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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—31']e2r 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: [tiP]
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+9W12W 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
Torafpoiver 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.2iJU.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—422ﬂ
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 = 82 = 16
The power absorbed va2 is P2 =v2i2 2162 =32 W. ...
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 Spring '08
 JUANG

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