Si la resistencia eléctrica de un tramo conductor de cobre es de 15 Ω a 30° C. ¿Cuál será el valor de la resistencia si la temperatura se incrementa a 90° C? T inicial + 234.5 °C R inicial = T final +234.5 °C R final R final = ( T final +234.5 °C T inicial +234.5 °C ) R inicial R final = ( 90 °C+ 234.5 °C 30 °C+ 234.5 °C ) ( 15 Ω ) R final =18.4026 Ω
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Si la resistencia eléctrica de un tramo conductor de cobre es de 15 Ω a 30° C. ¿Cuál será el valor de la resistencia si la temperatura se incrementa a 90° C?
T inicial+234.5 °CRinicial
=T final+234.5 ° C
Rfinal
R final=( T final+234.5 °CT inicial+234.5 ° C )RinicialR final=( 90 ° C+234.5 ° C
30 °C+234.5 °C )(15Ω)
R final=18.4026Ω
Lecturas:
VO =127 V
IO = 0.075 A
PO=14 watts
Admitancia:
Y exc=I excV n
=0.075 A127V
=0.0005905
fp=POV n I exc
= 14watts(127V )(0.075 A )
=1.4698
θ=cos−1POV n I exc
=cos−1 (1.4698 )=¿¿
Lecturas:
Vcc =18 V
Icc = 13.5 A
Pcc=140 watts
a= 2 a 1
Zcc=V cc
Icc= 18V13.5 A
=¿
fp=PccV cc I cc
= 140watts(18V )(13.5 A)
=0.5761
θ=cos−1PccV cc I cc
=cos−1 (0.5761 )=¿54.8232 °¿
Zcc=V cc
Icc∨θ∨¿1.333Ω[54.8232 ° ]
Zcc=RT+J X T=0.76794Ω+J (1.08956Ω)
Por lo tanto:
R x' =RT2
=0.76794Ω2
=0.38397Ω
R x=0.38397Ω
0.52=1.53588Ω
X x' =XT2
=1.08956Ω2
=0.54478Ω
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