 # Physics exercises_solution: Chapter 30

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5 ## Physics exercises_solution: Chapter 30

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 30

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## Nội dung Text: Physics exercises_solution: Chapter 30

1. 30.1: a)  2  M (di1 / dt )  (3.25  10 4 H) (830 A /s)  0.270 V, and is constant. b) If the second coil has the same changing current, then the induced voltage is the same and  1  0.270 V. 30.2: For a toroidal solenoid, M  N 2  B2 / i1 , and  B2   0 N 1i1 A/ 2r. So, M   0 AN 1 N 2 / 2r. 30.3: a) M  N 2  B2 / i1  (400) (0.0320 Wb) / (6.52 A)  1.96 H. b) When i2  2.54 A,  B1  i 2 M / N 1  (2.54 A) (1.96 H) / (700)  7.11  10 3 Wb. 30.4: a) M   2 / (di / dt )  1.65  103 V / (0.242 A /s)  6.82  103 H. b) N 2  25, i1  1.20 A,   B2  i1M / N 2  (1.20 A) (6.82  10 3 H) / 25  3.27  10 4 Wb. c) di 2 / dt  0.360 A / s and  1  Mdi2 / dt  (6.82  10 3 H) (0.360 A / s)  2.45 mV. 30.5: 1 H  1 Wb / A  1 Tm 2 / A  1 Nm / A 2  1 J / A 2  1 (J / AC)s  1 (V / A)s  1 Ωs. 30.6: For a toroidal solenoid, L  N B / i   / (di / dt ). So solving for N we have: (12.6  10 3 V) (1.40 A) N  i /  B ( di / dt )   238 turns. (0.00285 Wb) (0.0260 A / s) 30.7: a)   L(di1 / dt )  (0.260 H) (0.0180 A / s)  4.68  10 3 V. b) Terminal a is at a higher potential since the coil pushes current through from b to a and if replaced by a battery it would have the  terminal at a. (500 μ0 ) (1800) 2 (4.80  10 5 m 2 ) 30.8: a) LK m  K m μ0 N A/ 2πr  2  0.130 H. 2π (0.120 m) 1 1 b) Without the material, L  LK m  (0.130 H)  2.60  10  4 H. Km 500
2. 30.9: For a long, straight solenoid: L  N B / i and  B  μ0 NiA / l  L  μ0 N 2 A/ l. 30.10: a) Note that points a and b are reversed from that of figure 30.6. Thus, according V V to Equation 30.8, dt  b L a  1.04 H   4.00 A / s. Thus, the current is decreasing. di 0.260 V b) From above we have that di  (  4.00 A / s)dt. After integrating both sides of this expression with respect to t , we obtain i  (  4.00 A / s)t  i  (12.0 A)  (4.00 A/s) (2.00 s)  4.00 A. 30.11: a) L   / (di / dt )  (0.0160 V) / (0.0640 A/s)  0.250 H. b)  B  iL / N  (0.720 A) (0.250 H) / (400)  4.50  104 Wb. 1 2 30.12: a) U  LI  (12.0 H) (0.300 A) 2 / 2  0.540 J. 2 b) P  I R  (0.300 A) 2 (180 )  16.2 W. 2 c) No. Magnetic energy and thermal energy are independent. As long as the current is constant, U  constant. 1 2 μ0 N 2 Al 2 30.13: U  LI  2 4πr 4πrU 4π (0.150 m) (0.390 J) N   2850 turns. μ0 AI 2 μ0 (5.00  10 4 m 2 ) (12.0 A) 2 30.14: a) U  Pt  (200 W) (24 h/day  3600 s / h)  1.73  10 7 J. 1 2 2U 2(1.73  10 7 J) b) U  LI  L  2   5406 H. 2 I (80.0 A) 2 30.15: Starting with Eq. (30.9), follow exactly the same steps as in the text except that the magnetic permeability  is used in place of  0 .
