Stephanie Carolina Moore 030826023

Paola Vargas Chocron 004189457

Problema #1

RBLA=RBL1+RBL2+RBL3

RA=R+R+R

RA=3 R

RBLB=RBL4+RBL5+RBL6

RBLB=R+R+R

RBLB=3 R

V T=V BLA=V BLB

V T=12V

IBLA=V BLARBLA

IBLA=12V3 R

IBLA=4R

=IBL1=IBL2=IBL 3

BL1=BL2=BL 3=BL4=BL 5=BL6

V BL1=IBL1 x RBL 1

V BL1=4Rx R

V BL1=4V

V BL1=4Rx R

PBL 1=(V BL1 )2

R

PBL 1=(4 )2

R

R=¿1

PBL 1=16W

Problema #2

E=150V

A + D

τ=RC

τ=(270k ) x (10 μF )

τ=2,7 s

A + C

τ=(270k ) x (50 μF )

τ=13,55 s

B + C

τ=(470k ) x (50μF )

τ=23,5 s

B + D

τ=(470k ) x (10μF )

τ=4,7 s

B) t=−[ ln(1−V cE )] x τt=−[ ln(1−100150 )] x 2,7

t=¿2,97s

Problema #3

R1=1RBL1

+ 1RBL2

R1=R2

R2=1RBL3

+ 1RBL 4

+ 1RBL5

+ 1RBL6

R2=R4

R3=1RBL7

+ 1RBL8

R3=R2

V=I x R

V 1=30 xR2

V 1=15VR

V 2=15 xR4

V 2=3,75VR

V 3=15 xR2

V 3=7,5VR

P=V2

R

P1=(15 R)2

R2

P1=450RW

P2=(3,75 R)2

R4

P2=56,3 RW

P3=(7,5 R)2

R2

P3=113RW

Potencia del limpia parabrisas

P=V x I

P=12 x20

P=240W

Potencia de las puertas

P=V x I

P=12 x30

P=360W

Corriente de la batería

60+20+30+30= 140A