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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$t_<_15ks
0 g t 3 4000s
2’ = 15 — 1.25 x 10*%
p = 135 — 8.25 x 1031: — 0.25 x 10—6712 4000 ml = A (135 — 8.25 x 10~3t  0.25 x 10612) dt = (540 — 66 — 5.3333)103 = 468.667 kJ
4000 S t _<_ 12,000 2' = 12 — 0.5 x 10311
p = 108 — 2.1 x 10%  0.1 x 10612
12,000
’LUg = [1000 (108  2.1 x 10~3t  0.1 x 106t2)dt == (864 ~— 134.4 —~ 55.467)103 = 674.133 kJ 12,000 S t 3 15,000 2 H H 30 — 2 x 103t
270 — 12 x 103t — 0.4 x 104%2 15,000
/ (270 — 12 x 10315 — 0.4 x 10‘6t2) dt
12,000 (810 — 486 — 219.6)103 = 104.4 kJ
wl + 1122 + 11);; = 468.667 + 674.133 + 104.4 = 1247.2 kJ ll ’U’L 400 x 103t2e~8°°t + 7000240“ + 0.25580” = e800t [400,0001:2 + 700:: + 0.25] = {e800t[800 >< 103t + 700] — 8006’8°°t[400,000t2 + 700t + 025]}
= [—3,200,000t2 + 24001: + 51100e—800t P 1.25 [a] p H @
dt Therefore :12 = 0 when 3,200,000t2 — 24001; — 5 = 0 ’ dt
so pmax occurs at t = 1.68 ms.
[b] pmax = [400,000(.00168)2 + 700(.00168) + 0.251e8°°<°°168)
= 666 mW
t
[c] w = 0 pdm
t t
w = 400,0002026 800$ dm + 700936 800$ dx + 0 0 25680” d3:
0 0
400,000e800m 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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 Summer '06
 Milenkovic
 PGA, Voltage source, Current Source, Pgov

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