so v R  v L  vC  v . Also, the sum of currents entering any junction must equal the current leaving the junction. Therefore, the sum of the currents in the branches must equal the current through the source: i  i R  i L  iC . b) i R  R is always in phase with the voltage. iL  v v ωL lags the voltage by 90 , and iC  vωC leads the voltage by 90 . c) From the diagram, 2 2 V   V  I  I R  I C  I L  2 2 2     VωC     R  ωL   2 1  1  d) From (c): I  V 2   ωC    But R  ωL  2 V 1 1  1  I   2   ωC   . Z Z R  ωL  1 1 V 31.51: a) At resonance, ω0   ω0C   I C  Vω0C   I L so I  I R LC ω0 L ω0 L and I is a minimum. V2 V2 b) Pav  rms cos   at resonance where R < Z so power is a maximum. Z R c) At ω  ω0 , I and V are in phase, so the phase angle is zero, which is the same as a series resonance.
  20. V 311 V 31.52: a) V  2Vrms  311 V ; I R    0.778 A. R 400  b) I C  VωC  311 V  360 rad s  6.00  10 6 F  0.672 A . I   0.672 A  c)   arctan  C I   arctan    0.778 A   40.8 , leading the voltage.   R    d) I  I R  I C  0.778 A   0.672 A   1.03 A . 2 2 2 2 e) Leads since   0 .
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