HW_3_Solutions

HW_3_Solutions - MIL/544% M 74/47. Lé/fl/flax/ I (7/ch W...

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Unformatted text preview: MIL/544% M 74/47. Lé/fl/flax/ I (7/ch W .33 3% = {km E flea. :3:/ofeh_ {/9 3 V, '3 (4.52.. DZ 1.! 10m.” E '5 Jr 5 ’/& 7': 0 ~‘z71;— zofo #15,) z: /z: ‘ /5/~/oz;——/a:§/;,z—/0=o f F a (9:0 /4 A92 /0[f/%a) /' a/75—/01;—/o§_:/o .. u-v are: “9.7: Z/C’ I, “90:5, "/0 = 7/ '1’, == f3 7’31; :— ‘bfléé %/. 77: , #Wfim @ s_ In the circuit below find the power dissipated in the 10S! resistor by using Thevenin’s Theorem. J; 2,115 55 %a'x%MWD—55cc> ‘Qfifigfifl Mac, 3 370 %5’5 5 2% % Z1 9%: .- m: fighéwfio 552. P095; /0296W (a A7052. 5/05314/ 02 44W $4;me . +— I ’ .52.. M 1 fl .5). 51. 4W? 0?— 6 J l0? % V~m=0217 —-3— (gag—9:023) 13-452) {Maxi/5V; Rpgcg—(i) 1.: 97.33.... .3ch “My? +0”— VD— /5"'a?_.Z7-"3flg‘79/ :: O ’“39’SQ”5'1;” 3.49971: 0 /s’va717’3(fzv‘é}70 #53527; - (gap—Tag): 0 /5"’ ant-51‘, #39:; T: 0 #5:: 3T Pb “75* “Ezr/szé = /S’ r I F “'8? ’_ ‘ F “7 g “‘3 xi .._ {’73-645'": wfl“gsp7/47 --- /“s* “3/ ér»? 59 {1/30 /°-7/ «$571; '35:; 76>;- ‘3':- its: /0.'7f fig.“ zagasfl 21/50 1/9 4‘ .iog 7 O W3 69mm 33’ Use Thcvenin’s Theorem to find V0 in the circuit below. Ewe \/%#Qo H/OCS'): O wag: go+30250v /\/OC~—60V W?RM 552’ 431. f9 ...
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HW_3_Solutions - MIL/544% M 74/47. Lé/fl/flax/ I (7/ch W...

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