2/14/2017

1

problem 1 solution

4Ω5Ω

8Ω

v1

3A6A20Ω

v2i2

i3

i4

i1

Homework #4 solution

5Ω

KCL at : 0;at : 3 6 0;⇒ 9

Express resistor currents:

5;

20;

8;

4Plug into KCL equations:

5 20 83;

8 49;

⇒3 24;

3 72024

problem 2 solution

8A

40Ω10Ω12Ω4Ω

i1

8A

i4i3i2

v1 v2

18A

KCL at , :

Resistor currents:

8 8 16;18 8 10;

4;

12;

10;

40

4 1216;

10 4010

48 ; 80 ;12 ; 4 ;8 ; 2

2/14/2017

2

A.Ai

Vv

Vv

vv

vv

vvv

vvv

v, i

v, i

vi

ivv

iii

66716

2636

4

26

1084

36014

6

3618

060156

3660156

36

18

0

0

2

1

21

21

121

211

22

11

10

021

210

problem 3 solution

+

+

36V

i0

6Ω

18i0

60Ω15Ω

v2v1

i2i1

Due to the floating dependent voltage source, a super-node need to be formed

KCL at super-node:

By the dependent voltage source:

Resistor currents:

Plug in:

1.667

1 2

2 3

1 1 0 0 11 2 3

1 1 1 0

1

01 01 0

3

6

2 1 1 36

3.62 3 13.6

1

x

x

x

i i i

i i i

v v v v vi , i , i , i

v v v vv V

v Vv vv v

problem 4 solution

+

3ix

ix

1Ω

3Ω

2Ω v0i3

i2i1

6V

v1

1Ω

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3

+

2Ω 4A

v1v2

5Ω

90V

4Ω

Figure 4.5

problem 5 solution

i1 i2

v3

There is an additional voltage which is different from ,By voltage division,

Due to the 90 V voltage source, form a super-node

KCL at super-node: 4 0; (1)By the voltage source : 90; (2) Resistor currents: ; , plug into (1)

4 0 (1)

90 (2)Solving (1) and (2):

6 5 120;90

30 ; 60

Then 13

20

Vv

Vv

Vv

vv

vv

vv

vv

vvvv

vvv

v, i

v, i

vv, i

vi

iii

ii

2

2

10

4053

5845

30

10

103

5

3520

230

10

10520

2

3

3

0

2

1

20

21

21

02

2021

211

24

03

212

11

432

21

problem 6 solution

+

+

3A

10V

30Ω5Ω

10Ω20Ω

V0

2V

i4

i3

i2i1

v2v1

10 ;2 ;2

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4

Av

i

Vv

Vv

vv

vv

vvv

v

vvvv

v, i

vv, i

v, i

vi

iii

iii

15

12

17

7

24

48519

412

2

12245

12242

12

5

12

5

10

2

1

21

21

121

2

1121

13

122

21

10

201

320

problem 7 solution

+

5Ω

2Ω

4Ω

2Ω

i0

5i012V

i1

i3

i2

v2

v1

7 ;17 ;1

problem 8 solution

2v0

0

5KΩ5mA

v1 v2

4KΩ

i2

i120KΩ

There is an additional voltage which is different from ,Due to the floating voltage source, form a super-node

: 20

;5

; 4

: 5 ;KCL : 0By the voltage source: 2 2

Put current expressions into KCL equations:

20 55 ;

20 5 40;

2

5 4 100;5 6 4 0;

08 ; 20 , 20

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5

problem 9 solution

+

i1 i2

2i2

15Ω

15Ω

30Ω

45i1

v1i3

15V

v

Assign , Then 45 (1)

Resistor currents: ; ;

KCL at : 2 0 ⇒ 0Put in current expressions: 0

Simplifytoobtain:3 2 15; 2 From (1): 45 3 (3)

Solving (2) and (3) : 5 ; 15 ;

15515

0.333

1530

0.333