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# ch08_ism - 8 Natural and Step Responses of RLC Circuits...

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8 Natural and Step Responses of RLC Circuits Assessment Problems AP 8.1 [a] 1 (2 RC ) 2 = 1 LC , therefore C = 500 nF [b] α = 5000 = 1 2 RC , therefore C =1 µ F s 1 , 2 = 5000 ± s 25 × 10 6 (10 3 )(10 6 ) 20 =( 5000 ± j 5000) rad/s [c] 1 LC =20 , 000 , therefore C = 125 nF s 1 , 2 = ± 40 ± q (40) 2 20 2 ² 10 3 , s 1 = 5 . 36 krad/s ,s 2 = 74 . 64 krad/s AP 8.2 i L = 1 50 × 10 3 Z t 0 [ 14 e 5000 x +26 e 20 , 000 x ] dx +30 × 10 3 = 20 ( 14 e 5000 x 5000 ³ ³ ³ ³ t 0 + 26 e 20 , 000 t 20 , 000 ³ ³ ³ ³ t 0 ) × 10 3 = 56 × 10 3 ( e 5000 t 1) 26 × 10 3 ( e 20 , 000 t 1)+30 × 10 3 = [56 e 5000 t 56 26 e 20 , 000 t +26+30] mA = 56 e 5000 t 26 e 20 , 000 t mA ,t 0 AP 8.3 From the given values of R, L, and C, s 1 = 10 krad/s and s 2 = 40 krad/s. [a] v (0 )= v (0 + )=0 , therefore i R (0 + 8–1

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8–2 CHAPTER 8. Natural and Step Responses of RLC Circuits [b] i C (0 + )= ( i L (0 + )+ i R (0 + )) = ( 4+0)=4 A [c] C dv c (0 + ) dt = i c (0 + )=4 , therefore dv c (0 + ) dt = 4 C =4 × 10 8 V / s [d] v =[ A 1 e 10 , 000 t + A 2 e 40 , 000 t ] V ,t 0 + v (0 + A 1 + A 2 , dv (0 + ) dt = 10 , 000 A 1 40 , 000 A 2 Therefore A 1 + A 2 =0 , A 1 4 A 2 =40 , 000; A 1 , 000 / 3V [e] A 2 = 40 , 000 / [f] v = [40 , 000 / 3][ e 10 , 000 t e 40 , 000 t ] V 0 AP 8.4 [a] 1 2 RC = 8000 , therefore R =62 . 5Ω [b] i R (0 + 10 V 62 . = 160 mA i C (0 + ( i L (0 + i R (0 + )) = 80 160 = 240 mA = C dv (0 + ) dt Therefore dv (0 + ) dt = 240 m C = 240 kV / s [c] B 1 = v (0 + )=10 V , dv c (0 + ) dt = ω d B 2 αB 1 Therefore 6000 B 2 8000 B 1 = 240 , 000 ,B 2 =( 80 / 3) V [d] i L = ( i R + i C ); i R = v/R ; i C = C dv dt v = e 8000 t [10 cos 6000 t 80 3 sin 6000 t ] V Therefore i R = e 8000 t [160 cos 6000 t 1280 3 sin 6000 t ] mA i C = e 8000 t [ 240 cos 6000 t + 460 3 sin 6000 t ] mA i L =10 e 8000 t [8 cos 6000 t + 82 3 sin 6000 t ] mA 0 AP 8.5 [a] ± 1 2 RC ² 2 = 1 LC = 10 6 4 , therefore 1 2 RC = 500 ,R = 100 Ω [b] 0 . 5 CV 2 0 =12 . 5 × 10 3 , therefore V 0 =50 V [c] 0 . 5 LI 2 0 . 5 × 10 3 ,I 0 = 250 mA
Problems 8–3 [d] D 2 = v (0 + )=50 , dv (0 + ) dt = D 1 αD 2 i R (0 + )= 50 100 = 500 mA Therefore i C (0 + (500 + 250) = 750 mA Therefore dv (0 + ) dt = 750 × 10 3 C = 75 , 000 V / s Therefore D 1 αD 2 = 75 , 000; α = 1 2 RC = 500 ,D 1 = 50 , 000 V/s [e] v = [50 e 500 t 50 , 000 te 500 t ] V i R = v R =[0 . 5 e 500 t 500 te 500 t ] A ,t 0 + AP 8.6 [a] i R (0 + V 0 R = 40 500 =0 . 08 A [b] i C (0 + I i R (0 + ) i L (0 + 1 0 . 08 0 . 5= 1 . 58 A [c] di L (0 + ) dt = V o L = 40 0 . 64 =62 . 5 A / s [d] α = 1 2 RC = 1000; 1 LC =1 , 562 , 500; s 1 , 2 = 1000 ± j 750 rad/s [e] i L = i f + B 0 1 e αt cos ω d t + B 0 2 e αt sin ω d t, i f = I = 1 A i L (0 + )=0 . i f + B 0 1 , therefore B 0 1 . 5 A di L (0 + ) dt . αB 0 1 + ω d B 0 2 , therefore B 0 2 = (25 / 12) A Therefore i L ( t 1+ e 1000 t [1 . 5 cos 750 t + (25 / 12) sin 750 t ] A 0 [f] v ( t Ł di L dt =40 e 1000 t [cos 750 t (154 / 3) sin 750 t ] Vt 0 AP 8.7 [a] i (0 + , since there is no source connected to L for t< 0 . [b] v c (0 + v C (0 15 k 15 k +9 k ! (80) = 50 V [c] 50+80 i (0 + )+ L di (0 + ) dt = 100 , di (0 + ) dt =10 , 000 A / s [d] α = 8000; 1 LC = 100 × 10 6 ; s 1 , 2 = 8000 ± j 6000 rad/s [e] i = i f + e αt [ B 0 1 cos ω d t + B 0 2 sin ω d t ]; i f ,i (0 + Therefore B 0 1 ; di (0 + ) dt , 000 = αB 0 1 + ω d B 0 2 Therefore B 0 2 . 67 A ; i . 67 e 8000 t sin 6000 t A 0

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8–4 CHAPTER 8. Natural and Step Responses of RLC Circuits AP 8.8 v c ( t )= v f + e αt [ B 0 1 cos ω d t + B 0 2 sin ω d t ] ,v f = 100 V v c (0 + )=50 V ; dv c (0 + ) dt =0 ; therefore 50 = 100 + B 0 1 B 0 1 = 50 V ;0 = αB 0 1 + ω d B 0 2 Therefore B 0 2 = α ω d B 0 1 = ± 8000 6000 ² ( 50) = 66 . 67 V Therefore v c ( t ) = 100
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ch08_ism - 8 Natural and Step Responses of RLC Circuits...

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