ch06_ism - 6 Inductance Capacitance and Mutual Inductance...

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6 Inductance, Capacitance, and Mutual Inductance Assessment Problems AP 6.1 [a] i g =8 e 300 t 8 e 1200 t A v = L di g dt = 9 . 6 e 300 t +38 . 4 e 1200 t V ,t > 0 + v (0 + )= 9 . 6+38 . 4=28 . 8 V [b] v =0 when 38 . 4 e 1200 t =9 . 6 e 300 t or t = (ln 4) / 900=1 . 54 ms [c] p = vi = 384 e 1500 t 76 . 8 e 600 t 307 . 2 e 2400 t W [d] dp dt when e 1800 t 12 . 5 e 900 t +16=0 Let x = e 900 t and solve the quadratic x 2 12 . 5 x x =1 . 45 = ln 1 . 45 900 = 411 . 05 µ s x =11 . 05 = ln 11 . 05 900 =2 . 67 ms p is maximum at t = 411 . 05 µ s [e] p max = 384 e 1 . 5(0 . 41105) 76 . 8 e 0 . 6(0 . 41105) 307 . 2 e 2 . 4(0 . 41105) =32 . 72 W [f] i max [ e 0 . 3(1 . 54) e 1 . 2(1 . 54) ]=3 . 78 A w max =(1 / 2)(4 × 10 3 )(3 . 78) 2 =28 . 6 mJ [g] W is max when i is max, i is max when di/dt is zero. When di/dt , v , therefore t . 54 ms. 6–1
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6–2 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance AP 6.2 [a] i = C dv dt =24 × 10 6 d dt [ e 15 , 000 t sin 30 , 000 t ] =[0 . 72 cos 30 , 000 t 0 . 36 sin 30 , 000 t ] e 15 , 000 t A ,i (0 + )=0 . 72 A [b] i ± π 80 ms ² = 31 . 66 mA ,v ± π 80 ms ² =20 . 505 V , p = vi = 649 . 23 mW [c] w = ± 1 2 ² Cv 2 = 126 . 13 µ J AP 6.3 [a] v = ± 1 C ² Z t 0 idx + v (0 ) = 1 0 . 6 × 10 6 Z t 0 3 cos 50 , 000 xdx = 100 sin 50 , 000 t V [b] p ( t )= = [300 cos 50 , 000 t ] sin 50 , 000 t = 150 sin 100 , 000 t W ,p (max) = 150 W [c] w (max) = ± 1 2 ² 2 max =0 . 30(100) 2 = 3000 µ J =3 mJ AP 6.4 [a] L eq = 60(240) 300 =48 mH [b] i (0 + )=3+ 5= 2 A [c] i = 125 6 Z t 0 + ( 0 . 03 e 5 x ) dx 2=0 . 125 e 5 t 2 . 125 A [d] i 1 = 50 3 Z t 0 + ( 0 . 03 e 5 x ) dx +3=0 . 1 e 5 t +2 . 9 A i 2 = 25 6 Z t 0 + ( 0 . 03 e 5 x ) dx 5=0 . 025 e 5 t 5 . 025 A i 1 + i 2 = i AP 6.5 v 1 . 5 × 10 6 Z t 0 + 240 × 10 6 e 10 x dx 10 = 12 e 10 t V v 2 . 125 × 10 6 Z t 0 + 240 × 10 6 e 10 x dx 3 e 10 t 2 V v 1 ( )=2 V 2 ( 2 V W = ³ 1 2 (2)(4) + 1 2 (8)(4) ´ × 10 6 µ J
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Problems 6–3 AP 6.6 [a] Summing the voltages around mesh 1 yields 4 di 1 dt +8 d ( i 2 + i g ) dt + 20( i 1 i 2 )+5( i 1 + i g )=0 or 4 di 1 dt +25 i 1 di 2 dt 20 i 2 = 5 i g di g dt ! Summing the voltages around mesh 2 yields 16 d ( i 2 + i g ) dt di 1 dt + 20( i 2 i 1 ) + 780 i 2 =0 or 8 di 1 dt 20 i 1 +16 di 2 dt + 800 i 2 = 16 di g dt [b] From the solutions given in part (b) i 1 (0) = 0 . 4 11 . 6+12=0; i 2 (0) = 0 . 01 0 . 99+1=0 These values agree with zero initial energy in the circuit. At infnity, i 1 ( )= 0 . 4 A ; i 2 ( 0 . 01 A When t = the circuit reduces to · .. i 1 ( ± 7 . 8 20 + 7 . 8 780 ² = 0 . 4 A ; i 2 ( 7 . 8 780 = 0 . 01 A From the solutions ±or i 1 and i 2 we have di 1 dt =46 . 40 e 4 t 60 e 5 t di 2 dt =3 . 96 e 4 t 5 e 5 t Also, di g dt =7 . 84 e 4 t Thus 4 di 1 dt = 185 . 60 e 4 t 240 e 5 t
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6–4 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 25 i 1 = 10 290 e 4 t + 300 e 5 t 8 di 2 dt =31 . 68 e 4 t 40 e 5 t 20 i 2 = 0 . 20 19 . 80 e 4 t +20 e 5 t 5 i g =9 . 8 9 . 8 e 4 t 8 di g dt =62 . 72 e 4 t Test: 185 . 60 e 4 t 240 e 5 t 10 290 e 4 t + 300 e 5 t +31 . 68 e 4 t 40 e 5 t +0 .
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This note was uploaded on 03/26/2008 for the course ELEN 214 taught by Professor Zou during the Spring '08 term at Texas A&M.

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ch06_ism - 6 Inductance Capacitance and Mutual Inductance...

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