Question

8.23 why ic(0+)=45mA not 50mA?

8.21 Assume the underdamped voltage response of the ocircuit in Fig. 8.1 is written as mit wv(t)= (A1 +A2)e cos wt +j(A - A2)e a sin wt en in moves -at -at 0. The initial value of the inductor current is Io, and the initial value of the capacitor voltage is Vo- Show that A2 is the conjugate of A1. (Hint: Use the same process as and A2.) outlined in the chapter to find A 8.22 Show that the results obtained from Problem 8.21- that is, the expressions for A1 and A2-are consis- tent with Eqs. 8.16 and 8.17 in the text. 400 8.23 The initial value of the voltage v in the circuit in Fig. 8.1 is zero, and the initial value of the capacitor current, ic(0*), is 50 mA. The expression for the capacitor current is known to be ased ased -200r + Ape00r tz0, ic)= Aje -800t 4 sing when R is 250. Find a) the values of a, wo, L, C.A1, and A2 res- se a t the di (0*) diR(O*) -0) 1 ic(0*) dic(0*) Hint: cuit. dt dt C L R ation b) the expression for v(t),t 0, c) the expression for iR(t)0, d) the expression for i(t)0. de- onse P 8.23 a 20 500 rad/s 1000; = 2 a2 - 400 rad/s 600; wo 4uF 2(500) (250) 2aR 1 1.5625 H L = (400)2(4 x 10-6) X ic(0*) AA2 = 45 mA dic di + dt dig 0 dt dt dir (0) dir(0) dic(0) dt dt dt dir(0) 1.56250A/s dt 1 dv(0) diR(0) 1 ic(0) 45 x 10-3 45 A/s dt R dt R C (250)(4 x 10-) dic(0) =0-45=-45 A/s dt 200A800A2 = 45; AA2 0.045 Solving, A -15 mA A2=60 mA t20 ic-15e 200+60e-800 mA b] By hypothesis -200tA.e-800 A3e t20 v(0)= A3+ A4= 0 dv (0) 45 x 10-3 = 11,250 V/s dt 4 x 10-6 -200A 800A4 A418.75 V A3 18.75 V 11,250; V 18.75e-800t =18.75e-200 t20 [c] iR(t) 75e-200 mA t0 75e-800 250 [d iR-ic 200 S00L 302 Natural and Step Responses of RLC Circuits We describe th ic + 71 L iR circuit shown in Fig sudden application Vo OHAHERS 27 not be stored in the Respo Circuit 8.1 Introdu Respo Figure 8.1 A circuit used to illustrate the natural response of a parallel RLC circuit. + t=0 CHATE CTONTITS R Jie 6 vd batsnanep ai eluswFbioaue Circuit Tistavi onepsilov beilddEn Figure 8.2 AA circuit used to illustrate the step response of a parallel RLC circuit. Parallel RLC Circult p. 306 L In To find the natural deriving the differen ep oals find the voltage beca w Isbioaun ert paiegorde RotioLos bns otnn ns the voltage, we can fi c90 CLeste W fons

Answer 1

yes there should be A1 + A2 = 0.050

and 200 A1 + 800 A2 = 50

so

A1 = -16.67 mA

A2 = 6.67 mA