3. B2 (0.560 T) 2 30.16: a) free space: U  uV  V (0.0290 m 3 )  3619 J. 2 0 2 0 B2 (0.560 T) 2 b) material with K m  450  U  uV  V (0.0290 m 3 )  8.04 J. 2K m  0 2(450)  0 U B2 2 0U 2 0 (3.60  10 6 J) 30.17: a) u    Volume    25.1 m 3 . Vol 2 0 B 2 (0.600 T) 2 2 0U 2 0 (3.60  10 6 J) b) B  2  3  141.4 T 2  B  11.9 T. Vol (0.400 m)  0 NI  0 (600) (2.50 A) 30.18: a) B    4.35 mT. 2r 2 (0.0690 m) B2 (4.35  10 3 T) 2 b) From Eq. (30.10), u    7.53 J / m 3 . 2 0 2 0 c) Volume V  2rA  2 (0.0690 m) (3.50  10 6 m 2 )  1.52  10 6 m 3 . d) U  uV  (7.53 J / m 3 ) (1.52  10 6 m 3 )  1.14  10 5 J.  N 2 A  0 (600) 2 (3.50  10 6 m 2 ) e) L  0   3.65  10 6 H. 2r 2 (0.0690 m) 1 1 U  LI 2  (3.65  10 6 H) (2.50 A) 2  1.14  10 5 J same as (d). 2 2 di   iR di 6.00 V 30.19: a)  . When i  0    2.40 A / s. dt L dt 2.50 H di 6.00 V  (0.500 A) (8.00 ) b) When i  1.00 A    0.800 A/s. dt 2.50 H ε 6.00 V c) At t  0.200 s  i  (1  e ( R / L )t )  (1  e (8.00  / 2.50 H) (0.250 s) )  0.413 A. R 8.00   6.00 V d) As t    i    0.750 A. R 8.00 
4. 30.20: (a) imax  1000V  0.030 A  30 mA, long after closing the switch. 30 b) i  imax (1  e t / ( L / R ) )  0.030 A1  e 10 μs  20 μs       0.0259 A VR  Ri  (1000 Ω) (0.0259 A)  26 V VL  ε Battery  VR  30 V  26 V  4.0 V (or, could use VL  L dt at t  20 s) di c) 30.21: a) i   / R(1  e t / ),   L / R imax  ε / R so i  imax / 2 when (1  e  t / τ )  1 , and e  t / τ  2 1 2 L ln2 (ln 2) (1.25  10 3 H)  t / τ  ln ( 1 ) and t  2   17.3 μs R 50.0  b) U  1 Li 2 ; U max  1 Li 2 max 2 2 U  1 U max when i  i max / 2 2 1  e  t / τ  1 2 so e  t /τ  1  1/ 2  0.2929 t   L ln(0.2929)/ R  30.7 μs
5. 1 2 2U 2(0.260 J) 30.22: a) U  LI  I    2.13 A 2 L 0.115 H    IR  (2.13 A) (120 )  256 V. 1 1 1 11  b) i  Ie ( R / L )t and U  Li 2  Li 2 e 2 ( R / L )t  U 0   LI 2  2 2 2 22  1  e 2( R / L )t  2 L 1 0.115 H 1 t  ln     ln    3.32  10  4 s. 2R  2  2(120 )  2   60 V 30.23: a) I 0    0.250 A. R 240  4 b) i  I 0 e  ( R / L )t  (0.250 A) e  (240  / 0.160 H) (4.00  10 s )  0.137 A. c) Vcb  Vab  iR  (0.137 A) (240 )  32.9 V, and c is at the higher potential. i 1 L 1 (0.160 H)  1  d)   e ( R / L ) t1 / 2  t1 / 2   ln     ln    4.62  10 4 s. I0 2 R 2 (240 ) 2 30.24: a) At t  0  v ab  0 and vbc  60 V. b) As t    v ab  60 V and vbc  0. c) When i  0.150 A  vab  iR  36.0 V and vbc  60.0 V  36.0 V  24.0 V. 2 (6.00 V) 2 30.25: a) P  i  I 0 (1  e ( R / L )t )  (1  e ( R / L )t )  (1  e (8.00  / 2.50 H ) t ) R 8.00  1  P  (4.50 W ) (1  e  (3.20 s )t ). ε2 (6.00 V) 2 b) PR  i 2 R  (1  e ( R / L ) t ) 2  (1  e (8.00  / 2.50 H ) t ) 2 R 8.00  1  PR  (4.50 W ) (1  e  (3.20 s )t )2. 2 di ε ε  ε c) PL  iL  (1  e ( R / L ) t ) L  e ( R / L )t   (e ( R / L )t  e 2( R / L ) t ) dt R L  R 1 1  PL  (4.50 W ) (e  (3.20 s )t  e  ( 6.40 s )t ). d) Note that if we expand the exponential in part (b), then parts (b) and (c) add to give part (a), and the total power delivered is dissipated in the resistor and inductor.
