Question

I need help solving for step responses; What is the difference between a step response and a natural response? And how would I go about solving these circuits?

23 k22 t=0 + 6 V 30 kN2 3 0 + 22uF For the circuit shown above, find the capacitor voltage for t> (i.e after the switch closes). The solution, since this a "step" response, should be of the form a exp(-b t) + c followed by appropriate units. Vo(t):

If possible write out the steps, and please do not round

Thanks

Answer 1

Natural response due to initial and inductor) is system's response (output) only conditions (Energy stored in capautor with all external sources set to a due to Step the response is step input the system's response (ut)) 23652 & Boks - vo 22u6! - By souru transformation, TE V - 0.260mA - 236 T 22uF 234 parallel 23kr and 3our ith equivalent is are connected in Reg = 23k 4304 23kt 30k Req=13.018kr

0.26MA ! 2.2 MFF 13.018/ke Again by source transformation 13.018kr me fict 22uFT V - IR - 0.26m x 13.018k = 3.384 v Applying KVL to the loop ~Vt ic Req tvo=o. we know ic=covo a de Req -v+cavo tvo zo - at Reqcdvo - V-VO. at duo V-Vo = dt Req Integrating on both sides

-In (v-vo) = 5 tk. 70 at t=0 vo=o. [no initial condition given] -inv =k substitute in b e - Inv -in (V-vo) -t RC env-en (w-ro) = 7 (2- C Re Vava = e v-vo=vlet/RC) Vo = V-vettes VO - V Ve vor vli-et/2c)

V=3.384/ Req = 13.018ur C= 224F Refl = 13.018 x 103 x 22x106 - D. 2863 sec. Volt)= u(e-e/Rc) = 3.384 (le 10.2863). 395 himH 20/ | կ By source from & formation. 4InH Zio I ☺ {39 { u {or I Ž = 28/1 A. Equivalent is 39 is parallel with air, its given by Rez = 3941 19.987552 39+41

4 IMH 02 { 19.98754 32or again By Sonu Transformation, 19.9875 41MH * V io Lort RL V= I Req = 8 x 19.9875 - Hoiv 39 Applying kVL to the circuit, -V+ io Reg + Vet lo RL = 0. we know = L dio dt. -v+io (R4+ Re] tadioo de . -4.1+ io (39.9875] + Ldio co de

Let Reg + R = R' [ just to simplify] R = 39.9875 r. i helt io Rt L dio Lodio : - H. - Ri dio 401-Ri.. de Integrating on Both sides - in (401 - Rlio) <?= at tk → at to 10 = • Intial condition -en (401) k substitute in ②

-rw-Pis) 1) - ali), It en[ul) - an [k1 - R:)_t rl enuoi) an ( 4.j - el;o) 미 I - e 1 2, ~ ?. -et. erl

hol-klio = hole ł - - E't 2 Rio = hol-hole na I t) to = hol (e-et e R = 39.9875 L="Limh. a = 39.9875 - 975.30 sec Lelm - 975-30t lo(t)= 0,1025 lire. Short cut current through incluctor is given by 1oE) - (2) +(160) – (we- → ® where i(0) = steady state inductor cuorent. i(o)= Initial current in the inductor.

in this problem i(o)= 0 to determine i(o), we know Inductor be names as under state condition, Replace short in the iscuit circuit. I 394 io 8v Y & zou uirs Il Rea Req = 52.44 I=v =0.1525A. lo - I x 41 lo current 2otul division nule) = 0.1525841 61 i = 0.1025 A substitute (ilo) Equation in ②

Poll 0.1025+ (0 - 0.1025) e te - = 0,025 -0,102se. Polt) = 0.1025 (1-et1t) similarly you can do for capacitor Velte ve(00) + (v. (O)- vo(o) e tra Capacitor behanes as open circuit under steady state condition,