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