SERIE DE EJERCICIOS

SERIE DE ejercicios

P 2.2

P 3.3

4

ia=(Vac+128)(1/8)=(Vac+128)/8

ib=Vac/48

(ic=Vac-Vcb)/18

I1=(Vac-Vcb)/18

id=Vbc/20

ie=(Vbc-70)/10

ia+ib+ic=0

I+id+ic=0

(Vac+128)/8+Vac/48+(Vac-Vcb)/18= 0

(Vac-Vcb)/18+Vbc/20+(Vbc-70)/10=0

1392Vac-348Vbc=-110592

-2001Vac+270Vbc=25200

Resolviendo ecuaciones

Vac=-75.697v

Vbc=60V

5

Malla 1

110=13I1-8I2-3I3

Malla 2

0=48i2-24i3-8i1

Malla 3

110=29i3-3i1-24i2

I1=(malla 1 sustituyendo voltaje en la columna 1)/(malla resistencia)=17.678ª

I2=(malla 2 sustituyendo voltaje en la columna 2)/(malla resistencia)=9.821A

I3=(malla 3 sustituyendo voltaje en la columna 3)/(malla resistencia)=13.75A

6

I1+I2+I3=0

(Vab-640)/5+Vab/50+(Vab+64)/10=I1

((Vab))/50=I2

(Vab+64)/10=I3

SUMANDO LAS TRES INTENSIDADES Y DESPEJANDO Vab ES IGUAL A ;

Vab=380v

Vth=In*Rn

Vth=12.8*5

Vth=64V

3.3.11

V1=6/(3+3+6) (12)=6v

V2=3/(3+3+6) (12)=3V

i=Vs/(R1+R2+R3)

i=12/(6+3+3)=1A

V3=3/(3+3+6) (12)=3V

Pt=It*Vt

(1┤ )(12┤)=12w

(1┤ )(6┤)=6

(1┤ )(3)=3

Pt=P1+P2+P3

Pt=12w

5.3.4

Vth=9x10 Y VTH=6x15

Vth=90v

MALLA 1

0=10I1

MALLA 2

-90=15I3

I1=0A

I2=-4.5A

I3=6A

MALLA 4

90=10I4 I4=9A

MALLA 5

-90=20I5 I5=-4.5

MALLA 6

0=15I6 I6=0

I2+I5=9A

V=IR

V1=9x20=180v

Ejercicio 10

Malla 1

180= 30(I1-I2)+60(I1-I2)

180=90I1-90I2

Malla 2

0=40I2+20(I2-I3)+80(I2-I3)+30(I2-I1)+60(I2-I1)

0=230I2-100I3-90I1

Malla 3

0=25I3+20(13-I2)+80(I3-I2)

0=125I3-100I2

[■(90&-90&0@-90&230&-100@0&-100&125)][■(180@0@0)]

I1=5

I2=3

I3=3

V0= (3)(80)= 240 V

Ejercicio 11

RA= (50*20)/(50+20+30)=10 Ω

RB=(20*30)/100=6 Ω

RC= (50*30)/100=15 Ω

RT= 40 Ω

DIVISOR DE CORRIENTE

P-3.4.13

IS=V/R1+V/R2

IS=24/10 + 24/R2

R2=V1/I = 24/1.6 = 15Ω

IS= 24/40 + 24/15

IS= 2.2 A

R1= 8/ 2.2= 6.63 Ω

P-5.3.6

Malla 1

-12= 2I1+4I1+12(I1-I2)

-12=18I1-12I2

Malla 2

0=6I2+12(I2-I1)

0=18I2-12I1

18I1-12I2=-12

-12I1+18I2=0

I1=■(12&-12@0&18)/■(18&-12@-12&18)=(216000-0)/(324000-144000)=12mA

I2=■(18&-12@-12&0)/18000=(0-144000)/18000=-8A

I=3.5Ma

P-5.4.1

TEOREMA DE THEVENIN

IN=VTH/RTH= 12/3= 4 A

R12= 1/(1/3 + 1/6)= 2Ω

IT= 4 – 3= 1 A

VTH=INRN

VTH=(1)(2)= 2 V

3.2.4

P_R1=V^2/R=(12)^2/6=24w

P_R2=I^2*R=(3)^2 (8)=72w

P_R3=V^2/R=(20)^2/4=100w

3.2.5

P_R1=I^2*R=(2)^2 (4)=16w

P_R2=I^2*R=(2)^2 (6)=24w

P_R3=V^2/R=(8)^2/8=8w

P 3.2.5 Determinar la potencia

3.2.13

Malla # 1

15=60(I_1-I_2 )

15=60I_1-60I_2

Malla # 2

I_2=0.75 Amp.

Malla # 3

I_3=0.25Amp.

〖60 I〗_1- 60(0.75A)=15

〖60 I〗_1=15+45

〖60 I〗_1=60

I_1=60⁄60

I_1=1Amp.

I_1-I_2=0.25A

V_R1=(0.25A)(60)=15v

V_R2= (0.75A)(10)=7.5v

I_2-I_3=1A

V_R3=(1)(20)=20v

3.2.14 Determine el voltaje y la corriente de cada elemento del circuito

Malla # 1

5-15=25I_1

I_1= 10⁄25=0.4 Amp.

I_2=1.5 Amp.

I_3=0.5 Amp.

V_R1=(0.4)(25)=10v

I_2-I_1=1 Amp.

V_R1=(1)(10)=10v

V_R1=(1.5)(10)=15v

V_R1=(0.5)(10)=5v

5.3.7

R_12=1/(1/100+1/100)=50Ω

R_123=50+100=150Ω

R_T=150Ω

I=V/R=10/150=66.66mA

V=(100)(66.666mA)=6.666 mA V_A=6.666v

R_12=1/(1/100+1/100)=50Ω

R_T=150Ω

I=V/R=8/150=53.333mA

V=(100)(53.333mA)=5.333v

V_TH=I_N*R_N

V_TH=(30mA)(100)=3v

R_12=1/(1/100+1/100)=50Ω

R_T=150Ω

I=V/R=3/150=0.02A

V=(100)(0.02)=2v

DIVISOR DE VOLTAJE---------3.3.1 Determinar los voltajes

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