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HW9sols - P 723[a ten lEIQ KVL equation at the top node Ha...

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Unformatted text preview: P 723? [a] ten lEIQ KVL equation at the top node: Ha Ha Ha “-E'E'n Multiply h}: 40 and solve: EDGE] = {5 — l — Qua; u: = 200V min—)1 = 1—0 = 2mm = 20A Use voltage division to ﬁnd the Thevenin voltage: 4|} 1/ = a=—son =2mv T“ 1’ 40+120l } Remove the voltage source and make series and parallel combinations of resistors to ﬁnd the equivalent resistance: RTh =10 +120||40 :10 +30 = 4am The simpliﬁed circuit is: 400 40m?! L ammo-3 1 2—2—21 - —=1no0 T R 40 "13‘ T 200 _ E — 3‘0 : 1:0{DC) + [io[0+} — iotm]]€—EIT = 5 + (21} — mfg—1mm = 5 +15e'1ﬂm‘A, t 3 a dip b 021ml,— [l'v +dt 1m + 15.24%) + 0.04{—1nm)(153-wm*] 5E} +15De‘lﬂm‘ — EDGE—1m 50 — 45De‘1ﬂm‘V, t 2: 0+ 5A iofco] = E II P 150 [a] 1.3!:{1 3.21:9 151;} 351.: uét: *n.a1JF —35 ia{E}+} = m = —T.2mA 1:] mm} = a c] r = RC = {BUDDHELS x1043}: 41115 [d] to = 0+ blag—35‘“ = —T.25‘25mmﬁ.l t2 0+ [2] Mo = —[35 + mum—i2 >< 1045—25”}] = —35 + 12.955—EEMV, :3 0+ P T52 [a] unﬁl‘} = 11:,{04'} = IEDV in 12.5Hﬁ 37-5141 5., = —155 + {125— {—155}}5-m 5.. = —155 + 2255-5th. 5:: 5 [5] 5., = 42.54 >< 15-ﬁ{—555}[2755—m] = 5454"”515. :2 5+ [5] 55, = 5.,— 12.5 x 1035., = —155 + 252.55-mv [d] 545+} = —155 + 252.5 = 52.52 Checks: ug{5+} = 5.,{5+}[27.5 X 103] — 155 = 252.5 — 155 = 52.52 - ”5r 4th = — = —3 4.55 515 2555 50k + 5 . U _ £15m“ = ﬂ = —l+1355 5mm 41: + £555. + £15115 + 4 = [1' {13k} P 153 For t a: 0.. mm} = aw t I} U: + 30 xllﬁIL "Th lnnv —l . _ 3- _ = a — hTh—SUXIUIﬁ 0.3{100} 30x10 (100x103 )—an=5nv UT = 30 x 10% — 15 >4: 10% = 30 x 103{u.a}iT — 15 x 10% = 40 X 10% RTh=i=4Ukn IT tin-U nu = 50 — {an — Ems—ﬂ" 1 r = RC = {an X IUEHE 3-: 104} = 200 X 10—5; — = 5000 '1" L": = 50 — SUE—WDMV1 t 3} E] F mg [a] D5 t 5 2.51m: 112,110+} = SUV; 1:01:90} = E} L T=E=2mﬁg lfT=EDD nag} = ans—5"” V: 0+ 5 t 5 25—11115 M25— nus} = sag—”5 = 22.9219“ {30 — 22.92} 20 ::.,{2.5+ mas} = 40112.35} = —5T.[}SV i425- 1115} = = 2.3m 1:111:06} = E}; '1" = 21115; 1;“? = 500 no = —5?.uag—5"”“—“-m5}v t2 2.5+ms [h] M WI Eéﬁnasaa t [ms] [c] 1:01:511115} = 45.3w . _ +1535 0_ =am33 A ‘ 20 m F TE] [51] DE 1: E 111125: 1140+} = E}; trcllioc} = 50V; RC = 400 >< 103(0le X 10-5} = 4m; mm: = 250 u: = 50 — Ema-25‘“ ya = ED— 50 + Eng—25‘“ = Eng—MU, E} 5 t 5 111125 lmEt-ﬁoc: v.11 111\$} = 50— 505—” = 11.0519“ Hebe} = UV '1" = 411125; UT = 25K} 11.; = 11.055—35”'1*—”-m”v “a 2 _u1: = —ll.DEE_25D[t—D-ml}v1 :3 11115 EEI ﬂEI 3E] gzu 2 1:1 -1E| —2|] I (ma: F 7.34 t 3:- D: i Bﬂkﬂ ur =12 x 10H“ 15 >< lﬂair 2E} ' _ __' _ -31 m IUD IT LT UT = —24 :r: IDZiT+ 15 X lﬂair RTh=£=—Skﬂ 1T "r 2 RC = {—3 X lﬂa}{2.5 x134}: —D.02 If? 2 —5[} 11.: = 2055““. 2055” = EQDDD 50:: 21111000 1:: 1331511115 P TEE 4RD 111' = 200% + 4000(1‘1‘ — 2 >1 10—3111} = 51:}an — 3:1,, = 50an — 3mm} 3 = 40.1101} 1T -1U}EQ 25\$ lUH IE} T=T1DDD=—lﬂlhll lll'rT'=—1DDD i = 25511100111115 111200 11141411 —3_ . _ _ ~ 255 X10 —5.. t— 1000 —5.3111:: ...
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HW9sols - P 723[a ten lEIQ KVL equation at the top node Ha...

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