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

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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

    Using MATLAB leads to

    ==