PS8_Solution[2] - Name: 45‘ Section: ; EE331 Homework PS8...

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Unformatted text preview: Name: 45‘ Section: ; EE331 Homework PS8 — spring 2011 Problems from Alexander & Sadiku: CH 10 0 10.55 (Just find the leerenin for parts a and b) CH 11 11.2 (Ans: Pavg 5 Ohm = 23.9 W, 0 for inductor and capacitor, why?) 11.3 11.6 (Ans: 69.22W) 11.9 Additional Problems (Instructor Option): 0 Any as assigned by instructor Page 1 of 1 Chapter 10, Problem 55. Find the Thevenin andNerten- equivalent circuits at terminals a-b for each of the circuits in Fig. 10.98. Hi 52 120 9 “WWW ‘ v“wwfiavxxva.WWMO a «j :0 s2 504m: \' """""" “"m—jWWWwwwo b (a) th) Figure10.98 a) 54295 P23 0 Ion 5.x 5T4)", m '{ECP'D 4 2L ‘5“) zm=2¢+il_“%c . \ fob = ((051+o'\+ (O HDQLD’M (v 905?: (0405) 2 lOSI-H-Dofl 2 — +3337. STEVE TH ’90 ‘I < a! %« ~°4 . + _ L50<30°V)’2L ,< ‘69 7 504» VTH— _ 56X 0 l/ 2TH: {2329],}!— -: (g.~SS)SL'(O§-[O:))n ‘35} —53 Hos 516135 I ' « + ,a- T— l ’?L LN?" m a} E - T“ l O . . 2. = 3H<~59 Mm BSCuW2n+DNR$DW 1 0 . j}: H<0°A'Z|Z ¥K+%c*%L ' ‘I<v°A- («WM STEM $51-33 H05 0 ___o 4 =— 3.'-| (-33 A been 33fi<s€v Chapter 11, Problem 2. Given the circuit in Fig. 11.35, find the average power supplied or absorbed by each element. , 9; , 4—51'57. Compkx ——=7 g : — 21% A) 25:} :@ 473$) VIA [hfims PM (Mews) I ...\ Chapter 11, Problem 3. A load consists of a 60-0 resistor in parallel with a 90—},LF capacitor. If the load is connected to a voltage source vs(t) = 40cos 2000t, find the average power delivered to the f load. peAK“ Ah“ ’A r 2 \% _,_._.~, N -_—_‘ l wgwar 5 42* fl / f / 6? SSH (+91? 5? Cowlex‘m CoWS‘3 —- 13.37; —- '43-‘13 \/A W T m, ‘4‘" \N A“ ?e% r 1 0 Is : 3 < 0 A For the circuit in Fig. 1 1.3 8, is = 3cos(103t) A. Find the average power absorbed by the 50-0 resistor. A Chapter 11, Problem 6. 20ix Figure 11.38 For Prob. 11.6. 'I ~ VA ' .3036 If“ 60-953 —3<o° + VA‘0 4. VA‘3‘0'x : 50-35;; IO+ 90A dome .7 VA: fi3.oo<q3fi‘i° \/ PMM‘AVkSsHAm) *h MASW‘V‘MAQ «X - 43-0? -L-D‘ - m»: A a, (le ‘ " " M 1 (50-9591 55"} om :2 1.5031: (01.9)“ 11 [IE1 .90 : (new Chapter 11, Problem 9. For the op amp circuit in Fig. 11.41, VS = 24300 V. Find the average power absorbed by the 20-kQ resistor. ah: Had < ASSuMzA 00% u dues C e, a o\ 6 ob“ Auswo‘ vats EMS) * I 94300 V Figure11.41 0 J_ = —— ' ‘1 LIL)?- (‘B’JHONA For Prob. 11.9. (gown-t H0003 n) ‘ _ 965v + I-(IOKSHWLD : art an)?" \/ o \/o __ . o ~ k (a o \IK_ (ofi‘KbZ? V 20 St 0 :4) 073 V onfL - [likfl ...
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PS8_Solution[2] - Name: 45‘ Section: ; EE331 Homework PS8...

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