chap07

# chap07 - Chapter 7 Solution 1 Applying KVL to Fig 7.1 1 t i...

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Chapter 7, Solution 1. Applying KVL to Fig. 7.1. 0 Ri dt i C 1 t - = + Taking the derivative of each term, 0 dt di R C i = + or RC dt i di = Integrating, RC t - I ) t ( i ln 0 = RC t - 0 e I ) t ( i = RC t - 0 e RI ) t ( Ri ) t ( v = = or RC t - 0 e V ) t ( v = Chapter 7, Solution 2. C R th = τ where is the Thevenin equivalent at the capacitor terminals. th R = + = 60 12 80 || 120 R th = × × = τ -3 10 5 . 0 60 ms 30 Chapter 7, Solution 3. (a) ms 10 10 2 10 5 , 5 10 // 10 6 3 = = = = = x x x C R k R Th Th τ (b) 6s 3 . 0 20 , 20 8 ) 25 5 //( 20 = = = = + + = x C R R Th Th Chapter 7, Solution 4. eq eq C R = τ where 2 1 2 1 eq C C C C + = C , 2 1 2 1 eq R R R R R + = = τ ) C C )( R R ( C C R R 2 1 2 1 2 1 2 1 + +

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Chapter 7, Solution 5. τ = 4) - (t - e ) 4 ( v ) t ( v where 24 ) 4 ( v = , 2 ) 1 . 0 )( 20 ( RC = = = τ 2 4) - (t - e 24 ) t ( v = = = 2 6 - e 24 ) 10 ( v V 195 . 1 Chapter 7, Solution 6. V e 4 ) t ( v 25 2 10 x 2 x 10 x 40 RC , e v ) t ( v V 4 ) 24 ( 2 10 2 ) 0 ( v v t 5 . 12 3 6 / t o o τ = = = = τ = = + = = Chapter 7, Solution 7. τ = t - e ) 0 ( v ) t ( v, C R th = τ where is the Thevenin resistance across the capacitor. To determine , we insert a 1-V voltage source in place of the capacitor as shown below. th R th R 8 i 2 i i 1 10 0.5 V + v = 1 + + 1 V 1 . 0 10 1 i 1 = = , 16 1 8 5 . 0 1 i 2 = = 80 13 16 1 1 . 0 i i i 2 1 = + = + = 13 80 i 1 R th = = 13 8 1 . 0 13 80 C R th = × = = τ = ) t ( v V 20 8 13t - e
Chapter 7, Solution 8. (a) 4 1 RC = = τ dt dv C i - = = → = C e -4) )( 10 ( C e 0.2 - -4t -4t mF 5 = = C 4 1 R 50 (b) = = = τ 4 1 RC s 25 . 0 (c) = × = = ) 100 )( 10 5 ( 2 1 CV 2 1 ) 0 ( w 3 - 2 0 C mJ 250 (d) () τ = × = 0 2t - 2 0 2 0 R e 1 CV 2 1 CV 2 1 2 1 w 2 1 e e 1 5 . 0 0 0 8t - 8t - = = or 2 e 0 8t = = = ) 2 ( ln 8 1 t 0 ms 6 . 86 Chapter 7, Solution 9. τ = t - e ) 0 ( v ) t ( v, C R eq = τ = + + = + + = 8 2 4 2 3 || 6 8 || 8 2 R eq 2 ) 8 )( 25 . 0 ( C R eq = = = τ = ) t ( v V e 20 2 t -

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Chapter 7, Solution 10. 10 10 mF + v i 15 i o i T 4 A 2 15 ) 3 )( 10 ( i i 10 i 15 o o = = → = i.e. if i , then A 3 ) 0 ( = A 2 ) 0 ( i o = A 5 ) 0 ( i ) 0 ( i ) 0 ( i o T = + = V 50 20 30 ) 0 ( i 4 ) 0 ( i 10 ) 0 ( v T = + = + = across the capacitor terminals. = + = + = 10 6 4 15 || 10 4 R th 1 . 0 ) 10 10 )( 10 ( C R -3 th = × = = τ -10t t - e 50 e ) 0 ( v ) t ( v = = τ ) e 500 - )( 10 10 ( dt dv C i 10t - 3 - C × = = = C iA e 5 - -10t By applying the current division principle, = = + = C C i -0.6 ) i - ( 15 10 15 ) t ( i A e 3 -10t Chapter 7, Solution 11. Applying KCL to the RL circuit, 0 R v dt v L 1 = + Differentiating both sides, 0 v L R dt dv 0 dt dv R 1 L v = + = + L Rt - e A v =
If the initial current is , t h e n 0 I A R I ) 0 ( v 0 = = τ = t - 0 e R I v, R L = τ = t - dt ) t ( v L 1 i t - t - 0 e L R I - i τ τ = τ = t - 0 e R I - τ = t - 0 e I ) t ( i Chapter 7, Solution 12. When t < 0, the switch is closed and the inductor acts like a short circuit to dc. The 4 resistor is short-circuited so that the resulting circuit is as shown in Fig. (a). 3 i(0 - ) + 12 V 2 H 4 (a) (b) A 4 3 12 ) 0 ( i = = Since the current through an inductor cannot change abruptly, A 4 ) 0 ( i ) 0 ( i ) 0 ( i = = = + When t > 0, the voltage source is cut off and we have the RL circuit in Fig. (b). 5 . 0 4 2 R L = = = τ Hence, = = τ t - e ) 0 ( i ) t ( i A e 4 -2t

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Chapter 7, Solution 13. th R L = τ where is the Thevenin resistance at the terminals of the inductor. th R = + = + = 37 16 21 20 || 80 30 || 70 R th = × = τ 37 10 2 -3 s 08 . 81 µ Chapter 7, Solution 14 Converting the wye-subnetwork to delta gives 16 R 2 80mH R 1 R 3 30 = = = = = = + + = 170 10 1700 , 34 50 1700 , 85 20 / 1700 20 10 50 50 20 20 10 3 2 1 R R x x x R 30//170 = (30x170)/200 = 25.5 , 34//16=(34x16)/50 =10.88 s x R L x R Th Th m 14 . 3 476 . 25 10 80 , 476 . 25 38 . 121 38 . 36 85 ) 88 . 10 5 . 25 //( 85 3 = = = = = + = τ
Chapter 7, Solution 15 (a) s R L R Th Th 25 . 0 20 / 5 , 20 40 // 10 12 = = =

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chap07 - Chapter 7 Solution 1 Applying KVL to Fig 7.1 1 t i...

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