Lap.akhir Wheat Stone
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Transcript of Lap.akhir Wheat Stone
8/9/2019 Lap.akhir Wheat Stone
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VI. PERHITUNGAN DAN SESATANNYA
• Menghitung Rx
Rb L
L Rx .
1
2= Rx Rb
Rb
L
L
L
L Rx
∆
+∆
+∆
=∆
1
1
2
2
Tabel 1Rb 500 Ohm Rx 1
=∆± )( 11 L L (57,4 cm)05,0± =∆± )( 22 L L (2,6 cm)05,0±
Rb L
L Rx .
1
2= Rx Rb
Rb
L
L
L
L Rx
∆
+∆
+∆
=∆1
1
2
2
Ω== 6,22500.4,57
6,2
Ω=
++= 7,26,22
500
50
4,57
05,0
6,2
05,0
Ω± )7,26,22(
=∆± )( 11 L L (56,2 cm)05,0± =∆± )( 22 L L (3,8 cm)05,0±
=∆± )( Rx Rx (33,8±3,8) Ω
=∆± )( 11 L L (57,1 cm)05,0± =∆± )( 22 L L (2,9 cm)05,0±
=∆± )( Rx Rx (25,3±3,0) Ω
=∆± )( 11 L L (57,2 cm)05,0± =∆± )( 22 L L (2,8 cm)05,0±
=∆± )( Rx Rx (24,4±3,0) Ω
=∆± )( 11 L L (57,6 cm)05,0± =∆± )( 22 L L (2,4 cm)05,0±
=∆± )( Rx Rx (20,8±2,5) Ω
=∆± )( 11 L L (57,0 cm)05,0± =∆± )( 22 L L (3,0 cm)05,0±
=∆± )( Rx Rx (26,3±3,1) Ω
Tabel 1Rb 500 Ohm Rx 2
=∆± )( 11 L L (55,5 cm)05,0± =∆± )( 22 L L (4,5 cm)05,0±
=∆± )( Rx Rx (40,5±4,5) Ω
=∆± )( 11 L L (55,4 cm)05,0± =∆± )( 22 L L (4,6 cm)05,0±
=∆± )( Rx Rx (41,5±4,6) Ω
=∆± )( 11 L L (55,3 cm)05,0± =∆± )( 22 L L (4,7 cm)05,0±
=∆± )( Rx Rx (42,4±4,7) Ω
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=∆± )( 11 L L (55,0 cm)05,0± =∆± )( 22 L L (5,0 cm)05,0±
=∆± )( Rx Rx (45,4±5,03) Ω
=∆± )( 11 L L (55,5 cm)05,0± =∆± )( 22 L L (4,5 cm)05,0±
=∆± )( Rx Rx (40,5±4,5) Ω
Tabel 1Rb 500 Ohm Rx 3
=∆± )( 11 L L (54,6 cm)05,0± =∆± )( 22 L L (5.4 cm)05,0±
=∆± )( Rx Rx (49,4±5,4) Ω
=∆± )( 11 L L (54,6 cm)05,0± =∆± )( 22 L L (5.4 cm)05,0±
=∆± )( Rx Rx (49,4±5,4) Ω =∆± )( 11 L L (55,2 cm)05,0± =∆± )( 22 L L (4,8 cm)05,0±
=∆± )( Rx Rx (43,4±4,8) Ω
=∆± )( 11 L L (54,8 cm)05,0± =∆± )( 22 L L (5,2 cm)05,0±
=∆± )( Rx Rx (47,4±5,2) Ω
=∆± )( 11 L L (55,3 cm)05,0± =∆± )( 22 L L (4,7 cm)05,0±
=∆± )( Rx Rx (42,4±4,7) Ω
Tabel 2 Rb 400 Ohm Rx 1
cm L L )05,06,51()( 11 ±=∆±
cm L L )05,04,8()( 22 ±=∆±
Rb L
L Rx .
