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Unformatted text preview: {a} The 1 cf and 2 uf in series an the left
have a cap cft1it2jitt1+21= 2.13. The 2.13
Lit in parallel with the 3 uf is (sews =
3.6666 uf. lGcing to the right sicle cf the
circuiL the 1 UT and 1 cf in parallel have an
equivalent capacitance cf 1+1 = 2 cf. The
2 cf in series with the 2 UT on the right has
an equivalent capacitance cf [2}{2}i(2+EJ =
1 cf. The remaining 36666 at and 1 cf
are in parallel sc their capacitance is the
sum cr4.6666 uf. {b} The 2 cf and 1 uf in parallel have 3 uf
tctal. The 4 tit and 2 cf in parallel have 6
uf total. The 6 cf in series with the 3 cf
have [6}{3)t(6+3] = 2 uf which is the
answer. P332. Find the equivalent capacitance between termi
nals x and y for each at the circuits shcwn in Figure P322.
2 .uF 2 ,uF
ﬁg “ lm
I “F 3 [LP 1 FF ' ptF
).
til]
I
1 HF I RF
4 .ch' 2 pF thi
Figure Fizz {a} The 1 h and 3 h are in series sc the
tctal fcr them is 4 h. The 4 h is in parallel
with the 2 h sc its equivalent inductance is
[4}(2}f[4+2}=315 = 1.3333 h. The 1.3333
h is in series with the remaining 1 h
inductance so the tctal is their sum or
2.3333 h. {h} The 2 h and 2 h are in parallel sc their
equivalent is {2}{2}i{2+2}=1 h. The 2 h
and 5 h are in parallel sc their equivalent
is {2}{5}f{2+5}=1215 = 1.5 h. The 1.5 h is
in series with the 1 h sc that inductance is
2.5 h that is to the right cf the 4 h. The 4 h
and the 2.5 h are in parallel so their
equivalent inductance is
{4}{2.5]i[4+2.51=1ﬂiﬁ.5 = 1.54 h. 'P3.53. Determine the equivalent inductance fcr each
of the series and parallel ccmhiriaticns shcwn in Figure P353. 1H 1H 2H 3H The initial ‘v'c is {J volts, therefore A+B=U
in the expression Vo=A+Bexp{—HRC}.
Note that expiDFt, thus the E! term is
times a 1. The finach = 100 volts fort
—r00. therefore A = tﬂﬂ. Note that exert03) = U thus the
E! term is times D. Therefore
a+s=o=1oc+s and Ei=—‘l DD.
RC=ituUe3]i.ﬂ1e—E}=.ﬂﬂ1 sec. The RC
time oonstant is .Dﬂ1 see. Therefore
‘v'c=1ﬂﬂ1 UGexpitiﬂﬂﬂ volts. The
graph is me i] t
u i] .ﬂﬂl sec t—rE‘O P43. Consider Lhe circuit shown in Figure P43. As
sume that the initial voltage across the capacitor
is LIL{UH = ﬂ. Find an expression for the volt
age across the capacitor as a function of time,
and sketch to scale versus time. The initial ‘v'o is 50 volts and A+B=5U.
The ﬁnal ‘v'o=1i2itl. A=1tltl and B=15i.‘i.
Vc=1DD15Dexpi1ﬂﬂﬂt) volts. ‘IDD 50 t
i] .ﬂﬂi see i ’00 “’44. Repeat Problem 4.3 if the initial voltage is
Ufiﬂi—h = —Sﬂ' The switch is closed so the current
source is shorted out before t=il This
means that the 1voltage on the capacitor
is initially.»r D volts. Therefore A+B=D.
After the switch has been open for a
long time, all the 1 me. current is going
through the 1D k ohm resistor and w:
00}={1e—3}(1ﬂe3}=1ﬂ volts. A=1Ei end
Et=—1ﬂ. RC={1De3}{1eE}=.D1 seconds.
tililt}ﬂ=1ﬂ‘i Dexpi—1 Dﬂt} volts. t] t
t] .01 sec t—ri‘ﬂ FIJI. Consider the circuit shown in Figure P43. Thez
switch opens at r = I]. Ftnd an expression for
um and sketch to scale versus time. 1 this The switch is clcsed fer a leng time and
l3=U because the 1ﬂﬂuf lccks like an
cpen circuit to DC. The 1 h is a short
circuit to DC. This leaves twc 'l k chm
resistcrs in parallel. Each has iﬂﬂ vclts
acrcss it sc the current in each is
1DGI1DUU=J amps and this means that
i2=.l and i4=.1 amps. i1=i2+i3+i4. i1=.2
amps. Vc:1U'U vclts. Figure I’d.14 III11:1. CensidertheeireuitshcwninFigure Pri. L4. Find
the steadystate values cf E]. i2. i3. 1'4, and u:
after the SWIICII has been closed for a leng time. The switch is cpen fer a lcng time so
‘v'cttsiJFED volts and A+Ei=5il After the
switch is clcsed at t=ﬂ, the steaelvr state
vcltage en the capacitcr is the value of
the vcltage divider cr Vciteeﬂl=25 vclts.
A=251 and 3:25.
RC=[{1eE}i1eEllileE+1eE}][2e—E}=
[.ﬁeEHEe—El = 1 seccnd time constant.
v§325+25expi4i volts. The graph is: 25 Egg? ELIE. Ccnsicier the circuit cf Figure PILLS in which
the switch has bBﬂl‘l cpen fer a long time timer
to r = ﬂ. Determine the 1values cf uctr] hefcre
r = {l and s lcng time aﬁer i' = I]. Also, deter
mine the time canstant after the switch closes and expressions for rich) . Sketch u: [r] to scale
versus time for —2 5 r 5 5. thin 
'5 r
SCI "J 2 pF tittil
t hm _ Figure Pillﬂ The switch is cpen far a lcng time so
iitsﬂi=20ii1ﬂ+1ﬂi=1 amp and A+B=1.
After the switch is clcsecl at t=D. the
steady state current is the current
thrcugh the lap 11'.) chm resistor and the
ccil. HOD): 2W1 D = 2 amps. A=2, and
=—1. The two 1U chm resistcrs lack like
they are in parallel tc the L so that the
time ccnstant UR = 1i5 = .2 seccntls.
iltJEZ—lexpl—tﬂi amps. IGraph is: 2 ‘i {l .2 .4 .B .3 11] P438. Re far to the circuit of Figu re P423. The current
through 1th: indUch is zem bet"an r = E]. De
termine cxprassimns for and sketch m} to scale
‘lr'ﬂl'ELIS Lime for —I:].2 5 r 5 1.0 .1 I09 lﬂ‘r‘ lﬂﬂ Figure ELIE ...
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This note was uploaded on 09/01/2010 for the course EE 331 taught by Professor Preston during the Spring '06 term at University of Texas.
 Spring '06
 Preston

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