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solution Sadiku - 5th- ch01, Resumos de Eletromagnetismo

solution Sadiku - 5th- ch01(eletromagnetismo)

Tipologia: Resumos

2021

Compartilhado em 29/11/2021

paulo-teixeira-90
paulo-teixeira-90 🇧🇷

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Baixe solution Sadiku - 5th- ch01 e outras Resumos em PDF para Eletromagnetismo, somente na Docsity! Chapter 1, Solution 1 (a) q= 6.482x107 x [-1.602x10" C]=-103.84mC (b) q= 1. 24x10!8x [-1.602x107ºC]=-198.65 mC (e) q = 2.46x10 x [-1.602x10"C]=-3.941 € (d) q= 1.628x10 x [-1.602x 107º C] =-26.08 C Chapter 1, Solution 2 (a) 1i=dq/dt=3 mA (b) i=dydt=(16t+4)A (c) i= dy/dt = (38! + 1082) nA (d) i=dq/dt = 12007 cos120xt pA (e) i=dq/dt = —e*' (80cos 50t + 1000 sin 50t) LA Chapter 1, Solution 5 44 = lidt=|-tdt= q h Chapter 1, Solution 6 (o att= ms, i= 10-20 0454 dt 2 . . dq Att-=6ms, i=—= 04 0) ar” LA . dg -30 (c) Att= 10ms, i=1=2 = 754 dt 4 — Chapter 1, Solution 7 4 25A, — O<t<2 ia -25A, I<t<6 25A, — 6<t<8 which is sketched below: Al 235 A tim sec T > ser Chapter 1, Solution 10 q=it= 10x10'x15x10º= 150 mC Chapter 1, Solution 11 qrit=90x102x 12 x 60x 60=3.888 kC E=pt=ivt=qv=3888x1.5=5.832 kJ Chapter 1, Solution 12 For O<t< 6s, assuming q(0) =0, ' ' a(o = fiat +aço) = fatdt+o = 151º o o Att, q(9)=1.5(6) =54 For 6<t< 10s, : : a(t) = Jia +a(o) =[usdt+ 54 =18t-54 6 6 Att-I0, q(10) = 180 -54= 126 For 10<t<15s, : : att) = fia +a(o) = Í (CI)dt+ 126 =121 4246 10 J0 Att=I5, q(15) = -12x15 +246= 66 For 15<t<20s, ' a) = fodt+aqs) =66 15 Thus, 15? CG 0<t<6s 18t-54 C, 6 <t< 105 -121+246C, 10 <t< 15s 66C, 15 <t <20s att) = The plot of the charge is shown below. Chapter 1, Solution 14 (a) a= fiat= [0.02(1-e"*Jat=0.02(1+200*)] =0.02(1+20ºº -2)=4.261 mc O p(D=vDi p(1) = 10cos(2)x0.02(1-e**) = (-4.161)(0.007869) =-32.74mW Chapter 1, Solution 15 ! -0.006 4)” q= fidt= ['0.00602at =" e? (a) 2 o =-0.003(e* -1)= 2.945 mC 10di a a : i (b) v To =-0.012e“(10)=-0.12e” V this leads to p(t) = v(bi(t) = (-0.12e(0.006€ 7) =—720€* uW 3 (9) w= fpdt=-0.72 fe! at=220 aos] = 180 3 - o Chapter 1, Solution 16 (a) 30tmA, 0<t<2 (O) = 120-30t mA, 2 < t<4 5V, O<t<2 vb) = -5V,2<t<4 150tmW, O<t<2 p(b) = -600+150t mW, 2 < t<4 which is sketched below. pfmw) 300 1 2 4 t(s) -300 (b) From the graph of p, 4 W= [pdt=03 õ Chapter 1, Solution 19 1=8-2=6A Calculating the power absorbed by each element means we need to find vi for each element. P$ amp source =—8x9 =—72 W Pelement with 9 volts across it = 2x9 = 18 W Pelement with 3 bolts across it = 3x6 =18 W Pó volt source = 6x6 = 36 W One check we can use is that the sum of the power absorbed must equal zero which is what it does. Chapter 1, Solution 20 P30 vott source = 30x(-6) =—180 W P12 volt element = 12x6 = 72 W P2s volt e.ement with 2 amps flowing through it = 28x2 = 56 W P2s volt element with 1 amp flowing through it = 28x1 = 28 W Pie STo dependent source = 5X2x(-3) =—30 W Since the total power absorbed by all the elements in the circuit must equal zero, or 0 =-1804+72+56+28-30+P into the element with Vo OT Pinto the element with Vo = 180-72-56-28+30 = 54 W Since p into the element with Vo = VoX3 =54 Wor Vo =18 V. Chapter 1, Solution 21 p-i —s PO lgsA v 120 q= it =0.5x24x60x60 = 43.2 kC N, = qx6.24210º = 2.696x10? electrons Chapter 1, Solution 24 W=pt=60x24 Wh= 0.96 kWh = 1.44 kWh C=8.2 centsx0.96 — 11.808 cents Chapter 1, Solution 25 Cost=1.5 we Sore30 x8.2 cents/kWh = 21.52 cents Chapter 1, Solution 26 * 08Ah = (b)p=vi=6x0.08=0.48W (0) w=pt=0.48 x 10 Wh = 0.0048 kWh =80 mA Chapter 1, Solution 29 w=pt= op CO torto, +stou(Eo =24+0.9=33kWh Cost = 12 cents x3.3 = 39.6 cents Chapter 1, Solution 30 Monthly charge = $6 First 250 kWh (O) $80.02/XWh = 85 Remaining 2,436-250 kWh = 2,186 kWh (O $80.07/kWh= $153.02 Total = $164.02 Chapter 1, Solution 31 Total energy consumed = 365(120x4 + 60x8) W Cost =$0.12x365x960/1000 = $42.05 Chapter 1, Solution 34 (a) Energy = > =200x6+800x2 +200x 10 + 1200x4 +200x2 =10 kWh (b) Average power = 10,000/24 = 416.7 W Chapter 1, Solution 35 energy = (5x5 + 4x5 + 3x5 + 8x5 + 4x10)/60 = 2.333 MWhr Chapter 1, Solution 36 . 160A-h a) i= =4 (a) mw Sê d) t= 160Ah - 160,000h = 6,667 days OO01A 24h/day ———