Chapter_31_Solutions1 - 31.1 a Vrms = = 31.8 V 2 2 b Since...

Info iconThis preview shows pages 1–17. Sign up to view the full content.

View Full Document Right Arrow Icon
31.1: a) V rms = V 2 = 45.0 V 2 = 31.8 V. b) Since the voltage is sinusoidal, the average is zero.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
31.2: a) I = 2 I rms = 2 (2.10 A) = 2.97 A. b) I rav = 2 π I = 2 (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.
Background image of page 2
31.3: a) V = IX L = I ω L I = V L = 60.0 V (100 rad s )(5.00H) = 0.120 A. b) I = V L = 60.0 V (1000 rad s = 0.0120 A. c) I = V L = 60.0 V (10,000 rad s )(5 .00H) = 0.00120 A.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
31.4: a) V = IX C = I ω C I = V C = (60.0 V) (100 rad s )(2 .20 × 10 6 F) = 0.0132 A. b) I = V C = (60.0 V) (1000 rad s )(2.20 × 10 6 = 0.132 A. c) I = V C = (60.0 V) (10,000 rad s × 10 6 = 1.32 A. d)
Background image of page 4
31.5: a) X L = ω L = 2 π fL = 2 (80 Hz) (3.00 H) = 1508 . b) X L = L = 2 fL L = X L 2 f = 120 2 (80 Hz) = 0.239 H. c) X C = 1 C = 1 2 fC = 1 2 (80 Hz) (4.0 × 10 6 F) = 497 d) X C = 1 2 fC C = 1 2 fX C = 1 2 (80 Hz) (120 ) = 1.66 × 10 5 F.
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
31.6: a) . 1700 Hz, 600 If . 170 H) Hz)(0.450 60 ( 2 2 = = = = = = L L X f L f L X π ω b) X C = 1 C = 1 2 fC = 1 2 (60 Hz) (2.50 × 10 6 F) = 1061 .If f = 600 Hz, X C = 106.1 . c) X C = X L 1 C = L = 1 LC = 1 (0.450 H) (2.50 × 10 6 Hz) = 943 rad s, so f = 150 Hz.
Background image of page 6
31.7: V C = I ω C C = I V C = (0.850 A) 2 π (60 Hz) (170 V) = 1.32 × 10 5 F.
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
31.8: V L = I ω L f = V L 2 π IL = (12.0 V) 2 (2.60 × 10 3 A)(4.50 × 10 4 H) = 1.63 × 10 6 Hz.
Background image of page 8
31.9: a) i = v R = (3.80 V) cos ((720 rad s )t) 150 = (0.0253 A) cos((720 rad s )t). b) X L = ω L = (720 rad s ) (0.250 H) = 180 . c) v L = L di dt =− ( L) (0.0253 A) sin((720 rad s )t) (4.55 V) sin((720 rad s
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
31.10: a) X C = 1 ω C = 1 (120 rad s )(4 .80 × 10 6 F) = 1736 . 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 from the capacitor’s voltage). D 90 i = v C X C = v cos( t) X C = (7.60 V) cos((120 rad s )t) 1736 = (4.38 × 10 3 A) cos((120 rad s )t) v R = iR = (4.38 × 10 3 A) (250 ) cos((120 rad s )t) = (1.10 V) cos((120 rad s )t).
Background image of page 10
31.11: a) If ω = 0 = 1 LC X = L 1 C X = L LC 1 CL C = 0. b) When > 0 X > c) When < 0 X < d) The graph of X against is on the following page.
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
31.12: a) Z = R 2 + ( ω L) 2 = (200 ) 2 + ((250 rad s ) (0.400 H)) 2 = 224 . b) I = V Z = 30.0 V 224 = 0.134 A c) V R = IR = (0.134 A) (200 ) = 26.8 V; V = I L L = (0.134 A) (250 rad s ) (0.400 H) V L = 13.4 V. d) φ = arctan v L v R ⎟ = arctan 13.4 V 26.8 V ⎟ = 26.6 ° , and the voltage leads the current. e)
Background image of page 12
31.13: a) Z = R 2 + (1 ω C) 2 = (200 ) 2 + 1 (250 rad s )(6.00 × 10 6 F) () 2 = 696 . b) I = V Z = 30.0 V 696 = 0.0431 A. c) V R = IR = (0.0431 A) (200 ) = 8.62 V; V C = I C = (0.0431 A) (250 rad s )(6 .00 × 10 6 = 28.7 V. d) φ = arctan V C V R ⎟ = arctan 28.7 V 8.62 V ⎟ =− 73.3 ° , and the voltage lags the current. e)
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
31.14: a) Z = ( ω L 1 C ) = (250 rad s ) (0.400 H) 1 (250 ad/s) (6.00 × 10 6 F) = 567 . b) I = V Z = 30.0 V 567 = 0.0529 A. c) V L = I L = (0.0529) (250 rad s ) (0.400 H) = 5.29 V V C = I C = (0.0529 A) (250 rad s ) (6.00 × 10 6 F) = 35.3 V. d) φ = arctan V L V C V R = arctan ( −∞ ) =− 90.0 ° , and the voltage lags the current. e)
Background image of page 14
31.15: a) b) The different voltages are: v = (30.0 V) cos(250t + 26.6 ° ), v R = (26.8 V) cos(250t), v L = (13.4 V) cos(250t + 90 ° ) At t = 20 ms:v = 20.5 V, v R = 7.60 V, v L = 12.85 V. Note v R + v L = v. c) At t = 40 ms: v =− 15.2 V, v R = − 22.49 V, v L = 7.29 V. Note v R + v L = v. Be careful with radians vs. degrees in above expressions!
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
31.16: a) b) The different voltages are: . Note . V 5 . 27 , V 45 . 2 , V 1 . 25 : ms 20 At ) 90 250 cos( ) V 7 . 28 ( ), 250 cos( ) V 62 . 8 ( ), 3 . 73 250 cos( ) V 0 . 30 ( v v v v v v t t v t v t v C R C R C R = + = = = = ° = = ° = c) At t = 40 ms: v =− 22.9 V, v R = − 7.23 V, v C = − 15.6 V. Note v R + v C = v. Careful
Background image of page 16
Image of page 17
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/01/2008 for the course PHYSICS 240 taught by Professor Davewinn during the Fall '08 term at University of Michigan.

Page1 / 71

Chapter_31_Solutions1 - 31.1 a Vrms = = 31.8 V 2 2 b Since...

This preview shows document pages 1 - 17. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online