Assignment1_Solution

# Assignment1_Solution - P1.9 P1.10 P1.12 Q =Tf"}df 2...

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Unformatted text preview: P1.9* P1.10 P1.12* Q =Tf(?"}df 2 P1.14* P1.18 dqir) 0’ —=—(2r+r2)=2+2m dr 0’? {U} = The positive reference for vis aT The head of The arrow, which is Terminal :5. The posiTive reference for ngls Terminal :5. Thus, we have vb” = V = —10 'v'. Also, 1' is The currenT enTering Terminal 0, and {gm is The currenT leaving Terminal :2. Thus, we have i 2 ﬁg, 2 —3 A. The True polariTy is posiTive aT Terminal a, and The True currenT direcTion is enTering Terminal 0. Thus, currenT enTers The posiTive reference and energy is being delivered To The device. [26"0’3‘ = —Ze" IS“ = 2 coulombs o E: The charge flowing Through The baTTery is Q = (5 amperes)x (24 x 3600 seconds) = 432 ><103 coulombs The sTored energy is Energy = (91/ = (432 x103}x(12) = 5.184 ><10é joules (a) EquaTing graviTaTional poTenTial energy, which is mass Times heighT Times The acceleraTion due To graviTy, To The energy sTored in The baTTery and solving for The heighT, we have 6 h = Energy = 5.184x 10 =17'6 km mg 30x 9.8 (b) EquaTing kineTic energy To sTored energy and solving for velociTy, we have ,— v = 5w = 537.9 m/s N m (c) The energy densiTy of The baTTery is 6 w 2172.8x103 J/kg which is abouT 0.384% of The energy densiTy of gasoline. Q = currenT >< Time = (2 amperes) x (10 seconds) = 20 coulombs Energy = Ql/ = (20} x (5) = 100 joules NoTice ThaT fab is posiTive. If The currenT were carried by posiTive charge, iT would be enTering Terminal 0. Thus, elecTrons enTer Terminal b. The energy is delivered To The elemenT. P1.20* P123 P1.24* P125 *P1.27 P1.36* (a) P 2 war}, = 30 W Energy is being absorbed by The elemenT. (b) P = vag, = 30 W Energy is being absorbed by The elemenT. (c) P = Wiggly = —60 W Energy is being supplied by The elemenT. The amounT of energy is W = QV =(4 C] x (15 V) = 60 J. Because The reference polariTy for m, is posiTive aT Terminal sand The volTage value is negaTive, Terminal :5 is acTually The posiTive Terminal. Because The charge moves from The negaTI've Terminal To The posiTive Terminal, energy is removed from The device. Energy = COST = i = 500 kWh RaTe 0.12 \$/kWh PZEnérgyzwzwmw I=£=ﬂ=5387 A Tlme. 30><24h V 120 60 6944 x100 /o = 8.64 /o ReducTion = NoTice ThaT The references are opposiTe To The passive conﬁguraTion. pH} = —V(T}I'(T) = —30e" W Energy = lp(f)df = 303* lg = — 3O joules I) Because The energy is negaTive, The elemenT delivers The energy. (a) P = 50 W Taken from elemenT A. (b) P = 50 W Taken from elemenT A. (c) P = 50 W delivered To elemenT A. AT The node Joining elemenTs A and B, we have T; +1}, 2 0. Thus, in = —2 A. For The node aT The Top end of elemenT 6', we have {5 + I, = 3. Thus, 3‘, :1 A. 3+ 1;, = Q. Thus, £0, = 4 A. ElemenTs A and Bare in series. Finally, aT The Top righT-hand corner node, we have P1.37* We are given I, = 2 A, 3‘, = 3 A, fa. = —5A, and I}, = 4 A. Applying KCL, we find {,=Q—r;,=14 g=¢+¢=5A f =I,—;,,=—7A ;,=r,,+ra,=—3A g P1.41* Summing voltages for The lower' lefT-hond loop, we have — 5 + v, +10 2 0, which yields 1/, = —5 V. Then for The Top—mosT loop, we have V, — 15 — V, = O, which yields v, =10 V. Finolly, writing KCL around The ouTside loop, we have — 5 + v, + Vb = 0, which yields vb = —5 V. P1.42* Applying KCL ond KVL, we have :;=r;,—{,=1A :‘b=—{,=—2A v,=v,—v,=—6v vczvd=4V The power‘ for' each elemen’r is a2—Va},,=—2OW P,=v,,.«,,=12w Pc=vce=4w e=erg=4w Thus,€,+ﬁ;+&+iﬁ) =0 P1.45 We are given V, =15 V, V, = —7 V,v, = 10 V, and 11., = 4V. Applying KVL, we find v,=v,+v,=8v —V,,—V,—V,,=—29V vs: V =—V,,—V,+VG,=22V V§=V8—Vh=18v 6 ...
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## This note was uploaded on 05/17/2011 for the course GENE 123 taught by Professor Dabbagh during the Spring '08 term at Waterloo.

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Assignment1_Solution - P1.9 P1.10 P1.12 Q =Tf"}df 2...

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