HW5answers - {a} The 1 cf and 2 uf in series an the left...

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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 exert-03) = 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=1flfl-1 UGexpi-tiflflfl 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-{U-H = 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=1DD-15Dexpi-1flflflt) volts. ‘IDD -50 t i] .flfli see i ’00 “’44. Repeat Problem 4.3 if the initial voltage is Ufifl-i—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}{1e-E}=.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. The-z 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 steady-state 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 b-Bfll‘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 tit-til 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.

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