6. 30.26: When switch 1 is closed and switch 2 is open: di di R i di  R t L  iR  0  i     dt  dt dt L I0 i L 0 R  ln(i / I 0 )   t  i  I 0 e t ( R / L ) . L 30.27: Units of L / R  H /   ( s) /   s  units of time. 1 30.28: a) ω   2πf LC 1 1 L   2.37  10 3 H. 4 f C 4 (1.6  10 ) (4.18  10 F) 2 2 2 6 2 12 1 1 b) C max    3.67  10 11 F. 4 f min L 4 (5.40  10 ) (2.37  10 H ) 2 2 2 5 2 3
7. 2π 30.29: a) T   2π LC  2π (1.50 H) (6.00  10 5 F) ω  0.0596 s, ω  105 rad s. b) Q  CV  (6.00  10 5 F)(12.0 V)  7.20  10 4 C. 1 1 c) U 0  CV 2  (6.00  10 5 F)(12.0 V) 2  4.32  10 3 J. 2 2 d) At t  0, q  Q  Q cos(ωt   )    0.  0.0230 s  t  0.0230 s, q  Q cos(ωt )  (7.20  10  4 C) cos   (1.50 H)(6.00  10 5 F)    4  5.43  10 C. Signs on plates are opposite to those at t  0. dq e) t  0.0230 s, i   ωQ sin(ωt ) dt 7.20  10 4 C  0.0230 s  i  sin    0.0499A. (1.50H)(6.00  10 H)5  (1.50 H) (6.00  10 5 H)    Positive charge flowing away from plate which had positive charge at t  0. q 2 (5.43  10 4 C) 2 f) Capacitor: U C   5  2.46  10 3 J. 2C 2(6.00  10 F) 1 1 Inductor: U L  Li 2  (1.50 H) (0.0499 A) 2  1.87  10 3 J. 2 2
8. 30.30: (a) Energy conservation says U L (max) = U C (max) 1 2 1 Li max  CV 2 2 2 18  10  6 F imax  V C L  (22.5V)  0.871A 12  10 3 H The charge on the capacitor is zero because all the energy is in the inductor. (b) 2 T  2 LC  1 1  at 1 4 period: T  (2 LC )  (12  10 3 H) (18 10 6 F) 4 4 2 4  7.30  10 s 3 at 3 4 period: T  3(7.30  10  4 s)  2.19  10 3 s 4 (c) q0  CV  (18F) (22.5 V)  405C Q 150  10 9 C 30.31: C    30.0 F V 4.29  10 3 V For an L-C circuit, ω  1 LC and T  2π ω  2 LC (T 2 ) 2 L  0.601 mH C
9. 1 30.32: ω   1917 rad/s (0.0850 H ) (3.20  10 6 F) imax 8.50  10 4 A a) imax  ωQmax  Qmax    4.43  10 7 C ω 1917 rad s 2  5.00  10  4 A  b) From Eq. 31.26 q  Q  LCi  (4.43  10 C)   2 2  1917 s 1  7 2    7  3.58  10 C. d 2q 1 di 30.33: a) 2  q  0  q  LC  (0.640 H)(3.60  10 6 F) (2.80 A s) dt LC dt  6.45  10 6 C. q 8.50  10 6 C b)     2.36 V. C 3.60  10 6 F imax 30.34: a) imax  ωQmax  Qmax   imax LC . ω  Qmax  (1.50 A) (0.400H)(2.50  10 10 F)  1.50  10 5 C. Q 2 max (1.50  10 5 C) 2  U max    0.450 J 2C 2(2.50  10 10 F) 2 1 1 b) 2 f     3.18  10 4 s 1 2  LC  (0.400 H ) (2.50  10 10 F) (must double the frequency since it takes the required value twice per period). 