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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 0339
= (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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 Spring '08
 Kamoua,R

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