hw01solution - P 1.17[a p = ’U'i = 306—500t ——...

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Unformatted text preview: P 1.17 [a] p = ’U'i = 306—500t —— 306—1500t - 4064000t + 506‘2000': —— 10e‘3000t p(1 ms) = 3.1 mW [b] w(t) : j:(308’5°0’” — 2’)()e"150°”c —— 403'1000“’+ 506400“ ~— 106”3°°0m)d$ : 21.67 * 6055‘” + 20e‘1500t + 40640”— 256—2000t + 3.338—3000tMJ w(1 ms) = 1.24MJ [C] wtotal 1: P122 [3] q area under 2' vs. t plot = [50x0 + (100) + 10%) + (8x6) + 20(6)] x 103 [10 + 40 + 16 + 48 + 91103 = 123,000 C [b] w = fpdtzfm’dt v = 0.2x 10‘3t+9 0$t_<_15ks 0 g t 3 4000s 2’ = 15 — 1.25 x 10*% p = 135 — 8.25 x 10-31: — 0.25 x 10—6712 4000 ml = A (135 — 8.25 x 10~3t - 0.25 x 10-612) dt = (540 — 66 — 5.3333)103 = 468.667 kJ 4000 S t _<_ 12,000 2' = 12 -— 0.5 x 10-311 p = 108 — 2.1 x 10-% - 0.1 x 10-612 12,000 ’LUg = [1000 (108 - 2.1 x 10~3t - 0.1 x 10-6t2)dt == (864 ~— 134.4 —~ 55.467)103 = 674.133 kJ 12,000 S t 3 15,000 2 H H 30 — 2 x 10-3t 270 — 12 x 10-3t — 0.4 x 104%2 15,000 / (270 — 12 x 10-315 — 0.4 x 10‘6t2) dt 12,000 (810 — 486 —- 219.6)103 = 104.4 kJ wl + 1122 + 11);; = 468.667 + 674.133 + 104.4 = 1247.2 kJ ll ’U’L 400 x 103t2e~8°°t + 7000240“ + 0.25580” = e-800t [400,0001:2 + 700:: + 0.25] = {e-800t[800 >< 103t + 700] — 8006’8°°t[400,000t2 + 700t + 025]} = [—3,200,000t2 + 24001: + 51100e—800t P 1.25 [a] p H @ dt Therefore :12 = 0 when 3,200,000t2 —- 24001; — 5 = 0 ’ dt so pmax occurs at t = 1.68 ms. [b] pmax = [400,000(.00168)2 + 700(.00168) + 0.251e-8°°<-°°168) = 666 mW t [c] w = 0 pdm t t w = 400,0002026 800$ dm + 700936 800$ dx + 0 0 256-80” d3: 0 0 400,000e-800m 4 2 t — _512X106 [64x10x +1600x+210+ 7006—8002: t 6—8009: t —-——— — — 1 .2 64x 101480037 )0“) 5—800 0 When t = 00 all the upper limits evaluate to zero, hence (400,000)(2) 700 0.25 _ 512 x 106 + 64 x 104 + 800 ‘ 2'97 mJ' w: P 2.1 [a] Yes, independent voltage sources can carry the 8 A current required by the connection; independent current source can support any voltage required by the connection, in this case 20 V, positive at the top, [b] 30 V source: absorbing 10 V source: delivering 8 A source: delivering [c] P30V = (30)(8) = 240 w (abs) PM,V .—_ —(10)(8)=~80W (del) PgA = —(20)(8)=—160W (del) ZPabs = 2Pdel 2 240 W [d] The interconnection is valid, but in this circuit the voltage drop across the 8 A current source is 40 V, positive at the top; 30 V source is absorbing, the 10 V source is absorbing, and the 8 A source is delivering Pgov = (30)(8) = 240 W (abs) P10V = (10)(8) z: 80 W (abs) PSA = ~(40)(8) = ——320 w (del) 2PM = 2PM : 320 W P 2.3 The interconnection is not valid. Note that both current sources in the right hand branch supply current through the 100 V source. If the interconnection was valid, these two current sources would supply the same current in the same direction, which they do not. ...
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hw01solution - P 1.17[a p = ’U'i = 306—500t ——...

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