HW1 - P 1.12 Assume we are standing at box A looking toward...

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Unformatted text preview: P 1.12 Assume we are standing at box A looking toward box B. Then, using the P 1.15 P 1.17 passive sign convention p = m', since the current i is flowing into the + terminal of the voltage '0. Now we just substitute the values for v and 2' into the equation for power. Remember that if the power is positive, B is absorbing power,,so the power must be flowing from A to B. If the power is negative, B is generating power so the power must be flowing from B to A. = (120)(5) = 600 W p 600 W from A to B p z: (250)(—8) = ——2000 W p P 2000 W from B to A 2400 W from B to A 4800 W from A to B bl [c] [d] [a] [ = (—150)(16) = —2400 w = (—480)(—*~10) = 4800 w [a] In Car A, the current i is in the direction of the voltage drop across the 12 V battery(the current i flows into the + terminal of the battery of Car A). Therefore using the passive sign convention, p = 112' : (30)(12) = 360 W. Since the power is positive, the battery in Car A is absorbing power, so Car A must have the ” dead” battery. [b] we): film; 60 360 d3: 0 w = 360(60 - 0) = 360(60) 2 21,600 J = 21.6 kJ 1min=605 w(60) z [a] p = ,m: : 306—500t __ 306—1500t _ 406—1000t + 5Oe~2000t _ 106—3000t p(1 ms) 2 3.1 mW [b] “’05) = /:(?)Oe‘5°°ac ~— 306‘15003 __ 406*100051: + 50840003; _ 108—3000m)d$ : 21.67 —— 60e’5‘Jot + 203~1500t + 406~1000t__ 2562000t + awe—30%,] w(1 ms) 2 1.24m [C] wtoml = P119 [a] Os£t<4s: v=2.5t V; izluA; p=2.5t,u,W 4s<t_<_8s: v=10V; i=0A; p=0W Ss_<_t<16s: v==——2.5t+30 V; i=v1pA; p=2.5t—30[1,W 16$<t$2082 v2~10V; i=0A; 2083t<36sz 'Uzt—BO V; i==0.4pLA; p=0.4t~l2,uW 368<t§4632 v=6V; i=0A; 46s_<__t<50s: v=~1.5t+75 V; i=—-0.6/1,A; p=0.9t——45uW t>5031 0W *5 || 0W *3 H [b] Calculate the area. under the curve from zero up to the desired time: 10(4) 2 —,f;(4)(10) = 20m M12) = «0(4) ~ 5(4)(10) = 0 J w(36) z M12) + —;~(4)(10) —- %(10)(4) + %(6)(2.4) 2: 7.2 “J w(50) = w(36)~§(4)(3.6)==0J P 1.20 [a] p = vi = (0.05e“10°0t)(75 _ 75.240006 x (3.755100% — 3.755200”) W % = —375oe*1°0°‘t + 75005200“ = 0 so 2620001" = e-lm‘” 2 = 6100‘” so ln2 = 1000i thus p is maximum at t = 693.15 us pmax == P(693.15 Ms) 2: 937.5 mW 3-75 eaoomfi 3-75 e—2000t __1000 -—2000 [b] w = /000[3.75e’1000t — 3.755200%] dt = [ 00 0 l 3.75 3.75 P 1.25 [a] p =2 m' = 400 X 103t2e“800t + 700te‘800t + 0256-800t = «2—800t[400,0001t2 + 7002: + 0.25] = {ta-80‘” [800 x 1031: + 700] — 8006—800t[400,000t2 + mm + 025]} 2 [—i’),200,000t2 + 2400t + 5]1006—800t Therefore, 2—}; = 0 when 3,200,000t2 - 2400t — 5 = 0 so pmax occurs at t = 1.68 ms. pmax 7— [400,000(.00168)2+700(.00168)+0-25]e—800(.00168) = 666 mW (LP dt 1‘. [c] w = / palm 0t t t w = / 400,000w2e-8001 dm+ / meme—800$ (133+ / 0.255800% dm 0 0 0 «meme—8%m t = _512 X 106 [64x 104x2+1600x+2] + 0 7006—8003: t 6*80032 t ————— —8 — 1 .25 64 x 104 ( 00$ )0 + 0 *800 0 When 75 = 00 all the upper limits evaluate to zero, hence zewgze 700 +92_5.=2.97m.1. w 512x106 +64x104 800 P 1.28 P 1.29 Ph Pj H H H II H H wag = ~(36)(250 x 1045) = —-9 mW Ubib : (44)(—250 x 10-6) = ~11 mW w; = (28)(—250 x 10-6) : ——7 mW vdid = (—108)(100 x 10-6) 2 —10.8 mW me = (432M150 x 10*6) ~_- ~4.8 mW Wm = —(60)(—350 x 10-6) = 21 mW vgz'g : (—48)(—200 X 10—6) 2 9.6 mW vhih = (80)(—150 x 10-6) = ~12 111W —vj.z'j = —(80)(—300 x 1045) = 24 mW Therefore, ZPabs : 21 + 9.6 + 24 = 54.6 mW 23131294—11+7+10.8+4.8+12=54.6W 2133.103 = Zpdel Thus, the interconnection satisfies the power check Pa Pb Po Pd Pe Pf 10:; 1% Pi ll H H ll —(1.6)(0.080) = ~128 mW wvbib = ——(2.6) (0.060) = -156 1111117 me = (“4.2)(—0.050) : 210 mW wvdid : ~(1.2)(0.020) = —24 mW we 2 (1.8)(0.030) = 54 mW ——'vfz'f = ~—(——1.8)(—0.040) :2 ~72 InW vgig = (~3.6)(—0.030) = 108 mW vhih = (3.2)(~0.020) 2 ~64 mW mrujz‘j : ——(-—2.4)(0.030) = 72 mW —va_ia, : 2121,31 = 128 + 156 + 24 + 72 + 64 = 444 mW Zpabs = 210+ 54+ 108+72 = 444 mW Therefore, ZPdel = ZPabS = 444 mW Thus, the interconnection satisfies the power Check ...
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HW1 - P 1.12 Assume we are standing at box A looking toward...

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