Question

R2 + i R V X is [3pts] In the circuit above, use R1 = 322, R2 = 62, and iz(t) = -36u(t) – 84u(=t), in A. Assume that element X is an inductor, with L = 25mH. What are the values of the following quantities at the specified times? a. current i at t = 0 : A b. voltage v at t = 0+: V c. current i at t = 6.3ms: A

Answer 1

As the inpect is inport is a emit step Voltage ult) is applied, the initial condition is zove. (a) X is capacitare curth c=53uf - the iken't R il Youth de of zu We know ů = Voust) u= с. dru at R At t70, uct) = | Vou г. du dt R

du det RC Voru we Integrate both side get I du. - Jet Voru RC + Re -log (Vo-u) tk ++ t=0, l=0 (initial condition = 0 - log (vo) tk k= log (Vo) t tlog Vo ) RC - log (Vo-u) + log (vo-u) - log (Vo) log (voll) RC + RC -ARC e Voll vo -the Voll = Vo e tire all = Vo-Vo

HRC) le = Voll- (1) At t= 650 les - 6 -650 Xlo u=18(1-e 158538106 u=10.05 v أط) x ês an înduefore with L=7.60H, the eikuid is R Vo uutt di + 2u krl By applying Vo ult) - eR - L di di co

Ayo ut) = 1 20 Vo - iR - L l Yo - ir th dy Lai alt z Vi - du dt Jenter the - tag (head) th = 14 Af d=0, i=0 -logy) th = 0 =log up - log no d) + lag (los) = 4 Lag (le - 1) - Lag (h) = -1 t Yo- ef that i/ VO t No - Vo e elpt R

Yo - -R/Lt e ? Yo -R/Lt Yo e Yo - = -R/Lt Vo A kle knew us L di dq _NL ty - R XL Y e Vox L -R/Lt -2) u Vo e Afit=650 us et 15 5x103 15 7.6x1 *650x106) u = 18 e = 4.99V u=5V * you use only the question eqh (1) 4 (2) to slove