Question

3. The switch has been open a long time before closing at t = 0. Find the initial and final energy stored in the inductor. Determine i(t) and v(t) fort > 0*. t = 0 1092 to i(t) 2A @ 500 FT VIC 30.4 mH 2.503 14 4. The switch has been closed a long time before opening at t = 0. Find il(t) and vc(t) fort > 0*. 2012 t = 0 vc(t) 4092 4uF 60V 3 10 mH 201} 2.54 İL(t)

Answer 1

to-) Swienis Upon, then circaut become fuco sopould Blu MO? * Both cor connected with cwront source * If cutant source is connected inductor Short circulod Capacitor opon circulod * 120 givee initial curent si inital Voltage 204 pon +3502 bt 100v1 ivoty 11 102 m 321512 valoj (VCO) = 100 volts By canent division swe P160 = 1AX 215 12185 co) = 0.20 t20Swith is closed [t-> ] Capacitor open cirewlod ? t-Jo cwe inductor short circuted Y 2A 20 2A 102 -M > OR 2A ) 4 OV * Vccoo Ovhillo) $2.52 41 9 - OV IcCool=0v I Apply Simpli Sourco transformulon Circuit convert int 2 piso + ON + Yils Vclosnov 215 YIt's Apply Nodal at Ov7 Node 0-215 til 10)-2 =0 12.5

- ticos -2=0 -5+5 12600)-10=0 5iL2003=15 riccool=3A stored Final Energy incapeitor - b c VCCOO = O Joules inducto r = YLicoo) = < 0.4x16x9 = 118m Joule Il fime constant IC while finding time constant Capa curent source open 10-2 SDS {5 m Roq o leoa Roq = 12152 I = Yeoarece TL = 0.032 x 103 To = 324 soch Roq=50 1. TC= Roqc = 5x1 HF (Tc = 5 x H 50C] vect) = velouſvelo) - VCCO)] Yee velt) = 0+100 - 070-4/5x10-6 Volt) = 1000? -t/5x106 ielts = illos & ficcos-illo = 3+ (0.2- 37 e 1/32x106 Pict) = 3-2,80 32x106 to switch is closed Capacitor - Open inductor = short e te

202 DA + VCO) op 40 O og insy 0 tor- /20 im + 2150 YiLO) TOU 20 GOD 100) = 6% PL00) = 30 Doply KUL au 16p 1 + volo) +50=0 VCO) = -5ovo (18 a - switch is opon t 2150 VCCO) Lol + - OD 500200 & 2.5D 116001 V I Lool=OA Vco) = -50v.