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Informe de Laboratorio N°3

Análisis de Datos Experimentales1.-Reconstrua el cuadro de la forma

!"#$ 9 7 6 5 4 3 2 1 0.1

t"s$ 24.8 34.

5

41.

1

47.

8

56.

7

69.

8

87.

8

117 228.

8

!%RI

R v/I

R1 0

 R2 0.1071428571428571428571428

5714286

R3 0.1079136690647482014388489

2086331

R4 0.1083032490974729241877256

3176895

R5 0.1077199281867145421903052

064632R6 0.1074498567335243553008595

9885387

R7 0.1075268817204301075268817

2043011

R8 0.1074718526100307062436028

6591607

V/R I0 0

2.7999999999999999999999999999999

2.8

5.56 5.56

8.3100000000000000000000000000

002

8.31

11.14 11.14

13.96 13.96

16.74 16.74

19.54 19.541 &ra'(ue los resultados de la tabla 1 ) !%f"t$ *+u, tipo de cur#acorresponde al ra'co Escriba su ecuaci/n emp0rica

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0 0.5 1 1.5 2 2.5

0

5

10

15

20

25

0

2.8

5.56

8.31

11.14

13.96

16.74

19.54f(x) = 9.31x - 0.02

R² = 1

Axis itle

Axis itle

I=mV+k -> I= MV

V=RI -> V/R=I

2.-Efectu, análisis de rá'cos *+u, representa la pendiente

 A=n∑ x i y i−∑ xi ∑ y i

n (∑ xi2 )−(∑ x i )

2  B=

∑ y i ∑ xi2−∑ xi∑ x i y i

n (∑ xi2 )−(∑ x i )

2

 A=8(117.129)−(8.4)(78.05)

8 (12.6 )−(8.4 )2  B=

(78.05 ) (12.6 )−(8.4 )(117.129)

8 (12.6 )−(8.4)2

 A=9.30595238 B=-0.015

I=BV+k -> I= BV

I=RI -> V/R=I

3.-&ra'(ue los tados de la tabla 2 *a (ue cur#a la recuerda su ra'coE4RI5A 6 E46A4I7N

Datos ti vi viti t i2

1 24.8 9 223.22 34.5 7 241.53 41.1 6 246.64 47.8 5 2395 56.7 4 226.86 69.8 3 209.47 87.8 2 175.68 117 1 1179 228.8 0.1 22.88

∑total

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I(mA) 0 5 10 15

V(v) 0 0.69 0.73 0.74

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0

2

4

68

10

12

14

16

0

5

10

15

f(x) = 14.48x - 0.32

R² = 0.65

Axis itle

Axis itle

I0=Idsev/0.05

4.-efectue e a!"isis de de #$afico . %&u' $e$ese!ta os a$"met$os esc$i*a su ecuaci+!

I(mA) 0 5 10 15

V(v) 0 0.69 0.73 0.74

 I 0= I ds e

v0

0.05ln I 

0=ln I 

dse

v0

0.05   ln I 0=ln I 

ds+

  v0

0.05 I 0= A v

0+ I 

ds  A=

  1

0.05

 I 0− A v

0= I 

ds  I 

ds

0 0

5-13.8 -8.8

10-14.6 -4.6

15-14.8 0.2

lnI 0

  lnI ds

1.60943791 0

2.30258509 2.174751722.7080502 1.5260563

1.60943791 -1.60943791

Datos ,i i ,ii  xi

2

11.60943791 0

1.609437

910

22.30258509 0

2.302585

090

32.7080502 0

2.708050

20

4

1.60943791 -1.609437911.609437

91

-

1.6094379

1∑total

8.22951111 -1.60943791

-

2.590290

39

17.816014

8

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 A=

n∑ x i y i−∑ xi ∑ y i

n (∑ xi2

)−(∑ x i )2   B=

∑ y i ∑ xi2−∑ xi∑ x i y i

n (∑ xi2

)−(∑ x i )2

 A=4 (−2.59029039 )−(8.22951111 ) (−1.60943791 )

4 (8.22951111 )−(17.8160148 )2

B=(−1.60943791 ) (17.8160148 )−(8.22951111 )(−2.59029039)

4 (8.22951111 )−(17.8160148)2

 A=0.81479453B=-2.07869964

5.-$afiue I=f(v) os $esutados de a ta*a 3 %&u' cue$va e $ecue$da e #$afico sc$i*a su

ecuaci+!

I(A) 0 0.03 0.04 0.05 0.06 0.07 0.0 0.09 0.1

V(v

)

0 1 2 3 4 5 6 7

0 1 2 3 4 5 6 7 8

0

0.02

0.04

0.06

0.08

0.1

0.12

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1f(x) = 0.01x + 0.03

R² = 1

Axis itle

Axis itle

6.- ! cua de os eeme!tos se cume a e de m %e! cua!o e,iue su $esuesta

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La ley de Ohm se cumple e todos los e!pe"#metos $=%&

7.-e,iue as o*se$vacio!es e,e$ime!taes

8A46LAD DE IN&ENIERIAELE4RI4A9 EE4R7NI4A9

:E4ANI4A ; :INA

4ARRERA <R78EI7NAL DE

IN&.ELE4R7NI4A

AI&NA6RA : I!I"# I$$

I%&RM'

%03

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AL6:N7 : *,M "

Docente =&$*

&rupo = 3>>-5

4?DI&7 = @13B>-5

4647 C <ER

2@11