Ch02p - Chapter 2 Exercise Problems EX2.1 iD ( peak ) = 30...

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Chapter 2 Exercise Problems EX2.1 () 30 12 0.6 87 0.2 max 30 12 42 D R ip e a k m A vV −− == =+= 1 1 12.6 sin 24.8 30 t ⎛⎞ =⇒ ⎜⎟ ⎝⎠ ω ° ° 2 By symmetry 180 24.8 155.2 t =− = 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 which yields 1 6.7 By symmetry, 2 180 6.7 173.3 =−= ° Then 173.3 6.7 % time 100% 46.3% 360 = (b) 1 14 sin 0 35 4 . . which yields 1 20 5 . By symmetry, 2 180 205 1595 .. = ° Then 159.5 20.5 % time 100% 38.6% 360 = EX2.3 3 24 2 260 10 0 .4 M r V C fRV or 500 CF μ = EX2.4 MM r r VV VR f RC f CV = or 6 75 60 50 10 4 R = × Then 6.25 Rk EX2.5 10 14 , 5.6 , PS Z ≤≤ = 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 IA I I I 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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or () max 591.5 L Im = A ( ) ( ) Power(min) max 0.5915 5.6 ZZ IV =⋅ = So Power(min) = 3.31 W Now ()() max 14 5.6 max min 0.5915 0.056 PS Z i ZL VV R II == ++ or 13 i R Ω EX2.6 ( ) ,max 13.6 9 For 13.6 , 0.2383 15.3 4 9 4 0.2383 9.9532 PS Z L vV I A = + =+ = ,min 11 9 For 11 , 0.1036 15.3 4 9 4 0.1036 9.4144 PS Z L I A = + = 9.9532 9.4144 Source Reg 100% 100% 13.6 11 L PS v v Δ = × Δ− or Source Reg 20.7% = , 13.6 9 For 0, 0.2383 15.3 4 9 4 0.2383 9.9532 LZ L noload A = + = For 100 , L I mA = 13.6 9 4 0.10 15.3 Z Z I I −+ ⎡⎤ ⎣⎦
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This homework help was uploaded on 02/10/2008 for the course ECE 138 taught by Professor N/a during the Spring '08 term at CSU Fresno.

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Ch02p - Chapter 2 Exercise Problems EX2.1 iD ( peak ) = 30...

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