Question

#2. (30p) Initially circuit was de energized + s (ma) 20122HR 10 a) Find v(1) for 0 <<<0.2 and 0.2 st so using sequential-switching method: b) Sketch y(t) using solution (a); c) Find () using the method of superposition of step responses; d) Sketch (1) using solution (c), showing details of how you build the graph; 0.2 (s)

Answer 1

Toat coas I = = = to = 0.1 =ool 3 loma I Lt = 0) = 10 mA I, Lts = I (o) [i-e 4e] = 10 mał - è e] = (10m) It-erot) oct_soud at t=002 Felt=vx) = (10mA) [l-e 1020-27 = 8.65 MA

loss current start - Decreesing due to ILIH = I (0.2) e Lt-0-2/2 resistive tsoa = (8-65mA) è lot ea tsoog

so Now V. ( = L di LoQ * E Tove Z VLMzm (10 me (i-ély] 10 mA 이아 Vilt = - 2 x 10mA ) X 10 eu Doa eyot v I v (0-2) = 0.axe = 0.027 V 0-2 < t < 0 Vilh = 2 d Tė e x 8-65 m 10 dt L =-2x ² x 8.65 m x 10 = -1.28 è lot v e lot v tutoret = =1:28 22 = -0.173 - 00027 -0.173 0:21

@ 39 3 1 [uw- uct-0-2)] mA (10 MA) ulty - (10 m 8) uct-0-3) using super positions Iolt t om-u- - Lomat too from 1st function Il !U = I110) [i- é 1077 - (OMA) [i-ě 10t] from Filo) Ti-a lotto toia and function In (H = Toton e ton = 410 mA) [ 1 - ēlot Lě 1017-0-237

V to t due H dine F 40mA) [1-erot If. = ë lot 2x(loma) x 10 0-2 e 10% v too Va due to la el dia - 041 (noma) (1- ē (to2] dr. e 2o (t-o.2) = 3 2x [-loma) x 1o --2 -io(-0-3} V. t > o-2 VLH = roul ēlotte Oct 20.2 ZOE - 70- t ab- Toode - 0.2 ēro e ospatao 0.2 è lot OTE 2008 o.2 Ttao aid e 1oz <taco gode tot [1 - éj Loa è tot ostao-d 1 -1.28 -107 ast <

tro-2 VLM = Looa xé z = 0.027 v Lajo28 é? -0.273 v

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