a. V = I . R

15 = I . 100

I = 0,15 Ampere

b. V = I . R

15 = I . 150

I = 0,1 Ampere

a. 6 branches and 5 nodes

b. 7 branches and 5 nodes

-V1 + 1+ 5 = 0 V1 = 6 V

-V2 – 2 + 5 = 0 V2 = 3 V

loop 1 -20 -25 +10 + V1 = 0 V1 = 35 V

loop 2 -10 +15 -V2 = 0 V2 = 5 V

loop 3 -V1 + V2 + V3 = 0 V3 = 30 V

1

2

3

I2 = - 7 + (-3) = -10 A

I1 + I2 - 2 = 0 I1 = 12 A

-I4 + (-4) + 2 = 0 I4 = -2 A

I3 = 7 + (-2) = 5 A

Jadi I1 = 12 A, I2 = -10 A, I3 = 5 A, I4 = -2 A

Rp1 8//12 = 8 𝑥 12

8+12 = 4,8 Ω

Rp2 3//6 = 3 𝑥 6

3+6 = 2 Ω

Rs1 = 4,8 + 1,2 = 6 Ω

Rs2 = 2 + 4 = 6 Ω

Rp3 6//6 = 6 𝑥 6

6+6 = 3 Ω

supernode,

2 = V1 + 2 (V1 - V3) + 8(V2 – V3) + 4V2

2 = 3V1 + 12V2 - 10V3 (1)

V1 = V2 + 2V0 V0 = V2.

V1 = 3V2 (2)

V3 = 13V (3)

Substitusi (2) dan (3) pada (1)

2 = 3V1 + 12V2 - 10V3

2 = 9V2 + 12V2 – 10(13)

2 = 21V2 – 130

132 = 21V2

V2 = 6,286 V

V1 = 3V2

V1 = 3 (6.286) = 18,858 V

jadi

V1= 18,858 V, V2 = 6,286 V, V3 = 13 V

Node 1

Node 2

Node 3

Jadi

V1 = 10 V, V2 = 4,9333 V, V3 = 12,2667 V

V1 V2 V3

V4

Vo = V1 – V4 = 3.8 – 2.2 = 1.6 V

Node 1

Node 2

Node 3

Node 4

[

80000

] = [

21 −20 0 −1−80 98 −8 0

03

−20

56

−2−11

] [

𝑉1𝑉2𝑉3𝑉4

]

B = A V V = A-1 B

V1 = 25,5247 V

V2 = 22,0480 V

V3 = 14,8420 V

V4 = 15,0569 V

Node 1

Node 2

Node 3

Jadi

V1 = 0,0625 V

V2 = 0,375 V

V3 = 1,625 V

Loop 1

Loop 2

Loop 3

I0 = I1 = 4,2857 mA

1

2

3

I0 = I3 = 1,7778 mA

Vab = 30I3 = 53.33 V

1

2

3

–12 + 4I1 – 3I2 – 1I3 = 0 (1)

–3I1 + 7I2 – 4I4 = 0

–3I1 + 7I2 = –12 (2)

–1I1 + 15I3 – 8I4 – 6I5 = 0

–1I1 + 15I3 – 6 = –24 (3)

I4 = –3mA (4)

–6I3 – 8I4 + 16I5 = 0

–6I3 + 16I5 = –24 (5)

I1 = 1.6196 mA

I2 = –1.0202 mA

I3 = –2.461 mA

I4 = 3 mA

I5 = –2.423 mA

-12 + 24 + 30I1 – 4I2 – 6I3 – 2I4 = 0 30I1 – 4I2 – 6I3 – 2I4 = –12

-24 + 40 + 4I1 + 30I2 – 2I4 – 6I5 = 0 –4I1 + 30I2 – 2I4 – 6I5 = –16

–6I1 + 18I3 – 4I4 = 30

–2I1 – 2I2 – 4I3 + 12I4 –4I5 = 0

–6I2 – 4I4 + 18I5 = –32

I1 = -0.2779 A

I2 = -1.0488 A

I3 = 1.4682 A

I4 = -0.4761 A

I5 = -2.2332 A

a. Tranformasi sub sirkuit Y ke ekivalen Δ

vs = 1V, vo = 0.5V, io = 0.5A

b. vs = 10V, vo = 5V, io = 5A

c. vs = 10V and R = 10Ω,

vo = 5V, io = 10/(10) = 500 mA

4.9 Use superposition to find vo in the circuit

Rek = 4||(3 + 1) = 2Ω

io = [20/(2 + 2)] = 5 A

i1 = io/2 = 2.5 A

Rek = 2||(1 + 3) = 4/3 Ω

vo’ = [(4/3)/((4/3) + 4)](-16) = -4 V

i3 = vo’/4 = -1 A

Rp = 2||4 = 4/3 Ω

Rek = 3 + 4/3 = 13/3 Ω

I2 = [1/(1 + 13/2)]2 = 3/8 = 0.375

I = 2.5 + 0.375 - 1 = 1.875 A

P = I2R = (1.875)2.3 = 10.55 W

Rth = 20 + 10||40 = 20 + 400/50 = 28 Ω

Node 1

(40 – V1)/10 = 3 + [(V1 – V2)/20] + V1/40, 40 = 7V1 – 2V2

Node 2

3 + (V1- V2)/20 = 0 V1 = V2 – 60

V1 = 32 V, V2 = 92 V, VTH = V2 = 92 V

4.37 Find the Norton equivalent with respect to terminals a-b in the circuit

R = 5//(14 + 6) = 4Ω = RN

RTH = RN = 4Ω

VTH = - 8 V

IN = -2 A

RN = (6 + 6)||4 = 3 Ω

IN = 4/2 = 2 A