Question

Problem 5 (20 points) No energy is stored in the 100 mH inductor or the 0.4 LF capacitor when the switch in the circuit shown in figure below is closed. 0.1 H -O 2800 0.4 F 50 V uc Fig. 5 a) Find the values of a and co b) What is the type of circuit response for t>0? c)What is the initial voltage across the capacitor at t=0- and at t=0+? d) Find an expression for the current through the inductor for 0. e) Find an expression for the voltage across the capacitor for 10.

Answer 1

Sole : Given Electrical ckt m t=0 o.lt 28012 sov ② е по-чру Initially the ckl is @ rest & No energy is stored in the 100 ml indector and ou juf capacitore The switch is closed @ 0=o. (@) @ tot ckt with laplace bransformation Mw (8) 0.18 280 Od4X106 25x105/8 5015 103) = 0.13 + 280+ 25 so 5018 Oxos 0.18+ 2803 + 28 X10% 0.15 + 2805 + 25X10 500 2 (8) = 50 Oo1(32+ 2800S +25x106) 8+ 28008 + 29 X 106 cm - 25 x106 28869 - 2800 [ & on = 1400 l Iwn = 5x103 rad/see. |

There force Resonance frequency Wo= L = wn= 5x103 rad/see. and Damping coefficient a= Gwn = 1400. (6 we have wn- 5000 & gwn - 1400 g = 1400 sooo = 0.28 & < 7 Therefore the cercent behaviour will be cenderdamped & the response produced is also conder damped response IT=21 wd (C) t where wd = damped frequency of oscillatio - en vlcea - 4800 rad/sel. beforce the switch was on there was no charge on the capacitor as stated by the question. Therefore beforce the switch was on the capacitare voltage was zero. We know that the voltage across capacitore doesn't chanze instantaneously but it changes when annars velat orolts there is an implese cerrent. so vc (6-) = 0 volt but velot) & ovolts

(a) We have I (8) = 500 82 + 2800S + 25 X106 1 (8) - 500 8 + 2.1400-8 +(1400) + ( 25X10%- (1400)") 500 (3 +400) 2304 xioa C+10??(480) 48 5 X 4800 ū8 * (+1400)? +(4800) Taking inverse laplace of 1 (8) &(t) = 5 e 4oot sin 4800t Al © Voltage across capacitors for to Vc = Ź / E (t) at = x 5x le lootsin 4800 t de o.ux106" 48 Té 1400€ sin 4800 t dt = 1 (usin} in tegration by parts) u= sin 4800 t vel" = usv at du fv det at = - Sin 4800 t x je 1400t at - (dsin usoot sin ustrot 1400t upco (casuacot se lo st) of Va e 1400€ -luoot Tat dt 1400

140ot. = Sin 4800t oot e + 4800 le 1900t cas usoot dt. 1400 14oo - I 2 = /e 14oot cas usoot at 1400t I cos 4800 t e Siousuunto Side 1400 4800 lasin usodt el 400t det 1400 2. - cos 4800 e u . U800 2 - P2 = 1900 Tubo 1400 - Sin 4800t - 14000 I= cos usoote - e tu 4800 146 (-1 1400 1400 1400 Cas usoot enoot - 4800 IL 1460+ 1400t e + - | sin usoot - = 12.75512= 48 14coxlu - cos usaste 4oot7 1400 L sin usuot e 140ot t 48 2) P -1 casusoo 12.7551 I luco 1400xly casusoot e luoot c - 14oot w2CH) = 181950 x inaasi xwen an veront cosyboot]o wet VE - 14.5834 [ Sin yapot + y cos uport Té i Ve(t) @ tot Ve Coty a - 14.5834 x 48 = -50 volts