Network theory miscellaneous


  1. In figure, the value of the source voltage is—











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    The given figure

    Applying KCL at node C, we get

    =
    VC - 0
    = 1 + 2
    10

    or
    VC = 30V
    and
    =
    Vs - VC
    = 2
    10

    or
    VS = 20 + Vc = 20 + 30
    = 50V
    Hence alternative (C) is the correct choice.

    Correct Option: C

    The given figure

    Applying KCL at node C, we get

    =
    VC - 0
    = 1 + 2
    10

    or
    VC = 30V
    and
    =
    Vs - VC
    = 2
    10

    or
    VS = 20 + Vc = 20 + 30
    = 50V
    Hence alternative (C) is the correct choice.


  1. In figure, the value of R is—











  1. View Hint View Answer Discuss in Forum

    The given circuit:

    Applying KCL at node C,

    =
    VC - VA
    +
    VC - VB
    +
    VC
    = 0
    1412

    or
    Vc
    1
    +
    1
    +
    1
    =
    VA
    +
    VB
    =
    100
    +
    40
    1412141141

    or
    Vc
    1 + 14 + 7
    =
    100 + 40 × 14
    1414

    or
    VC = 30V
    Now, current in 14Ω resistor,
    I14Ω =
    100 - VC
    =
    100 - 30
    1414

    = 5A
    So,
    IR = 10 – 5 = 5A
    and
    R =
    VA - VB
    =
    100 - 40
    IR5

    or
    R = 60/5 ≈ 12Ω
    Hence alternative (D) is most correct choice.

    Correct Option: A

    The given circuit:

    Applying KCL at node C,

    =
    VC - VA
    +
    VC - VB
    +
    VC
    = 0
    1412

    or
    Vc
    1
    +
    1
    +
    1
    =
    VA
    +
    VB
    =
    100
    +
    40
    1412141141

    or
    Vc
    1 + 14 + 7
    =
    100 + 40 × 14
    1414

    or
    VC = 30V
    Now, current in 14Ω resistor,
    I14Ω =
    100 - VC
    =
    100 - 30
    1414

    = 5A
    So,
    IR = 10 – 5 = 5A
    and
    R =
    VA - VB
    =
    100 - 40
    IR5

    or
    R = 60/5 ≈ 12Ω
    Hence alternative (D) is most correct choice.



  1. Assuming ideal elements in the circuit shown below, the voltage Vab will be—











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    The given circuit:

    Applying KVL in the loop, we get,
    Vab – 2 × 1 + 5 = 0
    or
    Vab = 2 – 5
    = – 3V.

    Correct Option: A

    The given circuit:

    Applying KVL in the loop, we get,
    Vab – 2 × 1 + 5 = 0
    or
    Vab = 2 – 5
    = – 3V.


  1. In the circuit shown in the figure, the value of the current i will be given by—











  1. View Hint View Answer Discuss in Forum

    The given circuit:

    Va = 5 ×
    1
    =
    5
    V
    1 + 12

    Vb = 4Vab ×
    1
    =
    Vab
    1 + 34

    Vab = Va – Vb = 5/2 – Vab
    or
    2Vab = 5/2
    or
    Vab = 5/4 V
    or
    Vab = 1.25 V
    and
    i =
    4 × Vab
    =
    4 × 1.25
    = 1.25A
    44

    Hence alternative (B) is the correct choice.

    Correct Option: A

    The given circuit:

    Va = 5 ×
    1
    =
    5
    V
    1 + 12

    Vb = 4Vab ×
    1
    =
    Vab
    1 + 34

    Vab = Va – Vb = 5/2 – Vab
    or
    2Vab = 5/2
    or
    Vab = 5/4 V
    or
    Vab = 1.25 V
    and
    i =
    4 × Vab
    =
    4 × 1.25
    = 1.25A
    44

    Hence alternative (B) is the correct choice.



  1. What is the Thevenin resistance seen from the terminals AB of the circuit shown above in the figure?











  1. View Hint View Answer Discuss in Forum

    The given circuit:

    The equivalent circuit for calculation of Rth is shown below—

    Now, from above figure

    1
    =
    1
    +
    1
    +
    1
    Rth4612

    or
    1
    =
    3 + 2 + 1
    Rth12

    or
    Rth =
    12
    = 2Ω
    6

    Hence alternative (A) is the correct choice.

    Correct Option: A

    The given circuit:

    The equivalent circuit for calculation of Rth is shown below—

    Now, from above figure

    1
    =
    1
    +
    1
    +
    1
    Rth4612

    or
    1
    =
    3 + 2 + 1
    Rth12

    or
    Rth =
    12
    = 2Ω
    6

    Hence alternative (A) is the correct choice.