Circuits 2 Summary Modules, Summaries of Electronic Circuits Analysis

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ACERIEL 8. VILLANUEYA HECTRICAL CIRCUITS 2 f PSEE - 12) Ergr. H- Panat) CIRCUITS2 REVIEW QUESTIONS NO. 1 PART | For Figure A below, let w = 10 rad/s, answer questions nos. 1-3. For Figure B below, answer questions nos. 4-5. Jn: DAD G50) Lp We LOrad/s Linz 2ABEAROA A) 9509 SAL HEM gy “an J * jot}lon420-|m0 , | 4n | 7 Zou Bins 2.708452 108.9 ~ - Sosn sa a to Jin: 511450.99 Jn i ; 30a a ba L f Wor oo = 12 te B.os asy Figure A Figure B 1. The capacitive reactance is ~P4X ;, 2, The inductive reactance is j$Jl_, 3. The Zinis > 2.7084} 2-508 Ly 4. TheZiais = SIS G99 7 5. TheYnis = 16-0 -j3.65 ms, 4a 6Q es 5 HAG: Toe a faye EROEES, 4. g34) 109.9 Wee “hen 4-je FB HAN Lage HA [° | Fe Gear a ak ° 120 40° @) | Tre es ROBEY. y9 ss pag? TT | % Gaaytoin “ | CURRENT DIVIDER: Tse (teasfoaat a) (UNS Sarat A, For Figure above, answer question nos. 6-10. 4-4 yet The (ass [Arjen 6. Findthe current hh. I= s.92¢,/90° ay (1 38/oe0'4 n) (SAK) - 17s FRA, 7. In terminal a and b, what is the impedance? 4.841092, 8. Findthe current. a= t9.1g5 £37-09° Ay ° 9. Determine the total impedance. 2%, = %.¢37j1070. KN 10. Compute the total current of the circuit. J; = 19-35 /o-49° Ay We 260 ne Tae (20 13480 E402) | & SES = 4 2 y 2mA 20454 ~ [102 5B jgcnia ' a wed pow 40. TSE ST H7A29 ms y 90 oy RIFF -WI4N For Figure above, answer question nos. 11-15. V7 Yon™ (0 LLLOUFAY ry) Varn > 4: hOto 2007 Ny