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1、Problem7>1ApplyingKVLtoFig.7.1•—fidt+Ri=0CJ-Takingthederivativeofeachterm,idi-+R—=0Cdt10/03/1999didtor——=-——iRCIntegrating,/、Kt)I【0丿In-tRCi(t)=I()e-t/RCv(t)=Ri(t)=RIoel/RCorv(t)=Voet/RCProblem7.2t=RlhCwhereR(histheTheveninequivalentatthecapacitorterminals.Rlh=1201180+12=600t=60x0.5x103=30msProble
2、m7.3t=ReqCeqwhereCeq(1)(3)1+33=4^FRcq=12114+5=8kGT=—X4AIO%(8xl03)=6msProblem7.4t=ReqCeqRcqR[R2R]+RoC.C0R1R2C1C2(RI+R2XG+C2)Problem7.5v(t)=v(4)e(l_4)^wherev(4)=24,t=RC=(20)(0.1)=2v(t)=24c«4"2v(10)=24e-6/2=1.195VProblem7.6v(t)=v(0)e't/T,t=RthCwhereRthistheTheveninresistanceacrossthecapacitor.Todet
3、ermineRth,weinserta1-Vvoltagesourceinplaceofthecapacitorasshownbelow.V5oIlfrOQV1-0.5I8=T6i=ii+i?1380Rih2_807=T3808T=RthC=-x0.1=-v(t)=20e43t/8VProblem7.7I(a)t=RC=—4・0.2e心二C(10)(・4)「i——>C=5mFR二丄=5()04C1(b)t=RC=—=0.25s4(c)wc(0)=
4、cV^=
5、(5xlO-3)(lOO)=250mJ(d)wR=
6、x
7、cV(?=
8、cV(?(l-e-21^)0.5=1-e'8l<)>e'8l°=
9、ore8l()=21t0=—In(2)=86.6ms8Problem7>8v(t)=v(0)et/T,t=RcaCRXI=2+8118+6113=2+4+2=8Qi=RcqC=(0.25)(8)=2v(t)=20e-t/2VProblem7.915i0=10ii.e.ifi(0)=3A,thenio(0)=2AiT(0)=i(0)+io(0)=5Av(0)=10i(0)+4iT(0)=30+20=50Vacrossthecapacitorterminals.Rlh=4+101115=4+6=1()。T=RthC=(10)(10xl0-3)=0」v(t)=v(0)e_t/T=50e'l0
10、tir=C—=(10xl0-3)(-500e10t)dtic二.5e-,otAByapplyingthecurrentdivisionprinciple,i(t)=1510+15(-ic)=-0.6icProblem7.10ApplyingKCLtotheRLcircuit,VVdt+R=°Differentiatingbothsides,v1dv—+=0LRdtv=Ae'Rt/LdvR——+——v=0dtLIftheinitialcurrentis1°,thenv(0)=I0R=Av=I0Rel/L1i=-I()Ret/Ti(t)=Ioe"Problem7>11Whent<0,the
11、switchisclosedandtheinductoractslikeashortcircuittode.The4Qresistorisshort-circuitedsothattheresultingcircuitisasshowninFig.(a).12V(b)12i(0-)=y=4ASincethecurrentthroughaninductorcannotchangeabruptly,i(0)=i(0')=i(0+)=4AWhent>0,thevoltagesourceiscutoffandwehavetheRLcircuitinFig.(b).L2R=Z0.5Hence,i(
12、t)=i(O)el/T=4e2tALProblem7.12t=RihwhereRthistheTheveninresistanceattheterminalsoftheinductor.Rlh=70II30+80II20=21+16=37Q2x237=81.08Problem7.13t=whereLcq=10+20=30mH,3OxlO3Rg=30ll6+10=15kQProblem7.14t=(“)Leq=LandReq=R2+R
