This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 The rootmeansquare voltage is always greater than the c) 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) 980611c9254828d04f3da1430005f3c67c7bc2a2.doc Page 1 of 27 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 To find the voltage across the resistor we need to know b) the current, which can be found from the capacitor (remembering that it is out of phase by o 90 from the capacitor’s voltage). 980611c9254828d04f3da1430005f3c67c7bc2a2.doc Page 2 of 27 ). ) 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: If a) . 1 1 1 = = ⇒ = ⇒ = = LC C LC L X ωC ωL X LC ω ω When b) . ⇒ X ω ω When c) . < ⇒ X ω ω The graph of d) X against ω is on the following page....
View
Full
Document
This note was uploaded on 02/25/2009 for the course PHY 127 taught by Professor Mr.yang during the Spring '08 term at SUNY Stony Brook.
 Spring '08
 MR.YANG

Click to edit the document details