30.35: [ LC ]  H  F  H  C V C Ω C 1  Ω  s     s2   A  s2  s2  V V s A   LC  s. d 2q 1 30.36: Equation (30.20) is 2  q  0. We will solve the equation using: dt LC dq d 2q q  Q cos(ωt   )   ωQ sin(ωt   )  2  ω 2Q cos(ωt   ). dt dt 2 d q 1 Q 1 1  2  q  ω 2Q cos(ωt   )  cos(ωt   )  0   2  ω . dt LC LC LC LC
10. 1 q 2 1 Q 2 cos 2 (ωt   ) 30.37: a) U C   . 2 C 2 C 1 1 1 Q 2 sin 2 (ωt   ) 1 U L  Li 2  Lω 2Q 2 sin 2 (ωt   )  , since ω 2  . 2 2 2 C LC 1 Q2 1 b) U Total  U C  U L  cos 2 (ωt   )  Lω 2Q 2 sin 2 (ωt   ) 2 C 2 2 1Q 1  1  2 2  cos 2 (ωt   )  L Q sin (ωt   ) 2 C 2  LC  1 Q2  (cos 2 (ωt   )  sin 2 (ωt   )) 2 C 1 Q2   U Total is a constant. 2 C 30.38: a) q  Ae  ( R / 2 L )t cos(ωt   ) dq R ( R / 2 L ) t   A e cos(ωt   )  ω Ae ( R / 2 L ) t sin(ωt   ). dt 2L 2 d 2q  R  R ( R / 2 L ) t  2  A  e ( R / 2 L ) t cos(ωt   )  2ω A e sin(ωt   ) dt  2L  2L  ω 2 Ae  ( R / 2 L )t cos(ωt   ). d 2 q R dq q   R 2 2   2   q      2  R  1   0 dt L dt LC   2L  2 L2 LC    2 1 R  ω 2   2 LC 4 L dq b) At t  0, q  Q, i   0: dt dq R  q  Acos   Q and  Acos   ωAsin   0 dt 2L Q QR R  A and   ωQ tan    cos  2L 2 Lω R  . 2 L 1 / LC  R 2 / 4 L2
11. 1 30.39: Subbing x  q, m  L, b  R, k  , we find: C d 2 x b dx kx d 2 q R dq q a) Eq. (13.41): 2    0  Eq.(30.27) : 2    0. dt m dt m dt L dt LC k b2 1 R2 b) Eq. (13.43): ω    Eq.(30.28) : ω   2. m 4m 2 LC 4 L c) Eq. (13.42): x  Ae ( b / 2 m )t cos(ωt   )  Eq.(30.28) : q  Ae  ( R / 2 L )t cos(ωt   ).  L  H s  V  L 30.40:       2     . C  F C V A  C 1 R2 1  1 1  1 1 30.41:   2   2   R 2  4 L2     R  2L  LC 4 L 6 LC  LC 6 LC  LC 6 LC 1 1  R  2(0.285 H)   45.4  . (0.285 H) (4.60  10 F) 6(0.285 H) (4.60  10  4 F) 4 1 1 30.42: a) When R  0, ω0    298 rad s. LC (0.450 H) (2.50  10 5 F) ω (1 LC  R 2 4 L2 ) R 2C b) We want  0.95   1  (0.95) 2 ω0 1 LC 4L 4L 4(0.450 H) (0.0975) R (1  (0.95) 2 )   83.8 . C (2.50  10 5 F) 30.43: a) b) Since the voltage is determined by the derivative of the current, the V versus t graph is indeed proportional to the derivative of the current graph.
