Realiza las siguientes actividades en tu cuaderno o mediante software, de tal manera que expongas el procedimiento:
Respuestas
Respuesta:
D) Realiza las siguientes actividades en tu cuaderno o mediante software, de tal manera que expongas procedimiento:
En el siguiente circuito, son seis resistores de
se conectan de forma mixta serie-paralelo con una pila de 120 volt. Hallar:
¿Cuál es la resistencia equivalente?
(1/(1/3+1/4+1/4 ))+4+4+2=R_eq=11.2Ω
¿Cuál es el valor de la corriente total?
I=V/R ∴I=120/11.2 →I=10.7142 Amp
c)¿Cuál es la diferencia de potencial de cada resistor
〖V_R〗_1=〖V_R〗_2=〖V_R〗_3=(1.2 Ω)(10.7142 Amp)=12.85704 v
〖V_R〗_4=(5 Ω)(10.7142 Amp)=53.571 v
〖V_R〗_5=(4 Ω)(10.7142 Amp)=42.8568 v
〖V_R〗_6=(2 Ω)(10.7142 Amp)=21.4284 v
d)¿Cuál es la intensidad de corriente en cada resistor?
〖I_R〗_4=〖I_R〗_5=〖I_R〗_6=10.7145 Amp
〖I_R〗_1=12.8504v/(3 Ω)=4.2834 Amp
〖I_R〗_2=〖I_R〗_3=12.8504v/(4 Ω)=3.2426 Amp
e)¿Cuál es la potencia consumida por cada resistor?
〖W_R〗_1=(12.85704 v)(4.2834 Amp)=55.0718 Watts
〖W_R〗_2=〖W_R〗_3=(12.85704 v)(3.2426 Amp)=41.6902 Watts
〖W_R〗_4=(53.571 v)(10.7145 Amp)=573.9864 Watts
〖W_R〗_5=(42.8568 v)(10.7145 Amp)=459.1891 Watts
〖W_R〗_6=(21.4284 v)(10.7145 Amp)=229.5945 Watts
f) ¿Cuál es la energía consumida por cada resistor?
〖E_R〗_1=(55.0718 Watts)(4.2834 Amp)=235.8945 J
〖E_R〗_2=〖E_R〗_3=(41.6902 Watts)(3.2426 Amp)=135.1846 J
〖E_R〗_4=(573.9864 Watts)(10.7145 Amp)=6.150 KJ
〖W_R〗_5=(459.1891 Watts)(10.7145 Amp)=4.920 KJ
〖W_R〗_6=(229.5945 Watts)(10.7145 Amp)=2.460 KJ
En el siguiente circuito, son siete resistores de
se conectan de forma mixta serie-paralelo con una pila de 110 volt. Hallar:
a) ¿Cuál es la resistencia equivalente?
(1/(1/1+1/4))+(1/(1/3+1/3))+8+12+16=R_eq=38.3 Ω
b) ¿Cuál es el valor de la corriente total?
I=(110 v)/(38.3 Ω)=2.8720 Amp
c)¿Cuál es la diferencia de potencial de cada resistor?
V_(R_1 )= V_(R_2 )=(0.8 Ω)(2.8720 Amp)=2.2976 v
V_(R_3 )= V_(R_4 )=(1.5 Ω)(2.8720 Amp)=4.308 v
V_(R_5 )= (8 Ω)(2.8720 Amp)=22.976 v
V_(R_6 )= (12 Ω)(2.8720 Amp)=34.464 v
V_(R_7 )= (16 Ω)(2.8720 Amp)=45.952 v
d)¿Cuál es la intensidad de corriente en cada resistor?
I_(R_5 )=I_(R_6 )=I_(R_7 )=2.8720 Amp
I_(R_1 )=((2.2976v))/((1 Ω) )=2.2976 Amp
I_(R_2 )=((2.2976v))/((4 Ω) )=0.5744 Amp
I_(R_3 )=I_(R_4 )=((4.308v))/((3 Ω) )=1.436 Amp
e)¿Cuál es la potencia consumida por cada resistor?
W_(R_1 )=(2.2976 v)(2.2976 Amp)=5.2789 Watts
W_(R_2 )=(2.2976 v)(0.5744 Amp)=1.3197 Watts
W_(R_3 )=W_(R_4 )=(4.308 v)(1.436 Amp)=6.1862 Watts
W_(R_5 )=(22.976 v)(2.8720 Amp)=68.9870 Watts
W_(R_6 )=(34.464 v)(2.8720 Amp)=98.9806 Watts
W_(R_5 )=(45.952 v)(2.8720 Amp)=131.9741 Watts
f) ¿Cuál es la energía consumida por cada resistor?
