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### Transcript of Chapter 5, Problem 1 - test bank and solution manual .Chapter 3, Solution 6 i1 + i2 + i3 = 0 0 2

• Chapter 3, Solution 1.

8

v2

2

40 v1

6 A 10 A

At node 1, 6 = v1/(8) + (v1 - v2)/4 48 = 3v1 - 2v2 (1) At node 2, v1 - v2/4 = v2/2 + 10 40 = v1 - 3v2 (2) Solving (1) and (2), v1 = 9.143V, v2 = -10.286 V

P8 = ( )

==8

143.98v 221 10.45 W

P4 = ( )

=4vv 221 94.37 W

P2 = ( )

==

=2286.10

2v 212 52.9 W

Chapter 3, Solution 2

At node 1,

2

vv6

5v

10v 2111 +=

60 = - 8v1 + 5v2 (1)

At node 2,

2

vv634

v 212 ++= 36 = - 2v1 + 3v2 (2)

Solving (1) and (2), v1 = 0 V, v2 = 12 V

• Chapter 3, Solution 3 Applying KCL to the upper node,

10 = 60v

230v

20v

10v 0oo +++0 + v0 = 40 V

i1 = =100v 4 A , i2 = =20

v0 2 A, i3 = =30v0 1.33 A, i4 = =60

v0 67 mA

Chapter 3, Solution 4 At node 1, 4 + 2 = v1/(5) + v1/(10) v1 = 20 At node 2, 5 - 2 = v2/(10) + v2/(5) v2 = 10 i1 = v1/(5) = 4 A, i2 = v1/(10) = 2 A, i3 = v2/(10) = 1 A, i4 = v2/(5) = 2 A Chapter 3, Solution 5 Apply KCL to the top node.

k4

vk6v20

k2v30 000 =

+

v0 = 20 V

i1 10 10

2A

i4 i3

v1 v2

i2

4 A 5 5 5 A

• Chapter 3, Solution 6

i1 + i2 + i3 = 0 0210v

6v

412v 002 =

++

or v0 = 8.727 V

Chapter 3, Solution 7 At node a,

babaaa VV

VVVV3610

10153010

=

+=

(1)

At node b,

babbba VV

VVVV72240

59

2012

10==

+

+

(2)

Solving (1) and (2) leads to Va = -0.556 V, Vb = -3.444V Chapter 3, Solution 8

i2

5

2

i13 v1 i3

1

+

4V0

3V+

+

V0

i1 + i2 + i3 = 0 05v4v

13v

5v 0111 =

+

+

But 10 v52v = so that v1 + 5v1 - 15 + v1 - 0v5

81 =

or v1 = 15x5/(27) = 2.778 V, therefore vo = 2v1/5 = 1.1111 V

• Chapter 3, Solution 9

+ v1

i2

6 i13 v1 i3

8 +

+ v0

2v0 12V

+

At the non-reference node,

6

v2v8v

3v 011112 += (1)

But -12 + v0 + v1 = 0 v0 = 12 - v1 (2) Substituting (2) into (1),

6

24v38v

3v 111 +=12 v0 = 3.652 V

Chapter 3, Solution 10 At node 1,

8v4

1vv 112 += 32 = -v1 + 8v2 - 8v0 (1)

1

4A 2i0

i0

v1 v0

8 2

v2 4

• At node 0,

00 I2

2v

+=4 and 8vI 10 = 16 = 2v0 + v1 (2)

At node 2,

2I0 = 4v

1v 212 +v and

8v1

0 =I v2 = v1 (3)

From (1), (2) and (3), v0 = 24 V, but from (2) we get

io = 624242

22v4 o

==

= - 4 A

Chapter 3, Solution 11

i3

6

vi1 i24 3

+

10 V 5 A Note that i2 = -5A. At the non-reference node

6v5

4v10

=+ v = 18

i1 = =4

v10 -2 A, i2 = -5 A

Chapter 3, Solution 12

i3

40

v1 v210 20

+ 5 A

50 24 V

• At node 1, 40

0v20

vv10

v24 1211 +

=

96 = 7v1 - 2v2 (1)

At node 2, 50v

20vv 221 =+5 500 = -5v1 + 7v2 (2)

Solving (1) and (2) gives, v1 = 42.87 V, v2 = 102.05 V

==40vi 11 1.072 A, v2 = =50

v2 2.041 A

Chapter 3, Solution 13

At node number 2, [(v2 + 2) 0]/10 + v2/4 = 3 or v2 = 8 volts But, I = [(v2 + 2) 0]/10 = (8 + 2)/10 = 1 amp and v1 = 8x1 = 8volts

Chapter 3, Solution 14

1 2

4

v0 v1

5 A

8

20 V +40 V

+

At node 1, 1

v405

2vv 001 =+

v1 + v0 = 70 (1)

At node 0, 8

20v4v

52

vv 0001 ++=+

4v1 - 7v0 = -20 (2)

Solving (1) and (2), v0 = 20 V

• Chapter 3, Solution 15

1 2

4

v0 v1

5 A

8

20 V +40 V

+

Nodes 1 and 2 form a supernode so that v1 = v2 + 10 (1) At the supernode, 2 + 6v1 + 5v2 = 3 (v3 - v2) 2 + 6v1 + 8v2 = 3v3 (2) At node 3, 2 + 4 = 3 (v3 - v2) v3 = v2 + 2 (3) Substituting (1) and (3) into (2),

