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Homework_6_Solutions - '1 5 5 fl 1 r 5 in ”m ‘5...

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Unformatted text preview: . '1 5 5; _ fl. 1' r 5 in ”m ‘5 5“" W (0.... I ' 5°“? 5 '5 E55? 535° 5. 5157" 55/ 5!- 2...... .5 ,5 52...! cf .5 w (Vi/51%”! 55 Ex 1&5 55mg @11110111511 11 011333161“ 1155101 5,5311% 5” J = K = nclqelvd 1.111111. 1‘ = 1 A5141 = 1 1111112, 719 = 3.45 x 1028 111—1114, and {qei 2 1.602 >< 10*19 C, the average fiflfi VBEOCTW 01" 17113 csflémfim 311501113111 1:10.11 5:16 0313511105311 (15 1’11 2 7.39 X 10—5 11134. 53523. $111701: vs @0160301119 (a) For 3111761 ((31% 2 6.17 x 107 S—m‘l), the 0111156115. (53113113! is V (6.17 x 107x11 J=UE=GE= 0.01 = 6.17 x 109 A—m'z 1113515: 03156115111101“: 511 §511~1{11} £531 10010131511171 1: a z 19"” 3.33%), we find J 2 10-12 a“ ° 5 m ; W5 a .2“ 1 1.: ‘ - ' ~ .1 1. 11111111013 3e11111111111mi01: (a) 1301 1115112011., 15113 111 1111319 02111161 cancemmimn N151 311011153 (301111114:- 11v1ty 03$ are gwen by 19 _ 1.124(1.602x10—l9) N13: = (2.8 x 10 )(104 x 101% m 2 6&3 X 109 6W3 (Tl-Si =1Q8K1u6 + MP)N1' 21.602 x 107190450 +4.50)(6.13 x 10912 1.87 x 1074 5-1114 and fespectfivefiy. (b) 311111101in for 13161111211111.1111, the intfinsic 0011161: concsntration N103 and the conducfivity or,-G can be calculated as a . 1' —19 1,56 = [(104 X 1019>(6 X 1018)ea:(16.63(8x6?::§1g)(300)) Z 2.25 X 3913 CHE-—3 N. and (Face Shem-Le "5’ i‘v‘p)Nz‘ 21.1102 >1 10755131511101 191112111225 >1 1515312 210521 31114 5.. 1%. 1115105311 1105150101: {:31} $013113: 51111131111 11.211 111:1de 11711121 011361119. 1513 flinging 05.111013111501103 is such 13101“. 5159 2 [55d 2 51535“ 121114 >> NP 2 IVE/Ne W§§8f6 Ni 2 5.13 X 18 mfg (1011110 111 Probism 5—9). As 0 1631151, the conducfivity of the doped 51110011 resistor can be approximaéed as 21116:”..Na 21.602, x 10—19 x 731 x 10“ 2 1171311171 aSidogcd “fl. gwfiffiigns Using tine mine :33 $330923 found, the tesistanee oi? tits 3 min tong stint Qitti 33332 siiieon tint enn tse eompnteti 33 10-3 W 2 33,432 (iimtooi x 13-6) Sidopeé 2 it ttse seine silicon bat is pni'e, then the conductivity at 300K is given by 05mm 2 1.87 X i0“4 S—ni‘i (found in Ptobiem 3-9). Using this value, the resistance oi the pure silicon bat“ toliows as it ~ N W193 333 339: ‘" “W M (is? 3 3343333 3 ttE‘W} "" " 7. integratettaeit‘enit {33:3} seniston Since this is an EC resistot made of n~type silicon with N6 = 2.5 x 1016 e314: 9336 3 3333 §m2~{V—§)~E, 3333 t :2 2 films its tesistnnee 333 3:33 written as l 5 ..————~ 2 _~_ to ten 3361363123 3.632. x to19 x (monies >< 331632 >< ion) R2 3?; testis :233 “on tonnt 33 wit 2‘: M f} M fiesistnnee of 3 33333313323333 tinge {3) Using the testiit oi Exempie 53, the resistance of the sentieit‘eniai’ ting stto‘svn in Eigtn‘e 5.20 can be written 3s 7T9, 7r R = atain(b/o) = otin(b/a) (b) For the straight conductor with iength l = «(a + b) / 2 and cross-sectional area A = (b — (1)1;a resistance is given by R _ i __ Ma 3- b) — 0A _ 20(1) — a)t (c) Substituting 0 = 7.4 x 194 8-in—1, b = 1.53 = tot = 3 cm into the R expression in part (3) yields R : 0.0349Q. Simiiasiys substituting the same values in the R expression in part (b) yields R r: 0.0354Q. (d) When a > b — a = C, the naturai iogarithm term in the resistance expression in part (a) can be approximated as is<a+c)=in(i+£)=§_€_2+5:_,,_2£ a UL: a a 20? 333 Using tttis approximation, time resistance expression in past (a) can be written as 773 R2 — 015g The resistance expression in part (b) can aiso be appi'oximated as 7r(a+ b) N 77(23) ,. 7m R: 20(b—a)t _. 23¢ _. Fa “which is tit 3 33333 espsessien o'tstsineti item past in} ‘8 ’“ 3333333333333 333 tiéeeeeeee? fie fieetfiee 533,, 33 the ieterfeee 33331333 333333; titefieett’éeet tee ween: eeeetietfi‘y‘ eeetfitéaen [$373 is m genera} ieeempetéeée with, the eieeteestette beenefiety @0333: ditéee cementing :33 eeetimity at" the manual eempeeent 0f eteeti‘fie flux density, emfiese a surface eherge Eeyer is essemed to exist. (a) At steady—state? the eteettic field between the mates meet eetfisfy the felfiowmg cendfitfitms; Egd+Egd:% $97-$30 -> J1n=J2n --> filEjg-‘ZGQEQ V-Efize ——+ p8362E2~61E§ v 50 teat we have 023 3/53 0‘1 Va 33-3 33m , —-» 3:33 Efiwm— €33: “Seth: a. $13652 d 31 (b) The 331mm to this 3333 33 $333135: emitted? eat at the em at? Seetfiett 5.69.2; we have M f 62 0‘1 .05: K51““52) E2: 5} “52“\ E1 0’} 0'2) 02‘ 62 \ or V 60 ~eao V fi3=§€3M—€23 B ”3:21 3‘3; \ £7.13 X” {3-5 4“ {3‘2 a, 7‘1””{37'3 d 542%, ieafieege 3e3§33333ee~ £33 Chaptet‘ 4»? the eepeeétance per unit Eength at a very long wee of 3363133 33 paraflei t0 3 pefi'eztiy coeduetmg greued pEme at a height of (23/2 aeeve it was given by {4.503 233 27r€o C: m ietd/a) Ueéng tee {teeth}; tefiatienehig RC 2 5/0, we can write the resistance per unit length between the glue of radms a suhmerged 331 a deep lake of conductivity 0' at a height h > a above its pianat ottem as e .. Em (Zh/a) __ in (2h/a) Rx”... W 0'0 0 Eire 27M" S 5 § E t ...
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