CH5 Answers - P5.1. Unitsofωareradians....

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Unformatted text preview: P5.1. Unitsofωareradians. Unitsoffarehertz. ω = 2πf P5.2. a DecreasingthepeakamplitudeV! ? 2. Compressthesinsoidalcurvevertically b Increasingthefrequencyf? 4. Compressthesinsoidalcurvehorzintally. c Increasingθ? 6. Translatesthesinsoidalcurvetotheleft d Decreasingtheangularfrequencyω? 3. Stretchesthesinsoidalcurvehorizontally. e Increasingtheperiod? 3. Strechesthesinsoidalcurvehorizontally. P5.4. v t = 10 cos 1000π − 60  π ω = 1000π, f = 500Hz, Angle =  − radians, T = 2ms, P = 1W, t !"#$ = 0.3333ms 3 P5.5. v t = 12 cos 400πt + 150  5π w = 400π, f = 200Hz, Angle = radians, T = 5ms, P = 1.44W, t !"#$ = 2.916ms 6 P5.6. v t = 28.28 cos 2π10! t − 72 V P5.9. P t = 5000 1 + cos 4000πt W, P!"# = 5000W P5.14. V!"# = 2V P5.17. V!"# = 16.58V P5.24. V! = 141.4∠ − 45, V! lagsV! by90, V! lagsV! by45, V! leadsV! by45. P5.25. V! t = 10 cos 400πt + 30 , V! t = 5 cos 400πt + 150 , V! t = 10 cos 400πt + 90 . V! lagsV! by120, V! lagsV! by60, V! leadsV! by60. P5.29. 24.51 cos ωt − 117.39 . P5.31. i! t = 14.14 cos ωt − 70 ! I! P5.33. V! = jωLI! , V! =  . jω C P5.34. Forpureresistance, currentandvoltageareinphase. Forpureinductance, currentlagsvoltageby90. For purecapictance, currentleadsvoltageby90. 1 1 P5.35. Z! = 200π∠90, V! = 10∠0, I! = ∠ − 90, i! t = sin 2000πt , I! lagsV! by90. 20π 20π P5.37. Z! = 15.92∠ − 90, V! = 10∠0, I! = 0.6283∠90, i! t = −0.6283 sin 2000πt . i! leadsv! by90. P5.42. I = 70.71∠ − 45mA, V! = 7.071∠ − 45V, V! = 7.071∠45V, IlagsV! by45. P5.44. I = 4.472∠63.43mA, V! = 4.472∠63.43V, V! = 8.944∠ − 26.57V, IleadsV! by63.43. P5.46. Z = 158.1∠ − 71.57, Z = 50∠0, Z = 158.1∠71.57. P5.49. I! = 10∠0mA, V = 10∠0V, I! = 10∠0mA, I! = 50∠ − 90mA, I! = 50∠90mA. P5.51. V! = 10∠45V, I = 0.2∠45A, V! = 100∠135V, V! = 10∠45V, V! = 100∠ − 45V. P5.53. I = 0.6202∠ − 82.88A, I! = 0.2774∠ − 146.31A, I! = 0.5547∠ − 56.31A. P5.54. V! = 100∠ − 90, V! = 100∠30, I = 0.7071∠ − 75A, V! = 70.71∠ − 75, V! = 70.71∠15V. IleadsV! by15. IlagsV! by90. P5.57. V! = 20∠126.89V. P5.58. V! = 10.84∠ − 167.47V. P5.59. PF = cos θ ∗ 100%. P5.60. Wattsforrealpower. Volt − amperesreactive VAR forreactivepower. Volt − ampers VA forapparentpower. P5.67. I = 15.11∠20.66, P = 10kW. Q = −3.770kVAR. Apparentpower = 10.68kVA. Powerfactor = 93.57%leading. P5.68. I = 16.01∠ − 27.95. P = 10kW. Q = 5.305kVAR.Apparentpower = 11.32kVA Powerfactor = 88.34%lagging. P5.71. Capacitiveload. S = 23.04 − j17.28kVA. P = 23.04kW. Q = −17.28kVAR. PF = 80%. apparentpower = 28.80kVA. P5.89. Z! = 7.0711∠ − 45. V! = 85.50∠63.07.I! = 12.09∠108.07. Z!"#$ = 5 + !5.I!"#$ = 8.55∠63.07A. P!"#$ = 182.75W. ForpureResistance:Z!"#$ = 7.0711. I!"#$ = 6.544∠85.57. P!"#$ = 151.4W. ...
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This note was uploaded on 12/03/2010 for the course EGN EGN3373 taught by Professor Ferekides during the Fall '10 term at University of South Florida - Tampa.

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