Informe Fisica 3 - Puente Unifiliar Wheatstone.docx 36yt

UNIVERSIDAD NACIONAL DE INGENIERIA FACULTAD DE INGENIERIA AMBIENTAL

ACCION BACTERIOSTATICA DEL CRISTAL VIOLETA – ACCION OLIGODINAMICA DE LOS METALES PESADOS

QUINTO LABORATORIO DEL CURSO DE MICROBIOLOGIA SANITARIA I – SA323

ALEXANDER JULIO LOPEZ CASTROMONTE – 20151358F

DOCENTE: ING. JORGE GILBERTO TELLO CEBREROS

Lima, Perú 2017

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental 2. Resumen

2

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental 3. Resultados Tabla 3.1: Resultados de las mediciones del circuito en serie. R (Ω) R12 R13 R14 R15 R16 R17 R23 R24 R25 R26 R27 R34 R35 R36 R37 R45 R46 R47 R56 R57 R67

Valor de R (Ω) 11.9 ± 0.1 23.8 ± 0.2 45.6 ± 0.4 67.4 ± 0.5 116.9 ± 0.7 216.2 ± 0.8 11.9 ± 0.1 33.7± 0.3 55.5 ± 0.4 105.0 ± 0.6 204.3 ± 0.7 21.8 ± 0.2 43.6 ± 0.3 93.1 ± 0.5 192.4± 0.6 21.8 ± 0.1 71.3 ± 0.3 170.6 ± 0.4 49.5 ± 0.2 148.8 ± 0.3 99.3 ± 0.1

a (cm) 58.35 ± 0.05 75.00 ± 0.05 87.00 ± 0.05 91.60 ± 0.05 96.50 ± 0.05 100.2 ± 0.05 57.2 ± 0.05 80.9 ± 0.05 89.2 ± 0.05 96.6 ± 0.05 99.6 ± 0.05 72.5 ± 0.05 86.4 ± 0.05 95.0 ± 0.05 99.2 ± 0.05 73.0 ± 0.05 92.8 ± 0.05 98.5 ± 0.05 87.3 ± 0.05 98.2 ± 0.05 95.1 ± 0.05

b (cm) 47.2 ± 0.05 30.0 ± 0.05 17.9 ± 0.05 12.9 ± 0.05 6.8 ± 0.05 3.7 ± 0.05 45.5 ± 0.05 21.4 ± 0.05 13.5 ± 0.05 7.2 ± 0.05 2.7 ± 0.05 30.5 ± 0.05 17.1 ± 0.05 11.7 ± 0.05 5.0 ± 0.05 28.9 ± 0.05 10.3 ± 0.05 3.2 ± 0.05 15.0 ± 0.05 4.8 ± 0.05 7.8 ± 0.05

Tabla 3.2: Mediciones de las resistencias conocidas y valor final. R

Intervalo de R (Ω)

Valor final (Ω)

R12

[ 11.8 - 12.0 ]

11.9 ± 0.1

R23

[ 11.9 - 12.0 ]

11.9 ± 0.1

R34

[ 21.7 - 22.0 ]

21.8 ± 0.2

R45

[ 21.7 - 21.9 ]

21.8 ± 0.1

R56

[ 49.4 - 49.7 ]

49.5 ± 0.2

R67

[ 99.2 - 99.4 ]

99.3 ± 0.1

3

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental 4. Cuestionario Cálculos y resultados para el informe 2 A. Determine el valor de Rx para todos los datos medidos: R (Ω) R12 R13 R14 R15 R16 R17 R23 R24 R25 R26 R27 R34 R35 R36 R37 R45 R46 R47 R56 R57 R67

Valor de R (Ω) 11.9 ± 0.1 23.8 ± 0.2 45.6 ± 0.4 67.4 ± 0.5 116.9 ± 0.7 216.2 ± 0.8 11.9 ± 0.1 33.7± 0.3 55.5 ± 0.4 105.0 ± 0.6 204.3 ± 0.7 21.8 ± 0.2 43.6 ± 0.3 93.1 ± 0.5 192.4± 0.6 21.8 ± 0.1 71.3 ± 0.3 170.6 ± 0.4 49.5 ± 0.2 148.8 ± 0.3 99.3 ± 0.1

