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# Electric Circuit Analysis ECE 280

Mason

GPA 3.94

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This 11 page Class Notes was uploaded by Antonina Wuckert on Monday September 28, 2015. The Class Notes belongs to ECE 280 at George Mason University taught by Staff in Fall. Since its upload, it has received 8 views. For similar materials see /class/215038/ece-280-george-mason-university in ELECTRICAL AND COMPUTER ENGINEERING at George Mason University.

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Date Created: 09/28/15

Pro l m 76gt 7 a The s domain form of the circuit is shown in Fig a 2s1s 2s2 1 Zia2ls1S2s1s 512s1 3 I 1 g s 3 2 5 2 lt2ls USP 1 a b b The sdomain equivalent circuit is shown in Fig 13 212s 25 2 2quot HZsquot 32s 352 55 6 12quot 12sm is Z Snss6s39 332 s5s6 352 S 3sz7s6 3s2 Problem f6 2 We apply nodal analysis to the s domain form of the circuit below 8 V 39 A 1539 10s1 F g 4 D 10 mVu i1 39 g 8 s 4 1o255szv 3915w 5 51 v AM Akgssg a w HUGS2 02551 31 52039253H 4o I s1v ul 7 15525 Ms2 255l13s2 s Cs1 Equa ngooe icients 52 0AB gt BA s 15025ABC075AC 5 2SAC A407 B407 0135 52 5332 55 Lg VBL7 2739340140 wyj 2 31 31 a3 7s1 It 9 53 s 432 8 w 9 w 5 v0 5714 e 5714 a coso882t2o2 e e sin0882 t v Problem Co We rst nd the initial conditions from the circuit in Fig a 1 Q 4 Q W a 5 V v n v40 i 0 5 A v40 o v v We now incorporate these conditions in the SLdeain cimuit as shown in Figb I 4 Va V W V Io b L 158 g 28 Sls 7 41s At node 0 Vo155V 5 vouo 1 25 s 44 45 15 5 1 s s 1254s1v 194524s252s2 Ssz s2 s 4ss1 quot 4ssl 40sl Ssz652 1 ng 25 s ssz12504 s Lg L 5 s s l2504 4s1 A52 12504Bs Cs Equatng coef cients s 404A gt A10 s 4l2AC gt C12A48 s OAB gt BAo10 5 10 lOs8 1 s2 12s04 15 10s06 1002 quot s quotso62o22 s062022 iot 15 10e cos02t sin02t ut A Problem 16 6 We rst need to nd the initial conditions For t lt O the circuit is shown in Fig a To do the capacitor acts like an open circuit and the inductor acts like a short circuit Hence 1 A v9 1V vc0 21 25 v We now incorporate the initial conditions for t gt 0 as shown in Fig b 2 Vo W J VW 1 lls 258 5s 2 gt I 12 Lg4 Formesh 1 5 I 1 25 V0 h Problem 1 011 25 12 3quot le quotquotquot T39 1 For mesh 2 1 1 1 1 39 quotquot 5 2 s1111 2 S 0 l 1 l 25 51 2sIZTl 2 Put 1 and 2 in matrix form 1111533 2 quot quot39quot 39 2 quot A quot s22522s3 quot39 s22s2s3 222 13 A252 23Bsz 25Cs2 Equating ooe icients s 22AB s 02AZBC 5 133A2C Solving these equations leads to A 07143 B 3429 C 5429 I a 07143 3429s 5429 07143 1714552714 52 252253 39 s2 szs15 07143 17145s05 3194Jl25 quot s2 s052125s052125 I i t 0714361 17145 c cos125t 31 sin125tj ut A Problem 16 l 9 We incorporate the initial conditions in the sdomain circuit as shownbelow v r u x I i 1 2 I 2 s r a huuu quot39 Wv I v 4s 2 15 7 2 11 s Jl At the supemode 4s2 V 2 Y 1sV s 3 2 2 11 1 5222S VlssVo b 1 v But vov 21 and 1151 2v1 VoZs sVoZ V quotV39 5 quot s2s39s2 2 Substituting 2 into 1 2 1 28II s 2 s22s s s2 Vquotus2 SV 2 1 22s1 281 522 sss2quot 52 3 V 25295259s 4s1V ss2 32 39 52 V 259 A B quots24s1 so2679 s3732 A 2443 B 04434 2443 04434 quotso2679 s3732 Therefore v t 2443 emsquot 04434e 3 m ut V Problem 42 2 For t gt 0 the circuit in the s domain is shown below I 9s zoos2 16 V 215 Applying KVL 2 2 6s lSO 452 32 quot39 s2 65 9s2 16 3 3632288 s ss323216 A B C DsE 8 36s2 288 As 653 2552 965144Bs 353 1632 483 Cs3 l6sDs 633 9sg1Es3 612 95 Equating coef cients s 288 2144A 1 s39 0 96A 483 16C9E 2 2 36ZSA16B9D6E 3 s3 06A3BC6DE 4 s 0AB D 5 Solving equations 1 2 3 4 and 5 gives A 2 B 47984 C 816 D 02016 E 2765 W3quot L 17984 816 020165 069124 5 53 33 2416 3216 vt 4111 17984 e 339 816t e393 02016 cos4t 06912 sin4t V Problem v 2 9 Consider the op amp circuit below But So R2 wv 113C I ll R1 0 V V v V o O Vo R1 R2 0 V sC 1 v KIR sCV V 1 f SRICRIR2 V3equott gt V3s5 3 V s2s5 3 l3 quots2s5 s2 s5 B1 v06 65 e um RC20x10350x103961 x 3 332 1 Problem 6 3 Hs XCS Xs S Y i 1 2s 3X4 s 2s3 s2 16 522 16 8 23 123 2s3 s2 4s20 s2 4s20 Hs s Ys a 4 Problem I Q 35quot Consider the following circuit I 28 8 Ir V1 L F At node 1 v1 21Ias3 where I 25 m MQL 23 s 3 v1 35 35 V1 s3quot 2 quot2 1 35 3s V 83 2 1 2 V 35s3 1 3s2 9s2 9s 3 V V13s9s2v 53 39vo 9s H V m 35 932 Problem V C 3 71 3 25 VW V Vx 7R 2s I 4Vx For 100p 1 2 2 V 3 s 1 23912 1 For 1001 2 quotWR 23 3 211 0 5 5 But V I 1963 8 2 2 So Il Iz25lz11 0 6 6 O SII SM 23 2 In matrix form 1 and 2 become V3 2s 2s II 0 39 6s 6323 I2 As 2sJ JJ A2 2 2 Ax39Az Vaquot51 quot12 s A V 2SV6SquotZSquot6s4v x A A I2 6sVs 3 Problem 1396 4 0 a Hs R 111 V3 RsL quot sRL ht 14m ut b Vtut gt VS1S RL V RL 21 B uquot5RL smssRLs 5RL V Equot m vot ut e m39 ut 1 e39m ut Supp mu YltsgtHltsXs ht 26 ut a Hs 25 via 5ut gt Vis Xs 55 yt 2 101 equot ut

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