Chap_7_Solut_Dorf

Chap_7_Solut_Dorf - Chapter 7 Energy Storage Elements...

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Chapter 7 Energy Storage Elements Exercises Ex. 7.2-1 () 22 4 11 4 0o t h e r w i s e Cs t d it vt t dt << == < 8 < 8 e and () () 242 4 18 4 t h e r w i s Rs tt t t = =− < < so 222 4 7 4 8 t h e r w i s CR i t t t −< < =+= < < e Ex. 7.3-1 2 2 4 1 2 1 0 1 0 0 1 J 2 2 0 0 100 V cc Cv vv +− × = W Ex. 7.3-2 a) b) () ( ) 0 44 00 0 F i r s t , 0 s i n c e 0 1 Next, 0 10 2 2 10 210 2 2 10 t t tv i d t v v t v i dt dt t C d t t =+ = = × 2 = × ∫∫ WW W W 4 1s 2 10 J = 20 kJ W ( ) 2 48 100s 2 10 100 2 10 J = 200 MJ W
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Ex. 7.4-1 Ex. 7.5-1 () 22 4 11 4 0o t h e r w i s e Ls t d vt it t dt << == < 8 < 8 e and () () 242 4 18 4 t h e r w i s Rs tt t t = =− < < so 2 4 7 4 8 t h e r w i s LR v t t t −< < =+= < e
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Ex. 7.7-1 Ex. 7.7-2
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Problems Section 7-2: Capacitors P7.2-1 () ( ) () t 0 1 0 v t v i d and q Cv C ττ =+ = In our case, the current is constant so t 0 id τ . 66 3 0 150 10 15 10 5 0 3 ms 25 10 Cv t Cv i t qC v t i −− ×− × == = × P7.2-2 ( ) ( 11 12cos 2 30 12 2 sin 2 30 3cos 2 120 A 88 dd it C vt t t t dt dt °° + = + = + ) ° P7.2-3 ( ) 3 3 10 cos 500 45 12cos 500 45 12 500 sin 500 45 6000 cos 500 45 d tC t C t dt Ct −° ° ° ×+ = = ° so 3 6 3 310 1 1 10 F 610 2 2 C μ × × = ×
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P7.2-4 () () () 3 12 00 11 01 21 0 tt vt i d v i d C ττ =+ = × ∫∫ 0 9 02 1 0 t <<× 33 12 0 1 0 0 10 10 0 t s it v t d τ =⇒ = = × 99 0 31 0 t −− ×< < × () ( 6 63 3 12 2ns 41 0 A 1 0 1 0 51 0 0 0 s t d t ⇒= × = × + × × ) 6 In particular, ( ) ( ) ( ) 93 6 9 3 10 5 10 2 10 3 10 10 v 3 ×= × + × 0 0 t < × 6 3 12 3ns 1 2 10 10 4 10 10 0 s t d t =− × × + = × × ) 6 In particular, ( ) ( ) ( ) 6 9 0 0 1 0 51 0 1 0 V v × 3 9 0 t 12 5ns 1 0 0 10 10 V 0 t s v t d = = × P7.2-5 (b) 2 42 3 03 t t d C t dt t 1 < < < < == −< < < (a) () () () 1 0 i d v i d C = V For 0 < t < 1, i ( t ) = 0 A so 0 0 t d = For 1 < t < 2, i ( t ) = (4 t 4) A so () ( ) 22 1 44 4 = 2 V 1 t t d t t =− + = + 2 (2) 2 2 4 2 2 2 V v + = . For 2 < t < 3, i ( t ) = ( 4 t + 12) A so 2 2 2 2 1 2 + 2 = 2 1 4 V 1 t t d t t + + = −+ −+− ( ) 2 (3) 2 3 12 3 14 4 V v + = For 3 < t , i ( t ) = 0 A so 0 04 4 t d = V
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P7.2-6 (a) 00 2 () 0 .1 2 6 06 t d it C vt t dt t < < == < < < (b) () () () 1 02 tt i d v i d C τ ττ =+ = ∫∫ For 0 < t < 2, i ( t ) = 0 A so 0 20 V t d = For 2 < t < 6, i ( t ) = 0.2 t 0.4 V so () ( ) 22 1 2 0.2 0.4 0 0.2 0.8 =0.2 0.8 0.8 V 2 t t d t t =− + = + ( ) 2 (6) 0.2 6 0.8 6 0.8 3.2 V v + = . For 6 < t , i ( t ) = 0.8 A so 6 2 0.8 3.2 1.6 6.4 V t d t = P7.2-7 43 6 6 0 66 0 1 5 2 . 5 1 0 6 1 0 25 150 1 2 5 150 50 25 V 6 t t t v i d e d C ed ee −− = + × × ⎡⎤ = ⎢⎥ ⎣⎦ P7.2-8 23 2 3 2 1 1 2 10 25 1 2 A 200 10 40 6 10 10 2 10 200 A 200 25 50 25 150 A R C RC t v ie e dv iC e e dt ii i e e e μ × = × ×− = = +−
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P7.2-9 () () () 0 2 3 23 34 2 1 1 for 0 2 2 2 1 4 2 1 4 7 for 2 3 2 2 2 2 1 4 4 3 1 4 17 for 3 4 2 2 2 2 1 1 for t t t vt it d t t dt t t t dt dt t t t dt dt =+ =≤ = + = + = + = + + = ∫∫ 4 t 2 3 In summary 10 47 2 41 73 4 14 t tt t ≤≤ = +≤ P7.2-10 () ( ) ss 00 1 0 4 1 0 v t v itd t t C = + For . 2 5 t s 8 for 0 0.25 it t t 2 2 0 0 08 48 0 44 0 2 t t d t τ ττ ⎛⎞ =− + ⎜⎟ ⎝⎠ For example 11 0 4, 3.375, 1.5 84 vv v =− For 0.25 0.5 t 0.25 1.5 10 2 1.5 20 0.25 20 6.5 t d
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Chap_7_Solut_Dorf - Chapter 7 Energy Storage Elements...

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