𝑅𝑥 =

a (cm) 58.35 ± 0.05 75.00 ± 0.05 87.00 ± 0.05 91.60 ± 0.05 96.50 ± 0.05 100.2 ± 0.05 57.2 ± 0.05 80.9 ± 0.05 89.2 ± 0.05 96.6 ± 0.05 99.6 ± 0.05 72.5 ± 0.05 86.4 ± 0.05 95.0 ± 0.05 99.2 ± 0.05 73.0 ± 0.05 92.8 ± 0.05 98.5 ± 0.05 87.3 ± 0.05 98.2 ± 0.05 95.1 ± 0.05

b (cm) 47.2 ± 0.05 30.0 ± 0.05 17.9 ± 0.05 12.9 ± 0.05 6.8 ± 0.05 3.7 ± 0.05 45.5 ± 0.05 21.4 ± 0.05 13.5 ± 0.05 7.2 ± 0.05 2.7 ± 0.05 30.5 ± 0.05 17.1 ± 0.05 11.7 ± 0.05 5.0 ± 0.05 28.9 ± 0.05 10.3 ± 0.05 3.2 ± 0.05 15.0 ± 0.05 4.8 ± 0.05 7.8 ± 0.05

Rx (Ω) 9.6 ± 0.1 9.5 ± 0.1 9.4 ± 0.1 9.5 ± 0.1 8.2 ± 0.1 8.0 ± 0.1 9.5 ± 0.1 8.9 ± 0.1 8.4 ± 0.1 7.8 ± 0.1 5.5 ± 0.1 9.2 ± 0.1 8.6 ± 0.1 11.5 ± 0.2 9.6 ± 0.2 8.6 ± 0.1 7.9 ± 0.1 5.5 ± 0.2 8.5 ± 0.1 7.3 ± 0.2 8.1 ± 0.1

Σ𝑅𝑖𝑗𝑥 , 𝑖 = 1,6 𝑗 = 1,7 , 𝑗 > 𝑖 21

𝑅𝑥 = [(9.6 ± 0.1) + (9.5 ± 0.1) + (9.4 ± 0.1) + (9.5 ± 0.1) + (8.2 ± 0.1) + (8.0 ± 0.1) + (9.5 ± 0.1) + (8.9 ± 0.1) + (8.4 ± 0.1) + (7.8 ± 0.1) + (5.5 ± 0.1) + (9.2 ± 0.1) + (8.6 ± 0.1) + (11.5 ± 0.2) + (9.6 ± 0.2) + (8.6 ± 0.1) + (7.9 ± 0.1) + (5.5 ± 0.2) + (8.5 ± 0.1) + (7.3 ± 0.2) + (8.1 ± 0.1)]/21 𝑅𝑥 = 8.52857142 ± 0.04619047 𝑅𝑥 = 8.53 ± 0.05

4

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental B. Resuelva las preguntas 2, 3, 4 y 5 de la guía utilizada en la práctica 2. Determine la resistencia total para: R23

R12

1

R34 3

2

R45 4

R56

5

R67 6

7

Req = R12 + R23 + R34 + R45 +R56 +R67 𝑅𝑒𝑞 = (11.9 ± 0.1) + (11.9 ± 0.1) + (21.8 ± 0.2) + (21.8 ± 0.1) + (49.5 ± 0.2) + (99.3 ± 0.1) 𝑅𝑒𝑞 = 216.2 ± 0.8 3. En el esquema siguiente determinar la resistencia total (Rxy) y comprobar este resultado mediante un procedimiento analítico utilizando los valores calculados para la esquena anterior.

R12

R23

R34

R67

R56

R45 2

𝑅𝑒𝑞1 = 𝑅12 + 𝑅23 + 𝑅34 𝑅𝑒𝑞1 = (11.9 ± 0.1) + (11.9 ± 0.1) + (21.8 ± 0.2) 𝑅𝑒𝑞1 = 45.60 ± 0.4 𝑅𝑒𝑞2 = 𝑅67 + 𝑅56 + 𝑅45 𝑅𝑒𝑞2 = (99.3 ± 0.1) + (49.5 ± 0.2) + (21.8 ± 0.1) 𝑅𝑒𝑞2 = 170.6 ± 0.4

5

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental 1 𝑅𝑒𝑞

=𝑅

1

𝑒𝑞1

+𝑅

𝑅

1

×𝑅

→ 𝑅𝑒𝑞 = 𝑅𝑒𝑞1 +𝑅𝑒𝑞2 … (𝑎)

𝑒𝑞2

𝑒𝑞1

𝑅𝑒𝑞1 = 45.6 ± 0.4 = 45.6 × (1 ±

𝑒𝑞2

0.4 ) 45.6

𝑅𝑒𝑞2 = 170.6 ± 0.4 = 170.6 × (1 ±

0.4 ) 170.6

0.4 0.4 𝑅𝑒𝑞1 × 𝑅𝑒𝑞2 = (45.6) × (170.6) [1 ± ( + )] 45.6 170.6 𝑅𝑒𝑞1 × 𝑅𝑒𝑞2 = 7779.36 ± 86.48 𝑅𝑒𝑒𝑚𝑝𝑙𝑎𝑧𝑎𝑛𝑑𝑜 𝑒𝑛 (𝑎): 𝑅

