Utilizar analisis nodal para determinar el valor de K que permite que Vy = 0 V.R1 = 5, R2 =4, R3 = 3, R4 = 4 , Vs = 6V, Is = 2A

−vx+vsR1

+ −vxR4

+ vy−vxR2

= 0

vx =(

vsR1

)(1R1

+ 1R2

+ 1R4

)−1

vx =(65

) (15

+ 14

+ 14

)−1

vx = 127

vx−vyR2

+ Is + −vy+kvxR3

= 0

k =(−Is − vx

R2

)R3

vx

k =(−2− 12

7∗4

)3∗712

k = −4,25

Calcular la potencia suministrada por la fuente Is2. Is1 = 4[A], Is2 = 3[A], Vs1 = 8[V], Vs2 =1[V], Vs3 = 3[V] R1 = 4, R2 = 6, R3 = 2, R4 = 8, R5 = 1, R6 = 3

vs1 − vs3 − Is1R4 + Is2R3 − IR4 − IR3 − IR5 = 0

I = vs1−vs3−Is1R4+Is2R3

R4+R3+R5

I = 8−3−4∗8+3∗28+2+1

= −2111

R1(Is1 + Is2) + vs2 + R2(Is1 + Is2) + R3(Is2 − I)− vs3 = vfvf = 4(4 + 3) + 1 + 6(4 + 3) + 2(3 + 21

11)− 3

vf ∼= 77,818P = vfIs2 ∼= 233,454

Utilizar analisis de malla para determinar la potencia generada por la fuente dependiente 1000i3.Donde i1 = 2 mA, R1 = 5 K, R2 =10 K, Vs1 = 4 V, Vs2 = 6 V

R1(I1 − I2)− vs1 − 100I3 = 0−R1I2 − 100I3 = vs1 −R1I1− 5000I2 − 1000I3 = −6

100I3 − vs2 −R2(I3 + 0,5I2) = 0−R20,5I2 + (1000−R2)I3 = vs2−5000I2 − 9000I3 = 6

I2 = 1,5[mA] ∧ I3 = −1,5[mA]

1

P = 1000I3 ∗ (I2 − I3)P = 1000 ∗ 1,5 ∗ 10−3 ∗ (3∗10−3)P = 4,5[mW ]

Determinar iA en el circuito de la figura. Vs1 = 20 V, R1 = 2.5 , R2 = 3.5 , R3 = 4 , R4 = 3 .

iA = I1 − I3−R4I1 − vs1 −R3I1 + R3I3 = 0− 7I1 + 4I3 = 20

vs1 −R1I2 − 2,5I1+2,5I3 = 0− 2,5I1 − 2,5I2+2,5I3 = −20

R3I1 −R3I3 + 2,5I1−2,5I3 −R2I3 = 06,5I1 − 10I3 = 0

I1 = −5011∧ I2 = 211

22∧I3 = −65

22

iA = −5011− −65

22= −35

22[A]

2