1
2= Rx Rb
Rb
L
L
L
L Rx
∆
+∆
+∆
=∆1
1
2
2
Ω== 1,65400.6,51
4,8 1,65
400
50
6,51
05,0
4,8
05,0
++= =8.6
± )6,81,65(
=∆± )( 11 L L (50 cm)05,0±
=∆± )( 22 L L (10 cm)05,0±
=∆± )( Rx Rx (65,1±8,6)Ω
=∆± )( 11 L L (50,3 cm)05,0±
=∆± )( 22 L L (9,7 cm)05,0±
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=∆± )( Rx Rx (77,1±10,1)Ω
=∆± )( 11 L L (51,7 cm)05,0±
=∆± )( 22 L L (8,3 cm)05,0±
=∆± )( Rx Rx (64,2±8,5) Ω
=∆± )( 11 L L (51,4 cm)05,0±
=∆± )( 22 L L (8,6 cm)05,0±
=∆± )( Rx Rx (66,9±8,8) Ω
=∆± )( 11 L L (51,2 cm)05,0±
=∆± )( 22 L L (8,8 cm)05,0±
=∆± )( Rx Rx (68,75±9,05) Ω
• Tabel 2 Rx 2
=∆± )( 11 L L (54,3 cm)05,0± =∆± )( 22 L L (5,7 cm)05,0±
=∆± )( Rx Rx (41,9±5.6) Ω
=∆± )( 11 L L (54 cm)05,0± =∆± )( 22 L L (6.00 cm)05,0±
=∆± )( Rx Rx (44,4±5,96) Ω
=∆± )( 11 L L (54,5 cm)05,0± =∆± )( 22 L L (5,5 cm)05,0±
=∆± )( Rx Rx (40,3±5,4) Ω
=∆± )( 11 L L (53,9 cm)05,0± =∆± )( 22 L L (6,1 cm)05,0±
=∆± )( Rx Rx (45,2±6,1) Ω
=∆± )( 11 L L (53,8 cm)05,0± =∆± )( 22 L L (6,2 cm)05,0±
=∆± )( Rx Rx (46,09±6,17) Ω
=∆± )( 11 L L (54,1 cm)05,0± =∆± )( 22 L L (5,9 cm)05,0±
=∆± )( Rx Rx (43,6±5,9) Ω
• Tabel 2 Rx 3
=∆± )( 11 L L (52,7 cm)05,0± =∆± )( 22L L (7,3 cm)05,0±
=∆± )( Rx Rx (55,4±7,3) Ω
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=∆± )( 11 L L (52,8 cm)05,0± =∆± )( 22 L L (7,2 cm)05,0±
=∆± )( Rx Rx (54,5±7,2) Ω
=∆± )( 11 L L (53,1 cm)05,0±
=∆± )( 22 L L (6,9 cm)05,0±
=∆± )( Rx Rx (51,9±6,9) Ω
=∆± )( 11 L L (53 cm)05,0± =∆± )( 22 L L (7,0 cm)05,0±
=∆± )( Rx Rx (52,8±7,02) Ω
=∆± )( 11 L L (52,9 cm)05,0± =∆± )( 22 L L (8,1 cm)05,0±
=∆± )( Rx Rx (61,2±8,1) Ω
=∆± )( 11 L L (53,2 cm)05,0± =∆± )( 22L L (6,8 cm)05,0±
=∆± )( Rx Rx (51,1±6,8) Ω
• Menghitung Rx rata-rata
a) N
Rx Rx
∑=
1
)( 22
−
−=∆ ∑
N
Rx N Rx Rx
Tabel 1Rb 500 Ohm Rx 1
N
Rx Rx
∑=1
)( 22
−
−=∆ ∑
N
Rx N Rx Rx
=25,53Ω =4.52
(25,53 Ω± )52,4
Tabel 1Rb 500 Ohm Rx 2
(42,12±1,72)Ω
Tabel 1Rb 500 Ohm Rx 3
(45,9±3,2) Ω
• Tabel 2 Rb 400 Ohm Rx 1
(69,34±7,63) Ω
• Tabel 2 Rb 400 Ohm Rx 2
(42,58±2,19) Ω
• Tabel 2 Rb 400 Ohm Rx 3
(54,48±3,72) Ω
b) Ω=+= 43,472
).400()500( 111
Rx Rx Rx
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=2 Rx 42,35Ω
3 Rx 50,19Ω
= 66,46 Rx
1
)( 22
−
−=∆∑
N
Rx N Rx Rx =14,55
(46,66 Ω± )55,14
• Menghitung 1 L dan 2 L untuk setiap Rx
o Untuk Rx 1
N
L L
∑= 1
1
1
)(2
1
2
1
1−
−=∆∑
N
L N L L
=12
7,648=
11
3468,111
=54,06 cm = 3,18
(54,06±3,18) cm
N
L L
∑=
2
2
1
)(2
2
2
2
2−
−=∆∑
N
L N L L
= 12
3,71
= 11
7468,113
= 5,94 cm = 3,21
(5,94±3.21) cm
o Untuk Rx 2
=1 L (54,72±0,25) cm
=2 L (5,28±0,71) cm
o Untuk Rx 3
=1 L (53,94±1,15) cm
• Membuat GRAFIK
Rb L
L Rx .
1
2= Rb L Rx
L ..1
21 =
xat y .=
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bt xat y += .
%100 graf
hit graf
Rx
Rx R KSR
−=
Tabel 1 Rx 1
%100 graf
hit graf
Rx
Rx R KSR
−=
= %10053,25
66,4653,25 −= 82,76%
Tabel 1 Rx 2
%88,10%10012,42
66,4612,42=
−= KSR
Tabel 1 Rx 3
%65,1%1009,45
66,469,45=
−= KSR
Tabel 2 Rx 1
%7,32%10034,69
66,4634,69=
−= KSR
Tabel 2 Rx 2
%58,9%10058,42
66,4658,42=
−=
KSR
Tabel 2 Rx 3
%35,14%10048,54
66,4648,54=
−= KSR
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VII. ANALISA
Dapat terlihat dari Konstanta Kesalahan Relatif (KSR) bahwa percoban ini pun
tidak benar-benar menghasilkan nilai resistansi (R) yang tepat benar. Hal ini
bisa disebabkan banyak hal seperti, kurangnya ketelitian dari praktikan, kalibrasi
alat yang kurang tepat, kurang teliti pada saat menghitung jarak dan kualitas alat
yang kurang baik.
VIII. KESIMPULAN
• Penentuan hambatan dengan menggunakan metode
Jembatan Wheatstone tidak benar-benar tepat terlihat dari nilai KSR yang
dihasilkan
• Kesalahan hasil akhir dapat disebabkan oleh kalibrasi
ataupun kualitas alat yang kurang baik
• Ketelitian dari praktikan juga sangat menentukan
keberhasilan percoban terutama saat menghitung jarak (L)
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