EECE 210 Homework, Spring 2008

Homework 7 Solution P7.14 t<0: i 100 240 mi J i,(0t) =8A t>0: (10)(40) R= “at 10 =182 L 0.072 1 seg os = = 250 i= 8e~ 250 A = —10i, — 0. oa = = —80¢e~75% + 144760 = 64e75% A ¢>0t P7.21 [a] v,(0-) = (0+) =40V v2(0+) =0 Coa = (1)(4)/5 = 0.8 uF 25kQ WW —i + 0.8uF 7 40v T = (25 x 10°)(0.8 x 107°) = 20ms; * = 50 i= 40 et — 1.6e5 mA, t>0* 25, 25k + — i + YET Vy V2 7 aur “= as [16 « 0" SoS dr + 40 = 32e™" +8V, t>0 50x _ —50t 8V > » = Tres fp 8 «108 dz +0 = —8e £20 [b] w(0) = 5(10-*)(40)? = 800 pJ [e] wrapped = 3(10-*)(8)? + i x 107*)(8)? = 160 J. The energy dissipated by the 25 kf resistor is equal to the energy dissipated by the two capacitors; it is easier to calculate the energy dissipated by the capacitors (final voltage on the equivalent capacitor is zero): Wain = 3(08 x 10~*)(40)? = 640 yJ. Check: Wtrapped + Weiss = 160 + 640 = 80043; — w(0) = 800. (2+1)6 _ +n v.(0) = —5 + 30 = 25V P7.30 [a] C= ‘7 = (2 x 10-6)(250 x 10°) = 0.55; 31H ll i] [b] w, = 33 x 107°)(30)? + eG x 107%)(5)? = 1425 pJ Weiss = 52 x 107®)(25)? = 625 pJ 1495 — 625 Lee eS gga le) fo = ox 19-8 = We HA 6-22 7 * 62 de = — sx Gps [100 x 10-%e* de — 5 = -16.67 fede — 5 ae —5 =8.33e-" —13.33V t>0 [d] v, + v2 = v9 Uz = Up — v1 = 25e~™ — 8.33e-** + 13.33 = 16.67e-* + 13.33V t>0 fe] wrapped = 3(6 x 10~®)(13.33)? + +(3 x 107®)(13.33)? = 800 pJ Weiss + Wrrapped = 625 + 800 = 1425 x3 (check) L P736 [a] v(0*)=—Ipas T= Be Uo(co) = 0 Uo(t) = —IpRoe"+RN/ty, = > OF [b] vp = —(10)(15)e~ Sere = —150e9** yV, tt > OF [c] vo(0*) — 00, and the duration of v,(t) — zero pie ae [d] vw = Roipy =o T= Rik R 1A =J,; ig(00) = Le ie Therefore i,(t) = we, + [le - #8] 7 ((Ri+Ra)/LIt ; J, aad io(t) = ey + ee ne Therefore Vew = wis + Ral; e7((Ri+Ra)/L 1+R1, [el |vsw(0t)| +00; duration > 0 t>0* P7.44 [a] t<0 25maA\ 15Q) 60mH OmH 10 15mA} t>0 vV3120Q @Ws50ma _ 24x 10-8 ;-(0-) = i7(0*) = : iz(0-) = i, (0*) = 25mA; f. 120 = 0.2 ms; 5 = 5000 T iz(co) = —50mA iz = —50 + (25 + 50)e5° — —50 + 75e5°™" mA, t>0 Uo = —120[75 x 10~Fe-50] = —ge“5O"' WY t > OF aoe ee Db) in = aps ff -9e te de + 10 x 10-* = (302 _99)ma, t>0 ee ae. lel in = Fos ff 96 de + 15 x 10-9 = (45e-** — 30) ma, t>0