12. di d 30.44: a) ε   L   L ((0.124 A) cos[(240 π s)t ] dt dt  ε   (0.250 H) (0.124 A) (240  ) sin((240  s)t )   (23.4 V) sin ((240 π s)t ). b)  max  23.4 V; i  0, since the emf and current are 90 out of phase. c) imax  0.124 A;   0, since the emf and current are 90 out of phase. b b  μ Ni  b μ Nih dr μ0 Nih  B   B (hdr )    0 (hdr )  0 2  r 30.45: a)  2π r   ln (b a ). a a   a 2 N B μ0 N h 2 b) L   ln(b a). i 2π b  a (b  a ) 2 μ N 2h  b  a  c) ln (b / a )  ln ( 1  (b  a ) / a )      L  0  . a 2a 2 2π  a  2 N2 N A N A μ N IA μ NN A μ N N πr 30.46: a) M   B2  2 2  B1  2 2 0 1 1  0 1 2 2  0 1 2 2 . I I A1 IA1 l1 l1 l1 d B2 2 μ0 N1 A2 di1 μ0 N1 N 2 π r2 di1 b) ε 2  N 2  N2  . dt l1 dt l1 dt 2 di di μ N N π r di2 c) ε1  M 12 2  M 2  0 1 2 2 . dt dt l1 dt
13. di 30.47: a) ε   L  L  ε /(di / dt )  (30.0 V) /( 4.00 A / s )  7.5 H. dt d b) ε    f   i  εt   f  (30.0 V)(12.0 s)  360 Wb. dt di c) PL  Li  (7.50 H)(48.0 A)(4.00 A /s)  1440 W. dt P PR  i 2 R  (48.0 A) 2 (60.0 )  138240 W  L  0.0104. PR di d 30.48: a)   L  (3.50  10 3 H) ((0.680 A) cos( t / 0.0250 s)) dt dt    max  (3.50  10 3 H)(0.680 A)  0.299 V. 0.0250 s Li max (3.50  10 3 H)(0.680 A) b)  B max    5.95  10 6 Wb. N 400 di c)  (t )   L   (3.50  10 3 H)(0.680 A)( / 0.0250 s)sin ( t / 0.0250 s). dt   (t )   (0.299 V) sin((125.6 s 1 )t )   (0.0180 s)   (0.299 V)sin ((125.6 s 1 )(0.0180 s))   (t )  0.230 V di1 di di 30.49: a) Series: L1  L2 2  Leq , dt dt dt di di di but i1  i 2  i for series components so 1  2  , thus L1  L2  Leq dt dt dt di1 di2 di b) Parallel: Now L1  L2  Leq , where i  i1  i 2 . dt dt dt di di1 di2 di Leq di di Leq di So   . But 1  and 2  dt dt dt dt L2 dt dt L2 dt 1 di Leq di Leq di 1 1      Leq    L  .  dt L1 dt L2 dt  1 L2 
14.   μi 30.50: a) B  di  μ0 I encl  B 2πr  μ0i  B  0 . 2πr μ0 i b) d B  BdA  ldr. 2πr b b μ il dr μ0il c)  B   d B  0   ln ( b a ). a 2π a r 2π N B μ d) L   l 0 ln(b a ). i 2π 1 1 μ μ li 2 e) U  Li 2  l 0 ln(b a)i 2  0 ln(b a). 2 2 2π 4π   μi 30.51: a)  B  dl  μ0 I encl  B 2π r  μ0i  B  0 . 2πr 2 B2 1  0i   i 2l b) u   dU  udV  u (l 2 rdr )    (l 2 rdr )  0 dr. 2 0 2 0  2 r    4 r b b μ0i 2l dr μ0i 2l c) U   dU  4π  r  ln (b / a ). a a 4π 1 2 2U μ d) U  Li  L  2  l 0 ln (b / a), which is the same as in Problem 30.50. 2 i 2π N1 B1 2 N1 A  μ0 N1i1  μ0 N1 A 30.52: a) L1     , i1 i1  2π r    2π r N 2  B2 N 2 A  μ 0 N 2 i2  μ 0 N 2 A 2 L2     i2 i2  2π r    2π r 2 2 2  μ N N A μ N A μ0 N 2 A b) M   0 1 2   0 1 2  2π r   L1 L2 .   2π r 2π r ε0 E 2 B2 30.53: u B  u E    B  ε 0 μ0 E 2  μ0 ε 0 E 2 2 μ0  ε0 μ0 (650 V/m)  2.17  10 6 T.
