Ex.14.pdf

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概要信息:

¢›o †6>xÿþ>{
Xhè 1300011329
±e5|11Ò
¢žm2014c1230F
wžm2014c1231F
1 8‡¦
(1) †6>xÄn
(2) †6>x(¯Ý9K§σ
(3) ²ï>xÿþ>{Ø5
2 ¤ì^ä
>{‡3‡(ZX96.),ªu6O(AC/5.),–ÿ>{3‡,†6­Ø>
,V
V•m',ü

V•m',> ì,êi^L,‚eZ.
3 ¢n
3.1 †6>x
n>´ãX㤫:
1
ã¥u6Oüà>³ƒÓž, u6O«", …k:
Rx = R0(
R1
R2
)
>x¦^'{?1ÿþ,é>
­½5‡¦Øp.…‡u6Ov
(¯,KRxҌ±ˆ†Ù
¦x:ƒÓ°Ý.ù¦>x¤O(ÿþ>{¤ì.
3.2 Ø©Û
Ø5
̇kü‡:
3.2.1 x:>{‘5Ø
R1, R2, R0Øò¬éRxE¤K.
éuù«Ø, ·‚kXeØ(½ÝO:
σRx
Rx
= [(
σR1
R1
)2 + (
σR2
R2
)2 + (
σR0
R0
)2]
1
2

žØR1/R2'ŠXÚØé(JK, Œ±†Rx†R0 ˜2ÿ˜g,Œ:
Rx = (Rx1 ·Rx2)
3.2.2 >x(¯Ý‘5Ø
½Â>x(¯ÝS
S =
∆n
∆Rx
Rx
§L«>x²ï,RxƒéUCþ¤Úåu6O =‚ê. Œ±^R0UC5O“Rx. §
‘5ØŒ±‰XeO:
δRx =
0.2R1 ·∆R0
∆n ·R2
Äþ¡ü‡Ïƒ,Œ
σRx = [(δRx)2 + (
R0
R2
σR1)2 + (
R0R1
R2
2
σR2)2 + (
R1
R2
σR0)2]1/2
4 ¢SN
¢>´X㤫:
2
4.1 ÿþ–ÿ>{±9ƒA>x(¯Ý
©Ok^^Loÿ–ÿ>{, ëÐ>´N!>x²ï,Öê.5¿,R0kêi ê
(½{´l$ m©úúCz,U¦u6O =0.1‡ =R0˜ .
²ï,N!R0¦>x l²ï,¦u6O =3.0 6.0‡‚,AedžR′
0,OŽ>x(¯ÝS.
4.2 &Äÿþ^‡é>x(¯ÝK
©O››Cþ,UC>
>ijE,R1/R2,> ì>{Rh,ÓOŽÑ(¯Ý.
5 êâL‚9?n(J
5.1 ÿþ–ÿ>{±9ƒA>x(¯Ý
–ÿ>{ R1/R2 (Ω/Ω) R0 (Ω) R′
0 (Ω) ∆R0 (Ω) ∆n (‚) Rx (Ω) S (‚)
Rx1 500/500 47.2 47.3 0.1 4.0 47.2 1.8× 103
Rx2 50/500 3586 3560 26 4.8 358.6 6.6× 102
Rx2 500/500 358.7 357.6 1.1 5.0 - 1.6× 103
Rx2(R2/R1) 500/500 358.7 357.7 1.0 4.2 - 1.5× 103
Rx3 500/500 4020 4060 40 3.4 4020 3.4× 102
R1, R2, R39ƒA(¯Ý®3L¥, 	éu†ü:ÿþ,k
Rx2 =
√
R
(1)
0 R
(2)
0 = 358.7 Ω
3
Ĥì#
σ
R
(1)
0
=
1√
3
(300× 0.1% + 50× 0.1% + 8× 0.5% + 0.7× 2%) Ω = 0.2 Ω
Ón
σ
R
(2)
0
= 0.2 Ω

σRx2 = Rx2
√
(
1
2
σ
R
(1)
0
R
(1)
0
)2 + (
1
2
σ
R
(2)
0
R
(2)
0
)2 = 0.1 Ω
Rx2 ± σRx2 = (358.7± 0.1) Ω
5.2 &Äÿþ^‡é>x(¯ÝK
E (V) R1/R2 (Ω/Ω) Rh (Ω) R0 (Ω) R′
0 (Ω) ∆R0 (Ω) ∆n (‚) Rx (Ω) S (‚)
4.0 500/500 0 358.7 357.6 1.1 5.0 358.7 1.6× 103
2.0 500/500 0 358.7 356.6 2.1 4.5 358.7 7.7× 102
4.0 500/5000 0 3587 3513 74 3.3 358.7 1.6× 102
4.0 500/500 3k 358.7 343.7 15.0 4.6 358.7 1.1× 102
ÏLnØíŒ
S =
Si · E
R1 +R2 +R0 +Rx +Rg(2 + R1
Rx
+ R0
R2
)
(¯Ý¬‘E¤',‘R1/R2,RhCŒ
C.duêâ',¤±é(J?1†'Œ±
Œ—wÑù˜½5'X.
6 gK
(1) 1.¬ 2.ج 3.¬ 4.¬ 5.¬
(2) 1.u6Om'™4Ü,™Ue>O,x:>{LpѬ—ù˜y–
2.éŒU´R1R0
R2
†Rxõ
4

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