Question

1. For the circuit shown below, we wish to find v(t) for t>0. 1 1 a. Find the governing equation for the voltage v using KCL at the top node using the following definitions: a = 0,W, = dr. This will get you a governing equation in the same form as that derived for the case we did in class where the R, L, and C were in series. b. What is the particular solution in this case? c. If R=50 2, is the system underdamped, critically damped, or overdamped? What is the equation that describes v(t)? d. Determine the final equation for v using the initial conditions v(0)=5V and i(0)=0. (Hint: use KCL and i(0)=0 to find my at t=0). e. Plot v(t) until the transient dies out. + d1H 1mFv 1

Answer 1

Apply Kel is title=0 IMF dv V=0 +! jvdt at sides derivative on both en een + ¥ u op to de la terre et + t 0. Mat=rompared extily 34366 stest i = 0 S., 2 nou pe ✓ Co to the St 5,2 = -x 2 ſaw x= 1 Bec Wolne There is no input, so particular solation in (a) = volt) = (A1 A2) eat (x= wo, critically dampedy Octs et tray coswot the finwidt) (arwo, overdamped) sit A en ne taze Calwon under damped hd - Word C) = 504 as been record on 310 Whave a = 31.62 XL Wn, under damped.

(d) wd a verdade = 29.99 Lt) = e Caico 29.996+A2 bin 29.99) At t=0 LCD) -5V de Apply Kel it iftico at t=ot icot) + ielot ) reel veot) = 0 0 + m +17% dvcot) 0 ducot) = -1 108 ducet) = -100 at D- at dvet) 7 dt = Caico2q.gatt Azsin 2.9.99€) toe Ag sin29.99€ € 29.99A2C29.996 uit At t-of 9V COT) 400 400 or renorut Just -100 = -10 A, +29.99 Az -100 = 50 +29.9942 Az = -1.67 ci ucts seloto scesf 29.9976-167 Sim29.99€) beti ſ