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Unformatted text preview: 31.1: a) V. 8 . 31 2 V . 45 2 rms = = = V V b) Since the voltage is sinusoidal, the average is zero. 31.2: a) A. 97 . 2 ) A 10 . 2 ( 2 2 rms = = = I I b) A. 89 . 1 ) A 97 . 2 ( 2 2 rav = = = π π I I c) The rootmeansquare voltage is always greater than the rectified average, because squaring the current before averaging, then squarerooting to get the rootmeansquare value will always give a larger value than just averaging. 31.3: a) A. 120 . ) H 00 . 5 ( ) s rad 100 ( V . 60 = = = ⇒ = = ωL V I L Iω IX V L b) A. 0120 . ) H 00 . 5 ( ) s rad 1000 ( V . 60 = = = ωL V I c) A. 00120 . ) H 00 . 5 ( ) s rad 000 , 10 ( V . 60 = = = ωL V I 31.4: a) A. 0132 . ) F 10 20 . 2 ( ) s rad 100 ( ) V . 60 ( 6 = × = = ⇒ = = C Vω I ωC I IX V C b) A. 132 . ) F 10 20 . 2 ( ) s rad 10000 ( ) V . 60 ( 6 = × = = C Vω I c) A. 32 . 1 ) F 10 20 . 2 ( ) s rad 000 , 10 ( ) V . 60 ( 6 = × = = C Vω I d) 31.5: a) . 1508 ) H 00 . 3 ( ) Hz 80 ( 2 2 Ω = = = = π πfL ωL X L b) H. 239 . ) Hz 80 ( 2 120 2 2 = Ω = = ⇒ = = π πf X L πfL ωL X L L c) . 497 ) F 10 . 4 ( ) Hz 80 ( 2 1 2 1 1 6 Ω = × = = = π π ϖ fC C X C d) F. 10 66 . 1 ) 120 ( ) Hz 80 ( 2 1 2 1 2 1 5 × = Ω = = ⇒ = π π π C C fX C fC X 31.6: a) . 1700 Hz, 600 If . 170 H) Hz)(0.450 60 ( 2 2 Ω = = Ω = = = = L L X f π πfL ωL X b) = = Ω = × = = = C C X f πfC ωC X , Hz 600 If . 1061 ) F 10 50 . 2 ( ) Hz 60 ( 2 1 2 1 1 6 π . 1 . 106 Ω c) rad/s, 943 ) Hz 10 50 . 2 ( ) H 450 . ( 1 1 1 6 = × = = ⇒ = ⇒ = LC ω ωL ωC X X L C Hz. 150 so = f 31.7: F. 10 32 . 1 ) V 170 ( ) Hz 60 ( 2 A) 850 . ( 5 × = = = ⇒ = π ωV I C ωC I V C C 31.8: Hz. 10 63 . 1 ) H 10 50 . 4 ( ) A 10 60 . 2 ( 2 ) V . 12 ( 2 6 4 3 × = × × = = ⇒ = π πIL V f L Iω V L L 31.9: a) ). ) s rad 720 (( cos ) A 0253 . ( 150 ) ) s rad 720 (( cos V) 80 . 3 ( t t R v i = Ω = = b) . 180 ) H 250 . ( ) s rad 720 ( Ω = = = ωL X L c) ). ) s rad 720 ( sin( ) V 55 . 4 ( ) ) s rad 720 (( sin A) 0253 . ( ) ( t t ωL dt di L v L = = = 31.10: a) . 1736 ) F 10 80 . 4 ( ) s rad 120 ( 1 1 6 Ω = × = = ωC X C 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 o 90 from the capacitor’s voltage). ). ) s rad 12 ( cos( V) 10 . 1 ( ) ) s rad 120 ( cos( ) 250 ( ) A 10 38 . 4 ( ) ) s rad cos((120 A) 10 38 . 4 ( 1736 ) ) s rad 120 cos(( ) V 60 . 7 ( ) ( cos 3 3 t t iR v t t X ωt v X v i R C C C = Ω × = = ⇒ × = Ω = = = 31.11: a) If . 1 1 1 = = ⇒ = ⇒ = = LC C LC L X ωC ωL X LC ω ω b) When . ⇒ X ω ω c) When . < ⇒ X ω ω d) The graph of X against ω is on the following page. 31.12: a) . 224 H)) 400 . ( rad/s) 250 (( ) 200 ( ) ( 2 2 2 2 Ω = + Ω = + = ωL R Z b) A 134 ....
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This homework help was uploaded on 02/26/2008 for the course PHYS 11 and 21 taught by Professor Hickman during the Spring '08 term at Lehigh University .
 Spring '08
 Hickman

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