15. V 12.0 V 30.54: a) R    1860 . i f 6.45  10 3 A Rt  Rt b) i  i f (1  e ( R / L )t )    ln (1  i / i f )  L  L ln (1  i / i f )  (1860 )(7.25  10 4 s) L  0.963 H. ln (1  (4.86 / 6.45)) 30.55: a) After one time constant has passed:  6.00 V i  (1  e 1 )  (1  e 1 )  0.474 A R 8.00  1 2 1  U  Li  (2.50 H)(0.474 A) 2  0.281 J. 2 2 Or, using Problem (30.25(c)): 3/ 7 U   PL dt  (4.50 W )  (e  (3.20)t  e ( 6.40)t )dt. 0  (1  e ) (1  e 2 ) 1   (4.50 W )   3.20  6.40   0.281 J    L/ R L L  b) U tot  (4.50 W )  (1  e ( R / L )t )dt (4.50 W )  (e 1  1)  0 R R  2.50 H 1  U tot  (4.50 W ) e  0.517 J 8.00  L/R c) U R  (4.50 W )  (1  2e ( R / L )t  e 2( R / L )t )dt 0  L 2 L 1 L   (4.50 W )  R  R (e  1)  2 R (e  1)  2    2.50 H  U R  (4.50 W ) (0.168)  0.236 J. 8.00  The energy dissipated over the inductor (part (a)), plus the energy lost over the resistor (part (c)), sums to the total energy output (part (b)).
16. 2 2 1 2 1   1  60 V  30.56: a) U  Li0  L   (0.160 H)  240    5.00  10 J.  3 2 2 R 2    ( R / L )t di R dU L di  2  2( R / L )t b) i  e   i  iL   Ri  2 e R dt L dt dt R dU L (60 V ) 2  2( 240 / 0.160 )( 4.0010 4 )   e   4.52 W. dt 240  c) In the resistor: dU R ε2 (60 V) 2 2( 240 / 0.160 )( 4.0010 4 ).  i 2 R  e 2( R / L )t  e  4.52 W. dt R 240  ε 2 2 ( R / L ) t d) PR (t )  i 2 R  e R  ε 2 2( R / L )t ε 2 L (60 V) 2 (0.160 H) R  UR  e    5.00  10 3 J, 0 R 2R 2(240 ) 2 which is the same as part (a). 30.57: Multiplying Eq. (30.27) by i, yields:  d 1 q  2 di q di q dq d 1 i 2 R  Li  i  i 2 R  Li   i 2 R   Li 2      dt C dt C dt dt  2  dt  2 C   PR  PL  PC  0. That is, the rate of energy dissipation throughout the circuit must balance over all of the circuit elements.
