Question

chapter 7 Fundamental of Electrical Circuits

1. In the circuit below find: a) v(0+) and i(0+) b) dv(0*) and di(0) c) v(@) and icon 332 dt 0.25 H 4001 - ) V ( 0.1F 40(1) A

Answer 1

3.0 0.251 + © HUL-H) Iv 352 4Utd A 0.15 @ veot) and icot). - ? 40 UE t) = 40it 20. 740Utt) - o t o. . ay . t 40 ult) - 4 t 30 20 to 14 0(t) bet Cate! 1 for tco.40 UE+) = 40 and uLt)=. → Under Gteady Gtate Conditions Enductor acts as Chort Circuit for de Gopply and capacitor acts as open Circuit for de Guppler. na die (110) = 40 = 5 Amps., V260)-OX T 40. Vcco) 52 VCCOY - 5*5 = 25V, iccom)=0A -> Inductor opposers the Godden change in Currente 160) iulot) Carpací tor opposers the Godden change in voltage Vco) = recot) i v clot) = 250, icot) = 5 Amps. . dvcor, dllot) = ? 40 u t-t) -0 and 4 uct) = 441 dt dt voltage at nodes arie cale: 2 at t = of tu m _140+) = 5 © Apply kol at noden 7 - Lot) +ic cot)+I2-4 =0 35 QUA > -incot) + ic Cot) + vot) -4 =0 1-5+ (ccot) + 35-450 a eclot) = 544-% = 4 Amp. - C drocot) c dreot) as der lot)) - tot 4 / 2 = 40 Ampli Gec. Camdt inner 3.2 + VLOD - ΥΙη 5 bus 2 riots – dt dt

Apply krzen Loop2 => 3 x Eliot) + VLC0+) + r20+) = 0 => 345+ VL40+) +95 +0. => Velot) = -25-15 - Mhlot) = – 40. Ldillot) - 1 dicot) idt di Lot) - VLLOT) at is L -40 0.25 -40x200 0 = 160 16 = -160 Amplsce © voos and 1000): - je !! cace 3 : At time. ..!!!. ule know that Inductor ac aG Ghort circuit and capacator. actes als open circuit. i n ylcool. =(4* ) Bih 33 lcoo) YI! Go l coo) =-4* VCO) = -2.5 amps. (4A 144..ico) -9.58 [because toj direction er opposite to reference direction Cersilarly v Coo) = 5xI ; I= 4-2.5 = 1:59. VIO) = 5x2.5 = 4-5 voltes Govinos = 7.50 and i Coo) 2-2.5 A CS Scanned with CamScanner