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chapter 07

# chapter 07 - Chapter 7 Energy Storage Elements Exercises Ex...

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Chapter 7 Energy Storage Elements Exercises Ex. 7.3-1 ( ) ( ) 2 2 4 1 1 4 0 otherwise C s t d i t v t t dt < < = = < 8 < 8 e and ( ) ( ) 2 4 2 4 1 8 4 0 otherwis R s t t i t v t t t < < = = < < so ( ) ( ) 2 2 2 4 ( ) 7 4 8 0 otherwis C R t t i t i t i t t t < < = + = < < e Ex. 7.3-2 ( ) ( ) 0 0 0 1 1 ( ) ( ) 12 1 3 t t s s t v t i d v t i d C τ τ τ τ = + = < < 0 ( ) 3 4 12 12 12 for 0 4 t v t d t t τ = = < In particular, v (4) = 36 V. ( ) 4 ( ) 3 2 36 60 6 for 4 10 t v t d t t τ = + = < < In particular, v (10) = 0 V. 10 ( ) 3 0 0 0 for 10 t v t d t τ = + = Ex. 7.4-1 ( ) ( ) ( ) ( ) 2 2 4 1 2 10 100 1 J 2 2 0 0 100 V c c Cv v v + = = × = = = W 7-1

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Ex. 7.4-2 a) b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 4 4 0 0 4 4 0 0 First, 0 0 since 0 0 1 Next, 0 10 2 2 10 2 10 2 2 10 t t t t t vi dt v v t v i dt dt t C t t dt t = + = = = + = = × 2 = × = × W W W W ( ) 4 1s 2 10 J = 20 kJ = × W ( ) ( ) 2 4 8 100s 2 10 100 2 10 J = 200 MJ = × = × W Ex. 7.4-3 a) b) ( ) ( ) 5 5 5 0 0 We have (0 ) (0 ) 3 V 1 ( ) (0) 5 3 3 3 1 3 3 V, 0 1 t t t t t c c v v v t i t dt v e dt e e t C + = = = + = + = + = < < ( ) ( ) ( ) ( ) 5 5 5 ( ) 5 15 3 18 V, 0 1 t t t R c c v t v t v t i t v t e e e t = + = + = + = < < ( ) ( ) ( ) ( ) 2 0.2 2 5 10 0.8 ( ) 6.65 J 1 1 0.2 3 0.9 J 2 2 2.68 kJ t s t t c t s t t Cv t e e t = = = = = × = = W W W Ex. 7.5-1 7-2
Ex. 7.5-2 1 2 1 2 1 1 2 1 1 2 2 dv dv i i C v v i i dt dt C C C = = = = 2 KCL: 1 2 1 2 2 2 2 1 1 C C i i i i i C C = + = + = + 2 i C Ex. 7.5-3 (a) to (b) : 1 1 mF 1 1 1 9 + + 1 1 1 3 3 3 = , (b) to (c) : 1 10 = mF 9 9 1+ , (c) to (d) : 1 1 1 1 10 = + + = mF 10 2 2 19 9 eq eq C C 7-3

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Ex. 7.6-1 ( ) ( ) 2 2 4 1 1 4 0 otherwis L s t d v t i t t dt < < = = < 8 e < 8 e and ( ) ( ) 2 4 2 4 1 8 4 0 otherwis R s t t v t i t t t < < = = < < so ( ) ( ) 2 2 2 4 ( ) 7 4 8 0 otherwis L R t t v t v t v t t t < < = + = < < e Ex. 7.6-2 ( ) ( ) 0 0 0 1 1 ( ) ( ) 12 1 3 t t s s t i t v d i t v d L τ τ τ τ = + = < < 0 ( ) 3 4 12 12 12 for 0 4 t i t d t t τ = = < In particular, i (4) = 36 A. ( ) 4 ( ) 3 2 36 60 6 for 4 10 t i t d t t τ = + = < < In particular, i (10) = 0 A. 10 ( ) 3 0 0 0 for 10 t i t d t τ = + = Ex. 7.7-1 ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 1 4 1 V 4 1 4 4 1 W 1 1 1 4 2 J 2 2 4 t t t t t t di d v L t e t e dt dt P vi t e t e t t e Li t e t e = = = = = = = = = W t 7-4
Ex. 7.7-2 ( ) ( ) ( ) ( ) 0 0 0 0 2 0 1 1 0 1 and 2 2 1 2 2 1 0 2 0 2 t t t t t di di v t L i t v t t t dt dt t t t < < < < 1 1 2 < < = = = = < < < < > > ( ) ( ) ( ) ( ) 0 2 0 1 2 2 1 0 t t t p t v t i t t t t < < < = = < < > 0 2 2 ( ) ( ) ( ) ( ) ( ) 0 0 0 ( ) 0 for 0 0 for 0 0 t t t t p t dt i t t p t t t = + = < = < W W W = ( ) ( ) ( ) ( ) ( ) ( ) 2 0 2 1 0 1: 2 1 2 : 1 2 2 4 2 : 2 0 t t t t t dt t t t t dt t t t t < < = = < < = + = + > = = W W W W W 4 Ex. 7.8-1 7-5

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Ex. 7.8-2 Ex. 7.8-3 ( ) ( ) ( ) ( ) 0 0 1 1 0 2 2 0 1 0 2 0 1 2 1 1 , but 0 and 0 t t t t i v dt i t i v dt i t i t i t L L = + = + = = 0 0 0 0 0 0 1 2 1 1 2 1 2 1 1 1 2 1 2 1 1 1 + 1 1 1 1 1 + t t t t t t t t P t t t t P i i i v dt v dt v dt v dt L L L v dt i L L L i L L v dt L L L = + = + = = 1 L = = = + 7-6
Problems Section 7-3: Capacitors P7.3-1 ( ) ( ) ( ) t 0 1 0 v t v i d and q Cv C τ τ = + = In our case, the current is constant so ( ) t 0 i d τ τ .

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