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chapter 10

# chapter 10 - Chapter 10 Sinusoidal Steady-State Analysis...

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Chapter 10 – Sinusoidal Steady-State Analysis Exercises Ex. 10.3-1 (a) 2 / 2 / 4 T π ω π = = (b) leads by 30 ( 70) 100 v i − − = ° Ex. 10.3-2 ( ) ( ) ( ) 1 2 2 4 = 3cos 4 4sin 4 (3) (4) cos 4 tan 5 cos(4 53 ) 3 v t t t t t ° + = + = Ex. 10.3-3 ( ) 1 12 2 2 5 cos5 12 sin5 ( 5) (12) cos 5 180 tan 13 cos(5 112.6 ) 5 i t t t t t = − + = + + = ° Ex. 10.4-1 ( ) ( ) ( ) ( ) ( ) m KCL: = cos s v t v t d d I i t C v t v t t R dt dt RC C ω = + + ( ) Try cos sin & plug into above differential equation to get f v t A t B t ω ω = + ( ) m 1 sin cos cos sin cos I A t B t A t B t R C C t ω ω ω ω ω ω ω + + + = Equating sin & cos terms yields t t ω ω 2 m m 2 2 2 2 2 2 and 1 1 R I R A B C I R C R ω ω ω = = + + C Therefore ( ) 2 1 m m m 2 2 2 2 2 2 2 2 2 cos sin cos tan ( ) 1 1 1 f R I R I R C I v t t t t RC R C R C R C ω ω ω ω ω ω ω = + = + + + ω 10-1

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Ex. 10.4-2 10 10 0 10 KVL : 10 3 2 0 56.3 A 2 3 13 56.3 13 j j + + = = = = ∠ − + I I I D D D Therefore 10 ( ) cos(3 56.3 ) A 13 i t t = D Ex. 10.5-1 45 45 10 4.24 3 3 2.36 j j e j e = = Ex. 10.5-2 90 (90-111) 21 111 32 32 32 3.75 3 8 8.54 8.54 j j j j j e e e j e = = = − + Ex. 10.6-1 (a) 80 80 4cos( 80 ) Re{4 } 4 4 80 A j t j j i t e e e ω ω ° ° = = = = ∠ − I D ° ° ° ° (b) 20 20 10cos( 20 ) Re{10 } 10 10 20 j t j j i t e e e ω ω ° ° ° = + = = = I (c) 110 110 8sin ( 20 ) 8cos( 110 ) 8Re{ } 8 8 110 A j t j j i t t e e e ω ω ω ° ° = = = = = ∠ − I D D Ex. 10.6-2 (a) 140 140 10 140 10 V ( ) Re{10 } 10cos( 140 ) V j j j t e v t e e t ω ω ° ° = ∠ − ° = = = ° V (b) 43.2 80 75 109.7 43.2 109.7 j j e ° = + = ° = V 43.2 ( ) Re{109.7 } 109.7cos( 43.2 ) V j j t v t e e t ω ω ° = = + 10-2
Ex. 10.6-3 ( )( ) 0.01 10 cos100 0.01 100 10 10 7.071 45 1 7.071cos100 V d v v t dt j j v t + = + = = = ∠− ° + = V V V Ex. 10.6-4 { } 100 40cos100 Re 4 j t s v t = = e KVL: 3 3 ( ) 1 ( ) 10 10 ( ) 5 10 t S di t i t dt v dt −∞ i t + × + × = 100 Assume ( ) where is complex number to be determined. Plugging into the differential equation yields j t i t Ae A = 100 100 100 100 45 4 ( 2 ) 4 2 2 1 j t j t j t j t j Ae j Ae j A e e A e j ° + + − = = = In the time domain: { } { } 100 45 (100 -45 ) ( ) Re 2 2 Re 2 2 2 2 cos(100 45 ) A j t j j t i t e e e t ° ° = = = ° + t Ex. 10.7-1 (a) 10(5cos100 ) 50cos100 v Ri t = = = (b) [ ] 0.01 5( 100)sin100 5sin100 5cos(100 90 ) V di v L t t t dt ° = = = − = + (c) 3 1 10 5cos100 50sin100 50cos(100 90 ) V v i dt t dt t t C = = = = ° Ex. 10.7-2 6 10 10 [100( 500)sin (500 30 )] 0.5sin (500 30 ) 0.5sin (500 210 ) 0.5cos(500 120 ) A dv i C t dt t t t = = × + ° = − + ° = + ° = + ° 10-3

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Ex. 10.7-3 From Figure E10.7-3 we get m m m m m ( ) sin cos( 90 ) A 90 A ( ) cos 0 V i t I t I t I v t V t V ω ω ω = = ° = ∠ − = = ∠ ° I V ° The voltage leads the current by 90 ° so the element is an inductor: m m eq m m 0 90 90 V V I I ∠ ° = = = ° ∠− ° V Z I Also m m eq m m 90 V V j L L L L I I ω ω ω ω = = ° = = Z Ex. 10.8-1 Z R = 8 , Z C = 1 2.4 2.4 2.4 1 5 12 j j j j j j = = = − × , Z L1 = j 5 (2) = j 10 , Z L2 = j 5 (4) = j 20 and V S = 5 -90 ° V. Ex. 10.8-2 Z R = 8 , Z C = 1 4 4 4 1 3 12 j j j j j j = = = − × , Z L1 = j 3 (2) = j 6 , Z L2 = j 3 (4) = j 12 and I S = 4 15 ° V.
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