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# Ch02p - Chapter 2 Exercise Problems EX2.1 iD peak = 30 12...

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Chapter 2 Exercise Problems EX2.1 ( ) ( ) 30 12 0.6 87 0.2 max 30 12 42 D R i peak mA v V = = = + = 1 1 12.6 sin 24.8 30 t = ω ° ° 2 By symmetry 180 24.8 155.2 t = = ω 155.2 24.8 % time 100% 36.2% 360 = × = EX2.2 (a) 1 12sin 1.4 0 O v = = θ or 1 1.4 sin 0.1166 12 θ = = which yields 1 6.7 θ = ° By symmetry, 2 180 6.7 173.3 θ = = ° Then 173.3 6.7 % time 100% 46.3% 360 = × = (b) 1 1 4 sin 0 35 4 . . θ = = which yields 1 20 5 . θ = ° By symmetry, 2 180 20 5 159 5 . . θ = = ° Then 159.5 20.5 % time 100% 38.6% 360 = × = EX2.3 ( ) ( ) ( ) 3 24 2 2 60 10 0.4 M r V C fRV = = or 500 C F μ = EX2.4 M M r r V V V R f RC f CV = = or ( ) ( ) ( ) 6 75 60 50 10 4 R = × Then 6.25 R k = Ω EX2.5 10 14 , 5.6 , PS Z V V V V = 20 100 L R Ω ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 5.6 max 0.28 , 20 5.6 min 0.056 100 max max min min (max) min 0.9 0.1 max min 0.9 0.1 max L L PS Z L PS Z L Z PS Z PS PS Z PS I A I A V V I V V I I V V V V V V = = = = = or ( ) ( )( ) ( )( ) ( )( ) ( )( ) 14 5.6 280 10 5.6 56 max 10 0.9 5.6 0.1 14 L I =

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