2 + 6v2 + 60 + 8v2 = 3v2 + 6 v2 = 1156

v1 = v2 + 10 = 1154

i0 = 6vi = 29.45 A

P65 = =

== 6

1154Gv

R

221

21v 144.6 W

P55 = =

= 5

1156Gv

222 129.6 W

P35 = ( ) == 3)2(Gvv 223L 12 W

• Chapter 3, Solution 16 2 S

+ v0

13 V +

i0 1 S 4 S

8 Sv1 v2 v3

2 A At the supernode, 2 = v1 + 2 (v1 - v3) + 8(v2 v3) + 4v2, which leads to 2 = 3v1 + 12v2 - 10v3 (1) But

v1 = v2 + 2v0 and v0 = v2. Hence

v1 = 3v2 (2) v3 = 13V (3)

Substituting (2) and (3) with (1) gives, v1 = 18.858 V, v2 = 6.286 V, v3 = 13 V Chapter 3, Solution 17

60 V

i0

3i0

2

10

4

60 V +

8

• At node 1, 2

vv8v

4v60 2111 += 120 = 7v1 - 4v2 (1)

At node 2, 3i0 + 02vv

10v60 212 =+

But i0 = .4v60 1

Hence

( )

02

vv10

v604

v603 2121 =++ 1020 = 5v1 - 12v2 (2)

Solving (1) and (2) gives v1 = 53.08 V. Hence i0 = =4

v60 1 1.73 A

Chapter 3, Solution 18

+ v3

+ v1

10 V

+ v1 v2

2 2

4 8

v3

5 A

(a) (b)

At node 2, in Fig. (a), 5 = 2

vv2

vv 3212 + 10 = - v1 + 2v2 - v3 (1)

At the supernode, 8v

4v

2vv

2vv 313212 +=

+

40 = 2v1 + v3 (2)

From Fig. (b), - v1 - 10 + v3 = 0 v3 = v1 + 10 (3) Solving (1) to (3), we obtain v1 = 10 V, v2 = 20 V = v3

• Chapter 3, Solution 19

At node 1,

32112131 4716

48235 VVVVVV

VV=+

+

+= (1)

At node 2,

32132221 270

428VVV

VVVVV+=

+=

(2)

At node 3,

32132313 724360

42812

3 VVVVVVVV

+==

+

+

+ (3)

From (1) to (3),

BAVVVV

=

=

360

16

724271417

3

2

1

Using MATLAB,

V 267.12 V, 933.4 V, 10267.12933.410

3211 ===

== VVVBAV

Chapter 3, Solution 20 Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence

040414 321

321 =++=++ VVVVVV (1)

. V1 . V2 2 V3 4 1 4

• Between nodes 1 and 3,

12012 1331 ==++ VVVV (2) Similarly, between nodes 1 and 2, (3) iVV 221 +=But i . Combining this with (2) and (3) gives 4/3V=

.V (4) 2/6 12 V+= Solving (1), (2), and (4) leads to

V15 V,5.4 V,3 321 === VVV Chapter 3, Solution 21

4 k

2 k

+ v0

3v0v1 v2v3

1 k

+

3 mA

3v0

+ v2

+ v3

+

(b) (a) Let v3 be the voltage between the 2k resistor and the voltage-controlled voltage source. At node 1,

2000

vv4000

vv10x3 31213

+

= 12 = 3v1 - v2 - 2v3 (1)

At node 2,

1v

2vv

4vv 23121 =

+

3v1 - 5v2 - 2v3 = 0 (2)

Note that v0 = v2. We now apply KVL in Fig. (b) - v3 - 3v2 + v2 = 0 v3 = - 2v2 (3) From (1) to (3), v1 = 1 V, v2 = 3 V

• Chapter 3, Solution 22

At node 1, 8

vv3

4v

2v 011012 ++=

24 = 7v1 - v2 (1)

At node 2, 3 + 1

v5v8

vv 2221 +=

But, v1 = 12 - v1 Hence, 24 + v1 - v2 = 8 (v2 + 60 + 5v1) = 4 V 456 = 41v1 - 9v2 (2) Solving (1) and (2), v1 = - 10.91 V, v2 = - 100.36 V

Chapter 3, Solution 23

At the supernode, 5 + 2 = 5v

10v 21 + 70 = v1 + 2v2 (1)

Considering Fig. (b), - v1 - 8 + v2 = 0 v2 = v1 + 8 (2) Solving (1) and (2), v1 = 18 V, v2 = 26 V

+ v1

5 10

v1 v2

8 V

+

5 A 2 A

+

v2

(b)(a)

• Chapter 3, Solution 24

6mA 1 k 2 k 3 k V1 V2 + io - 30V 15V - 4 k 5 k + At node 1,

212111 2796

246

130 VVVVVV =++= (1)

At node 2,

211222 311530

253)15(

6 VVVVVV +=+=+ (2)

Solving (1) and (2) gives V1=16.24. Hence io = V1/4 = 4.06 mA Chapter 3, Solution 25 Using nodal analysis,

1

v0

2

4

2

10V +

+

40V +

i0

20V

40v

2v10

2v40

1v20 0000 =

+

+

v0 = 20V

i0 = 1

v20 0 = 0 A

• Chapter 3, Solution 26 At node 1,

32121311 24745

5103

2015 VVVVV

VVV=

+

+=

(1)

At node 2,

554

532221 VVVIVV o =

+

(2)

But 10

31 VVIo

= . Hence, (2) becomes

321 31570 VVV += (3) At node 3,

32132331 52100

5510

103 VVV

VVVVV+==

+

+

+ (4)

Putting (1), (3), and (4) in matrix form produces

BAVVVV

=

=

10045

5213157247

3

2

1