×𝑅

𝑅𝑒𝑞 = 𝑅𝑒𝑞1 +𝑅𝑒𝑞2 = 𝑒𝑞1

𝑒𝑞2

7779.36±86.48 216.2 ± 0.8

𝑅𝑒𝑞 = 35.98 ± 0.53

4. En el siguiente esquema determine la resistencia equivalente Rxy y comprobar analíticamente:

R12

R23

R45 R34

2

𝑅´ = 𝑅12 + 𝑅23 = (11.9 ± 0.1) + (11.9 ± 0.1) = 23.8 ± 0.2 1 𝑅𝑝

1

1

= 𝑅′ + 𝑅

34

𝑅´×𝑅

→ 𝑅𝑝 = 𝑅´+𝑅34 … (𝑏) 34

0.2

0.4

𝑅´ × 𝑅34 = (23.8) × (21.8) [1 ± (23.8 + 21.8)] 𝑅´ × 𝑅34 = 518.84 ± 9.12 𝑅𝑝 =

518.84±9.12 45.6±0.4

=

518.84 × 45.6

0.12

0.4

[1 ± (518.84 + 45.6)] = 11.4 ± 0.3

𝑅𝑝 + 𝑅45 = (21.8 ± 0.1) + (11.4 ± 0.3) = 33.2 ± 0.4 6

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental 5. ¿Cuál es la influencia de la f.e.m y de la resistencia interna en este método? La f.e.m al ser mayor da más energía al fuente da mayor diferencia de potencial entre los puntos M y N y dado que la resistencia entre esos puntos no varía la corriente aumenta, pasa lo contrario si la f.e.m. disminuye. Al aumentar la resistencia interna la corriente que pasa por él, disminuye, y si la resistencia disminuye la corriente aumenta, siendo constante la f.e.m.

7

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental 5. Conclusiones

8

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental 6. Fuentes de información

9

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental 7. Anexos

10

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental 8. Apéndice 8.1. Hoja de datos

11

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental 8.2. Procedimiento y cálculo 8.2.1. Calculo de las resistencias conocidas y su asociación en serie

𝑅12 = [11.8 − 12.0 ] 𝑅12 =

11.8+12.0 =11.9 2

𝐸𝑟𝑟𝑜𝑟 =

12−11.8 2

𝑅12 = 11.9 ± 0.1

= 0.1

𝑅23 = [11.9 − 12.0 ] 𝑅23 =

11.9+12.0 =11.9 2

𝐸𝑟𝑟𝑜𝑟 =

𝑅23 = 11.9 ± 0.1

12 − 11.9 = 0.1 2

𝑅34 = [21.7 − 22.0 ] 𝑅34 =

21.7+22.0 2

𝐸𝑟𝑟𝑜𝑟 =

= 21.8

𝑅34 = 21.8 ± 0.2

22.0 − 21.7 = 0.2 2

𝑅45 = [21.7 − 21.9 ] 𝑅45 =

21.7+21.9 2

𝐸𝑟𝑟𝑜𝑟 =

= 21.8

𝑅45 = 21.8 ± 0.1

21.9 − 21.7 = 0.1 2

𝑅45 = [49.4 − 49.7 ] 𝑅45 =

49.4+49.7 2

𝐸𝑟𝑟𝑜𝑟 =

= 49.5

𝑅56 = 49.5 ± 0.2

49.7 − 49.4 = 0.2 2

𝑅67 = [99.2 − 99.4 ] 𝑅67 =

99.2+99.4 2

= 99.3

𝑅67 = 99.3 ± 0.1 𝐸𝑟𝑟𝑜𝑟 =

99.4 − 99.2 = 0.1 2 12

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental 8.2. Calculo de las Rx en la tabla adjuntada. Observación: al colocar Rij, quiere decir que se utilizó dicha resistencia para determinar su respectivo Rx. 30.50 ± 0.05 𝑅34 = ( )(21.8 ± 0.2) 72.50 ± 0.05 30.50

0.05

0.05

= (72.50) [1 ± (30.50 + 72.50)]

