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Homework1 - P1.3—2 a 45>< 1c—9...

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Unformatted text preview: P1.3—2 _ a 45 >< 1c—9 I:—q:—_3:9><10_5= 9145 a: 5 x 10 P1.3—3 1' = 10billion €13“th 1.602x10'” C = 10x109 deem” 1.6{32x'10'19 C s electron s electron = 10” x1 sumo—1" Elam} C I s electron c ==1502x104——= L602 Int 5 _______P1.5—fi p(2‘)=v(a‘)i(£‘) Here is a IVIATLAB program to plot pfi‘): (Ssn13f)(25n13f)==8(cosO-—co56f)==3-—3co56r VF clear t0=0; % initial time tf=2: % final time dt=0.02; % time increment t=t0:dt:tf; % time v=8*sin(3*t}; % device voltage i=2*ein(3*t}: % device current for k=1:length(t} pik}=v{k}*i{k}: % power end plotttrp} xlabelt‘timer 5‘}: ylabelt‘pcwer, W‘} PLS—T pp} =v{r)i(r}= fine-”F 23-” = 3(1—3-1']e-2r w Here is a LEA'ILAB program to plctpfl}: clear tfl=fl: % initial time tf=2: % final time dt=D.DE; % time increment t=tfl=dt:tf; % time v=4*[l—exp{—2*t]}; % device voltage i=2*exp[—2*t}; % device current for k=l:length[t} Pik}=V{k}*i{k}i % pcwer end p101: [ti-P] xlabeli'timeJr 5'}; ylabeli'pcwer, W'} Plj—fl P =FI=3>= [12:13.6 W wzanrz flfixfixfiflflflfl] P LT—l Notice that the element voltage and current of each branch adhere to the passive convention. The sum of the powers absorbed by each branch are: (-2 V)(2 A)+(5 WC! A)+(3 V)(3 A)+(4 V)(-5 A)+{1 V)(5 A) = —4 W + 10 W + 9 W -20 W + 5 W = {l W The element voltages and currents satisfy conservation of energy and may be correct. P 13—2 Notice that the element voltage and current of some branches do not adhere to the passive convention. The sum of the powers absorbed by each branch are: {3 V)(3 AJ+(3 WC! A)+ (3 V30 A)+(4 V)(3 A)+(-3 V)(-3 A)+(4 V)(-3 A) =-9W+6W+6W+12W+9W-12W iUW The element voltages and currents do not satisfy conservation of energy and cannot be correct. P 1.7—3 Let’s tabulate the power received by each element. We‘ll identify each element by its nodes. nodes Power received, W a b —(1)(4)=—4 b c (2)(—2)=—4 a 9' -(5)(?)=—35 3’ 0' (-6)(2)=-12 r: 0’ (—s)(—5)=4a So Torafpoi-ver received = —(9+4+4+35 +12)+4{} = —24 i 0 Changing the current reference direction for a particular element will change the total power by twice the power of the particular element. Since the element connected between nodes b and (1 receives -12 W, changing the reference direction of its current will increase the total power received by 24 W, as required. After making that change Totofpower received = —(9 + 4+ 4+ 35) +(12 + 40) = I] We conclude that it is the reference direction of the element connected between nodes and b that has been reversed. P224 Letr=1A,thenv=3r+ 5=3V.Next2r' =2A but 16=2vi3(2r)+5=11.. Henee, the property of homogeneity is not satisfied. The element is not linear. I’ll—5 (a) [1421+i:E :> v=3.2v 10 4D 3 r=i=nnss 40 s 0421+"; : v2+E—G.S=D (b) It] 2 5 . . —U.2il.3 Using the quadratie formula 1:27:03, —l.{lV _ 0.32 _ (—1)2 Whenv=fl.3Vtheni=T=O.32A.“U1env=-l.flVtheni=T=fl.5A. v v2 v (e) 0.4=—+0.3+— : v2+—+0.8={] 10 2 5 _ , —D.2i-JU.U4—3.2 Using the quadratic formula 1? = f So there is no real solution to the equation. PEA—5 v1: v2 2 v5 =150V; 312509322259 v1 and i1 adhere to the passive eonvention so 150 El :v—lz—zfl £1 50 _ . . _ v2 150 v1 and :2 do not adhere to the passive convention so :2 = —n— = —E = A The power absorbed by RI is P; 2 v1 171 = 150 - 3 = 450 W The power absorbed byR2 is P2 = —v2.r'2 = —15{}(—6) = M PEA—6 £12132 =1} 22A; R1=4 Q and R2 =3 9 v1 and 1'1 do n_ot adhere to the passive convention so vlz—Rl i12—4-22fl The power absorbed by}? 1 is P 12‘“ 13 1 =_(‘3){2) = fl v2 and]?2 do adhere to the passive eonvention so v2 = R2 i2 = 8-2 = 16 The power absorbed va2 is P2 =v2i2 216-2 =32 W. ...
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