acoplar en laboratorio.docx - 5 Calcular el valor de la resistencia equivalente de los circuitos 01 y 02 del paso 11 En serie R2 R5 y R6 Req 1 =R 2 R 5

acoplar en laboratorio.docx - 5 Calcular el valor de la...

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5.- Calcular el valor de la resistencia equivalente de los circuitos 01 y 02 del paso 11. En serie R2, R5 y R6: R eq 1 = R 2 + R 5 + R 6 R eq 1 = 18 + 330 + 820 R eq 1 = 1168 En paralelo R eq 1 y R4 R eq 2 = R eq 1 ×R 4 R eq 1 + R 4 R eq 2 = 1168 × 47 1168 + 47 R eq 2 = 45.18 En serie R1, R eq 2 y R7 R eq = R 1 + R eq 2 + R 7 R eq = 1000 + 45.18 + 18 R eq = 1063.18
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Aplicando la conversión de triangulo a estrella en las resistencias R4, R7, R9 y R5, R8 y R10. Ambos triángulos tienen los mismos valores de resistencias, por lo tanto: R A = 47 × 18 47 + 18 + 330 = 2.14 R B = 18 x 330 47 + 18 + 330 = 15.03 47 + 18 + 330 ¿ = 39.26 R C = 330 × 47 ¿
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Para R5, R8 y R10 R A 2 = 1000 × 18 1000 + 18 + 1200 = 8.11 R B 2 = 18 x 1200 1000 + 18 + 1200 = 9.73 R C 2 = 1200 x 1000 1000 + 18 + 1200 = 541.02 En serie R B 1 , R6 y R B 2 R eq 1 = R B 1 + R 5 + R B 2 R eq 1 = 15.03 + 1200 + 9.73 R eq 1 = 1224.76 En serie R C 1 , R2 y R C 2 R eq 2 = R C 1 + R 2 + R C 2 R eq 2 = 39.26 + 56000 + 541.02 R eq 2 = 56580.28 En paralelo
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Unformatted text preview: R eq 1 y R eq 2 R eq 3 = R eq 1 ×R eq 2 R eq 1 + R eq 2 R eq 3 = 1224.76 × 56580.28 1224.76 + 56580.28 R eq 3 = 1198.81 Ω En serie R eq = R 1 + R eq 3 + R 3 R eq = 1000 + 1198.81 + 12000 R eq = 14198.81 Ω = 14.19 kΩ 6.- Calcular el valor de la resistencia equivalente de los eléctricos 01 y 02 con la resistencia calculada y completar en la tabla 05 cuadro de divergencias. R eq 1 R eq 2 R T ( Ω ) 1063.18 Ω 14.19 kΩ R exp ( Ω ) 1000.54 Ω 13.97 kΩ E A 62.64 Ω 0.22 kΩ E R 0.05 0.015 E R % 5 % 1.5% TABLA 05 : Circuitos en el protoboard...
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