= 0.42068965 ± 0.00097978 Ω

= (0.421 ± 0.001)(21.8 ± 0.2) 0.001 0.2 = [(0.421)(21.8)] [1 ± ( + )] 0.421 21.8 = 9.1778 ± 0.106 = 9.2 ± 0.1 Ω 𝑅35 = (

17.10 ± 0.05 ) (43.6 ± 0.3) 86.40 ± 0.05

17.10 0.05 0.05 ) [1 ± ( + )] 86.40 17.10 86.40

=(

= 0.19791666 ± 0.00069323 Ω

= (0.198 ± 0.001)(43.6 ± 0.3) 0.001 0.3 = [(0.198)(43.6)] [1 ± ( + )] 0.198 43.6 = 8.6328 ± 0.103 = 8.6 ± 0.1 Ω 𝑅36 = (

11.70 ± 0.05 ) (93.1 ± 0.5) 95.00 ± 0.05

11.70 0.05 0.05 ) [1 ± ( + )] 95.00 11.70 95.00

=(

= 0.12315789 ± 0.00059113 Ω

= (0.123 ± 0.001)(93.1 ± 0.5) 0.001 0.5 = [(0.123)(93.1)] [1 ± ( + )] 0.123 93.1 = 11.4513 ± 0.1546 = 11.5 ± 0.2 Ω

13

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental 𝑅37 = (

5.00 ± 0.05 ) (192.4 ± 0.6) 99.20 ± 0.05

5.00

0.05

0.05

= (99.20) [1 ± (5.00 + 99.20)]

= 0.05040322 ± 0.00052943 Ω

= (0.050 ± 0.001)(192.4 ± 0.6) 0.001 0.6 = [(0.050)(192.4)] [1 ± ( + )] 0.050 192.4 = 9.62 ± 0.222 = 9.6 ± 0.2 Ω

𝑅45 = (

28.90 ± 0.05 ) (21.8 ± 0.1) 73.00 ± 0.05

28.90

0.05

0.05

= (73.00) [1 ± (28.90 + 73.00)]

= 0.39589041 ± 0.00095608 Ω

= (0.396 ± 0.001)(21.8 ± 0.1) 0.001 0.1 = [(0.396)(21.8)] [1 ± ( + )] 0.396 21.8 = 8.6328 ± 0.0614 = 8.6 ± 0.1 Ω

𝑅46 = (

10.30 ± 0.05 ) (71.3 ± 0.3) 92.80 ± 0.05

10.30

0.05

0.05

= (92.80) [1 ± (10.30 + 92.80)]

= 0.11099137 ± 0.00059859 Ω

= (0.111 ± 0.001)(71.3 ± 0.3) 0.001 0.5 = [(0.111)(71.3)] [1 ± ( + )] 0.111 71.3 = 7.9143 ± 0.1046 = 7.9 ± 0.1 Ω

14

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental 𝑅47 = (

3.20 ± 0.05 ) (170.6 ± 0.4) 98.50 ± 0.05

3.20

0.05

0.05

= (98.50) [1 ± (3.20 + 98.50)]

= 0.03248730 ± 0.00052410 Ω

= (0.032 ± 0.001)(170.6 ± 0.4) 0.001 0.4 = [(0.032)(170.6)] [1 ± ( + )] 0.032 170.6 = 5.4592 ± 0.1834 = 5.5 ± 0.2 Ω

𝑅56 = (

15.00 ± 0.05 ) (49.5 ± 0.2) 87.30 ± 0.05

15.00

0.05

0.05

= (87.30) [1 ± (15.00 + 87.30)]

= 0.17182130 ± 0.00067114Ω

= (0.172 ± 0.001)(49.5 ± 0.2) 0.001 0.4 = [(0.172)(49.5)] [1 ± ( + )] 0.172 49.5 = 8.514 ± 0.0839 = 8.5 ± 0.1 Ω

𝑅57 = (

4.80 ± 0.05 ) (148.8 ± 0.3) 98.20 ± 0.05

4.80

0.05

0.05

= (98.20) [1 ± (4.80 + 98.20)]

= 0.04887983 ± 0.00053405 Ω

= (0.049 ± 0.001)(148.8 ± 0.3) 0.001 0.3 = [(0.049)(148.8)] [1 ± ( + )] 0.049 148.8 = 7.2912 ± 0.1635 = 7.3 ± 0.2 Ω

15

UNIVERSIDAD NACIONAL DE INGENIERIA Facultad de Ingeniería Ambiental 𝑅67 = (

7.80 ± 0.05 ) (99.3 ± 0.1) 95.10 ± 0.05

7.80

0.05

0.05

= (95.10) [1 ± (7.80 + 95.10)]

= 0.08201892 ± 0.0005688 Ω

= (0.082 ± 0.001)(99.3 ± 0.1) 0.001 0.1 = [(0.082)(99.3)] [1 ± ( + )] 0.082 99.3 = 8.1426 ± 0.1075 = 8.1 ± 0.1 Ω

16