E_(R_1 )=(5.2789 Watts)(2.2976 Amp)=12.1288 J
E_(R_2 )=(1.3197 Watts)(0.5744 Amp)=0.7580 J
E_(R_3 )=E_(R_4 )=(6.1862 Watts)(1.436 Amp)=8.8833 J
E_(R_5 )=(68.9870 Watts)(2.8720 Amp)=198.1306 J
E_(R_6 )=(98.9806 Watts)(2.8720 Amp)=284.2722 J
E_(R_5 )=(131.9741 Watts)(2.8720 Amp)=379.0296 J
Explicación:
1) a) 1/R₁-₃ = 1/3Ω+1/4Ω+1/4Ω ⇒ R₁-₃ = 6/5Ω
Requiv = 6/5Ω+ 5Ω+4Ω+ 2Ω = 61/5 Ω
b) It= V/Requiv = 120volt/(61/5)Ω = 9.83 A
It= I₁₋₃ = I₄ = I₅ = I₆ = 9.83 A
c) V₁-₃ = It* R₁₋₃ = 9.83 A*6/5Ω = 11.8 volt
V1 = V2 = V3 = 11.8 volt
V4 =I4*R4 =9.83 A*5Ω= 49.15 volt
V5= I5*R5= 9.83 A*4Ω= 39.32 volt
V6= I6*R6= 9.83 A*2Ω= 19.66 volt
d) I1= V1/R1 = 11.8v/3Ω = 3.93 A
I2= V2/R2= 11.8v/4Ω = 2.95 A
I3= V3/R3 = 11.8v/4Ω = 2.95 A
I₄ = I₅ = I₆ = 9.83 A
e) P1 = V1*R1 = 11.8v*3Ω = 35.4 watts
P2 = V2*R2 = 11.8v*4Ω = 47.2watts
P3 = V3*R3 = 11.8v*4Ω = 47.2 watts
P4 = V4*R4 = 11.8v*5Ω = 59 watts
P5 = V1*R1 = 11.8v*4Ω = 47.2 watts
P6 = V1*R1 = 11.8v*2Ω = 23.6 watts
f) W = P*t ; se sustituye cada valor de potencia y se multiplica por el tiempo de funcionamiento t.
2) 1/R₁₋₂ = 1 /1Ω+1/4Ω ⇒ R₁₋₂ = 0.8 Ω
1/R₃₋₄ = 1 /3Ω+1/3Ω ⇒ R₃₋₄ = 1.5 Ω
a) R equiv = 0.8 Ω+ 1.5 Ω +8Ω +12Ω+16Ω = 38.3 Ω
b) It= V/Requiv = 110volt/38.3Ω = 2.87 A
It= I₁₋₂ = I₃₋₄ = I5 = I6 = I7 = 2.87 A
c) V₁₋₂ = It*R₁₋₂ = 2.87 A *0.8 Ω= 2.296 v V1 = V2 = 2.296 v
V₃₋₄ = It*R₃₋₄ = 2.87 A *1.5 Ω= 4.305 v V3 = V4 = 4.305 v
V5= I5*R5= 2.87 A*8Ω= 22.96 volt
V6= I6*R6= 2.87 A* 12Ω= 34.44 volt
V7= I6*R6= 2.87 A*16Ω= 45.92 volt
d) It= I₁₋₂ = I₃₋₄ = I5 = I6 = I7 = 2.87 A
I1= V1/R1 = 2.296 v/ 1Ω = 2.296 A
I2= V2/R2 =2.296 v/ 4Ω = 0.574 A
I3= V3/R3 = 4.305v/ 3Ω = 1.435 A
I4= V4/R4 = 4.305v/ 3Ω = 1.435 A
e) P1 = I1*V1 = 2.296 A*2.296v = 5.27 watts
P2 = I2*V2 = 0.574 A*2.296v = 1.31 watts
P3 = I3*V3= 1.435 A*4.305v = 6.17 watts
P4= I4*V4 = 1.435 A*4.305v = 6.17 watts
P5 = I5*V5 = 2.87 A*22.96v = 65.89 watts
P6= I6*V6= 2.87 A*34.44v = 98.84 watts
P7 = I7*V7 = 2.87 A*45.92v = 131.79 watts
f) W = P*t ; se sustituye cada valor de potencia y se multiplica por el tiempo de funcionamiento t.