Matemáticas para ingeniería
Enviado por Eduardo Hernández Díaz • 19 de Marzo de 2020 • Biografías • 331 Palabras (2 Páginas) • 177 Visitas
[pic 1]
Matemáticas para ingeniería
Prof. Fernando Uriel Velázquez Delgado
880502
Hernandez Diaz barriga Eduardo
Examen 2.
Ecuaciones a utilizar:
[pic 2]
Resistencia | Valor teórico | Valor real | Desviación | Tolerancia |
1 | 220 Ω | 216 Ω | 1.81% | ∓5Ω |
2 | 220 Ω | 216 Ω | 1.81% | ∓5Ω |
3 | 220 Ω | 217 Ω | 1.36% | ∓5Ω |
Rt | 73.333 Ω | 72 Ω | 1.81% |
𝑑𝑅 =
2202(2202)
2 (−1.818) +[pic 3]
2202(2202)
2 (−1.818) +[pic 4]
2202(2202)
2 (−1.364)[pic 5]
(220(220)(3))
2204
(220(220)(3))
2204[pic 6]
(220(220)(3))
𝑑𝑅 = 2 ([pic 7]
2) (−1.818) +
2 (−1.364) = −0.88602𝛺
(2202(3))
2204
(2202(3))
2204[pic 8][pic 9]
𝑑𝑅 = 2 ([pic 10]
2) (5) +
2 (5) = 1.6667𝛺
(2202(3))
2204
(2202(3))
2204
𝑑𝑅 = 2 ([pic 11]
2) (−5) +
2 (−5) = −1.6667𝛺
(2202(3)) (2202(3))[pic 12]
𝑅 =
𝑅 =
𝑅 =
220
[pic 13]
3
220
[pic 14]
3
220
[pic 15]
3
𝛺 − 0.88602𝛺 = 72.4473𝛺
𝛺 + 1.667𝛺 = 75𝛺
𝛺 − 1.667𝛺 = 71.6667𝛺
Resistencia | Valor teórico | Valor real | Desviación | Tolerancia |
1 | 220 Ω | 219 Ω | .45% | ∓5Ω |
2 | 330 Ω | 331 Ω | .30% | ∓5Ω |
3 | 22000 Ω | 218000 Ω | .90% | ∓5Ω |
Rt | 116.6 Ω | 115 Ω | .60% |
𝑑𝑅 =[pic 16]
3302220002
2 (.45) +
2202220002
2 (.30)
(220(330) + 220(22000) + 330(22000))[pic 17]
22023302
+[pic 18]
(220(330) + 220(22000) + 330(22000))
2 (.90) =
𝑑𝑅 =
(330(220) + 330(1000) + 220(1000))
3302220002
2 (5) +[pic 19][pic 20]
100023302
2 (5)
(220(330) + 220(22000) + 330(22000))
33022202
+[pic 21]
(220(330) + 220(22000) + 330(22000))
2 (5) =
𝑑𝑅 =
(220(330) + 220(22000) + 330(22000))
220210002
2 (−5) +[pic 22][pic 23]
100023302
2 (−5)
(220(330) + 220(22000) + 330(22000))
33022202
+[pic 24]
(220(330) + 220(22000) + 330(22000))
2 (−5) =
(220(330) + 220(22000) + 330(22000))
𝑑𝑅 =.1600+.0474+.000032 = .2074𝛺𝑅+ = 131.21+.2074= 131.41𝛺[pic 25]
R-=131.21-.2074=131.0074
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