17. 3T  2 3T   3  Q 30.58: a) If t   q  Q cos( t )  Q cos   Q cos  8 T 8   4  2 1 1 Q2 i  ( Q2  q2 )  ( Q2  Q2 2)  LC LC 2 LC 2 1 1 Q 1 Q2 q2  U E  Li 2  L    UB. 2 2 2 LC 2 2C 2C Q 5π 5T b) The two energies are next equal when q   ωt  t  . 2 8 8 30.59: VC  12.0 V; U C  1 CVC2 so C  2U C / VC2  2(0.0160 J ) /(12.0 V) 2  222 μF 2 1 1 f  so L  2π LC (2πf ) 2 C f  3500 Hz gives L  9.31μ H Q 6.00  10 6 C 30.60: a) Vmax    0.0240 V. C 2.50  10 4 F 1 2 Q2 Q 6.00  10 6 b) Limax   imax    1.55  10 3 A 2 2C LC 4 (0.0600 H)(2.50  10 F) 1 2 1 c) U max  Limax  (0.0600 H)(1.55  10 3 A) 2  7.21  10 8 J. 2 2 2  3   Q 1 1 3  4  2 d) If i  imax  U L  U max  1.80  10 8 J  U C  U max    q 2 4 4 2C 2C 3 q Q  5.20  10 6 C. 4 1 1 q2 U max  Li 2  for all times. 2 2 C
18. 30.61: The energy density in the sunspot is u B  B 2 / 2 μ0  6.366  10 4 J /m 3 . The total energy stored in the sunspot is U B  u BV . The mass of the material in the sunspot is m  ρV . 1 2 1 K  U B so mv  U B ; ρVv 2  u BV 2 2 The volume divides out, and v  2u B / ρ  2  10 4 m /s 30.62: (a) The voltage behaves the same as the current. Since VR  i, the scope must be across the 150  resistor. (b) From the graph, as t  , VR  25V, so there is no voltage drop across the inductor, so its internal resistance must be zero. V R  Vmax (1  e  t / r ) when t   , VR  Vmax (1  1 )  0.63Vmax . From the graph, when e V  0.63 Vmax  16V, t  0.5 ms  τ L / R  0.5 ms  L  (0.5 ms) (150)  0.075 H (c) Scope across the inductor:
19. 30.63: a) In the R-L circuit the voltage across the resistor starts at zero and increases to the battery voltage. The voltage across the solenoid (inductor) starts at the battery voltage and decreases to zero. In the graph, the voltage drops, so the oscilloscope is across the solenoid. b) At t   the current in the circuit approaches its final, constant value. The voltage doesn’t go to zero because the solenoid has some resistance RL . The final voltage across the solenoid is IRL , where I is the final current in the circuit. c) The emf of the battery is the initial voltage across the inductor, 50 V. Just after the switch is closed, the current is zero and there is no voltage drop across any of the resistance in the circuit. d) As t  , ε  IR  IRL  0 ε  50 V and from the graph I RL  15 V (the final voltage across the inductor), so I R  35 V and I  (35 V) /R  3.5 A e) I RL  15 V, so RL  (15 V) / (3.5 A)  4.3 ε  VL  iR  0, where VL includes the voltage across the resistance of the solenoid. ε R VL  ε  iR, i  (1  e t / ), so VL  ε[1  (1  e t / τ )] Rtot Rtot ε  50 V, R  10 , Rtot  14.3 , so when t  τ , Vl  27.9 V From the graph, VL has this value when t = 3.0 ms (read approximately from the graph), so τ  L / Rtot  3.0 ms. Then L  (3.0 ms)(14.3 )  43 mH.
20. 30.64: (a) Initially the inductor blocks current through it, so the simplified equivalent circuit is ε 50 V i   0.333 A R 150  V1  (100 )(0.333 A)  33.3 V V4  (50 )(0.333 A)  16.7 V V3  0 since no current flows through it. V2  V4  16.7 V (inductor in parallel with 50  resistor) A1  A3  0.333 A, A2  0 (b) Long after S is closed, steady state is reached, so the inductor has no potential drop across it. Simplified circuit becomes 50 V i  ε /R   0.385 A 130  V1  (100 )(0.385 A)  38.5 V ; V2  0 V3  V4  50 V  38.5 V  11.5 V 11.5 V i1  0.385 A, i2   0.153 A 75  11.5 V i3   0.230 A 50  