HW 6 Solution - W1 obtain L = qfiflnrfl...

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Unformatted text preview: W1: obtain L_ = qfiflnrfl =..|'(25){0.1x1fl'5 :1 13:15.31” f.fl11;p=(11'(10}(_111x111 :1 1351111111 25 fl_95=¢ 25 111 N, _+__ _ 15.3 111 N, whichyields ' = 01133 N d' 9.11 (:1) We find fl" 1 D; = 114:] = (4311](1111259) = 12.4 cm 1'3 and L, = 11113,: = {12.4)[11.1x111*] =1 LP =11_1,um A1511 1 2 1.5 111‘“ pfl="—‘=—( 1:“ ) =2_25x111’cm" N, 111 Then J 211er (1.6x111"’)(12.4}(225x111’) "5— LP _ (11.11111?) 01' IF = 4.112x111‘“ Aim: For A =111“‘ cm! , them —I4 IN = 411211111 A (b) We have 5:2“ 31. = pn[:] = (13511)(1111259) = 35 cm‘ 1: and L_ = 1.111“ 1", = ., I'(35}(11_4x111“) :> L,- = 1.4 m A1511 "2 (1.5::111” 11' = —" = — = 4.5x1fl‘ arm—J ’0 Nd 5x111” 11:11:11 I = aflflnfi = (1.6x111‘“][35}(4.5x111‘) L (31.4;111“) GI .19 = 5.?4x1u‘" A £ch For A =11)" cmz, than In = mum” A [c] NN I'll-“=p: II!" "J [51:11?“ (10“) = (0.0259) In 1 (1.5::10'”) DI I; = {161? V Then for 1 F =—I’;I =I13fl9i’ “ 2 W: find 2%: P, =19” H 03-3-9 = (2.25x1u’ «4:4 J 0.0259 pi = 3.42::10'“ cm" GI 01' r = 113110" A The hale current is :2wa .F‘ 'I'heelectron cmrent i5 givenhy In =r—rp = 113x10" — 41:23:10“ ( ) [ [1309 J —(x—x_) X — EXP 0.0259 EXP LP 1 At 1: = I“ +E L1- —1 r, = 1131:1111" — (filflxlfl4)r:xp[?] 01' It = 3.4km" A _ + " _ (10'1)(1.6x10"’)(4sn)(m") R =2fifl. F pal. L L A _ at)! _ A[ep_N£) 0.113 R =— ' (1n‘*](1_fixm'“ ){1350}(m" R, = 46.3 n The 131131 suits resistance is R :19; +3! =26+4fi3 2} R = T13 D. (b) I? = m :> 0.1: I023) DI I = 1.33 Hui 9.43 Fur tbs: breakduwn voltage, we med N4 5:: 31cm“ (rm—J andforfljis dupingwefind fl! m43flcm1fF—L1hm I 1:]! =(4311)(11.11259)=11.14 cm 1': PM the fin jlmctiun, ..:"=£e.i'111-—I—fl‘a Ty: (1:51:10 1"'](1j:1:111"' 1114 3x111” 111" GI .15 = 1213:1111“ 1:1,?ch 'I'hen V I=J,A —" ‘1 I165 2x111" = (1.2?x111‘”“].4exp[ J 0.0259 Finally .4 = 1.991111? c1111 ...
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