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Solution HW_2

# Solution HW_2 - Chapter 4 Snlutlun 4 Sim V'm Van V we apply...

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Unformatted text preview: Chapter 4. Snlutlun 4? Sim V'm - Van - V; we apply KCL al ﬁle node a and obTain 30—? V n =—“+2Vn —> rm=isn.-'ucs=1.19v 12 {so - To ﬁnd Ru. mmidn the circuit Inﬂow. llﬂ v, a At node a. KC'L gives V V = 3V, ——‘.——’ —; K = 6G.-'1'26 =U.4‘.|62 50 L2 V 15" Rn=—‘=04‘-'62_ i" =—“=1.19«'0.4762=2.5 1 RD. "[1111; VD =1_1EW. RI." = R‘. = 0.47630. i'K = 2 wt Chapter 4. Solution 43. To get Rn. con-tidy The circuit in Fig. (2a). 3 {>51 from Fig. (a). L. - 1. 5— 10 —R' - (I. or V - -—1 Tu gar V15. conﬁde-1' the circuir in Fig. [b]. In = 3. 1-11, = -1Ehl.,+41ﬂ = -1.‘1.-' 13' = VTb’RTh = .13. Chapter 4.. Solution 5-1 To ﬁnd ‘u'n. =‘..-". Emmi-521' rhe left loop. —3—1CIDU.'_, —.'V, =0 —; 3=L043m', —3}’_. (1) For the night loop. V, = —5 0.1 Info: —.2|3I|3I0.fo [:2 Combining (1) and f2]. 3=ICICICII"_I - 4-0001: = -3{I'Clﬂfp —} f = -11JJ.A 2:40-00:52 —> 331:: To ﬁnd Rm insert a 1-“: source at Temﬂnals a-b and remove the 3-? indepmdenl source. as shown below. lkﬂ V Ix = 401}, + " = —\$OmA+ iA = {A}!le 50 SD RT = — = —U0.060 = —16.5TQ n- . 1 {I Chapter 4, Solution 59. R3: - (ID -2I’}j Isa—40' RU'FJ'CI - 22.5 ohms To find V3,. cmsirlm' thc cutuit brlow. i| = i: = SEE = 4. lﬂil—Vm-ZOL: = O. or ‘3'“: = JDiE-Iﬂil = [011 = 10x4 V11, = wt. and I3: = vme = 4031.5 = lJI-‘TE .4. Chapter 4, Solution 71. We need R1}. and Y1” a1 [ermirmls a and b. To find Rm. we insert a l-mA solute in The tannin-ale. a and b as shown below. Assume That all resismnces are in 1: 0111115. all cmrems are in 1113. and all voltage-5 are in volts. Al node a. 1 = (Ky-101+ [{vg - llﬂvn'ylﬂ]. or 40 = 5",. - 4301'0 (1] The loop an the left side has 110 voltage source. Hence. m I]. From [1]. v, = B Y. El = \‘Jl ma = 8 kolnua Tu get Yr“. consider the misinal cimuit. Fer [ht left loop. xi: = {If-HS = 31' Forﬂleﬁghlluop. \-'R = YT”. = {4O-"EO}(A1201'0} = -192 The 1'e=;1' stance at the reqlﬂived reel-nor is R = Rm = S Izoth p =VTE.-'(4Rn} = {-192f.=(4x3x103} = 1.15211an Chapter 5, Solution 5. -\',-A¥d+(R.+Rg}I=0 [1) But vd= [11-1. -1'i—(P~a—Rn—RJL}1=0 ="'—i RU—(1+A}R, '.‘ I duh - RnI — v0 = {I R R-A v. 1'n=A‘-'d—RnI=(R0—R4A)l =—( °+ ' 3' Rg—{I+A]R, it RU+RiA _100+10‘x10’_m, Y. R:+(1+A}R__ 100—c1+10‘) g I 10 _ 00*: mam: 019999000 5 [1+ 10? J} 100.001 Chapter 5. Suluiion 3. [:2] Eva and n are the voltages at the inverting and noujm'ertmg tennimls ofthe op amp. v1 — v1: — CI 0— lu].-’L= 2‘" —% 10=-2\' [b3 lﬂkﬂ 2 kn 4' 1V . "a i To la] {b} Since r. = 1:1, = 1'..-'amc1ja = 0. :10 CLmu flows TlLr0ugl1 The 10 m mat-3101'. From H g. (“0]. -\'=-2-'r'n=0 —"' \‘n=vn-2=1-.1=-1 ' Chapter 5, Solution .0. [a] Let 1.}. and u. be respeclively Illa wllages 31 the Eli-emu; and nomuvenmg terllumls of the op amp v, = n. = 4? At the m‘L'eItiug Imminal. 4 — Va 1 1A: ' =2? u 2k —r Va _ {b'l II' + 4- Since 1;, = vb = 3‘6. -1'.,—1—1L-'°=U —+ x'p=t'1,- |=2\-' Chaer- 5, Solution 13. ﬂ 4. By mllagc divisiul‘:1 90 T = — 1‘ =03? ' loo ( " 50 T H = _‘.-= = _° 150 3 Buha=vb —- %=09 —r wo= .7". . . ._ Y... Yo _ . L =11-I:= + =D.2.-1nA1-0.0L8nm=28-EE '5 10k LSDk Chapter 5, Solution 19. We convert the uncut source and back to a voltage BOLII'CC. 2H4 = i 3 (2 I3 11' lﬂk r! =— y @5313? " ' I I : 421 k ‘ ’ \ 3} -.-' r — 1 =—°’+ ° = —|}.3'.u'5mA ...
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Solution HW_2 - Chapter 4 Snlutlun 4 Sim V'm Van V we apply...

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