Network Elements and the Concept of Circuit


Network Elements and the Concept of Circuit

  1. Reciprocity theorem is valid for—









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    Reciprocity theorem is valid for all linear, passive and bilateral networks.

    Correct Option: D

    Reciprocity theorem is valid for all linear, passive and bilateral networks.


  1. An ideal voltage source and ideal current source are connected in parallel this circuit has—









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    NA

    Correct Option: B

    NA



  1. For the circuit shown in figure, the voltage Vs will be











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    Current in branch EF = 1 1 amp
    1

    Current in Branch AB =
    VAB
    =
    2
    = 2 amp
    1Ω 1

    Current in branch CA = 1 + 2 = 3 amp.
    Current in CD =
    VCD
    =
    ACA + VAB
    1 1

    = 3 × 1 + 2 × 1
    1

    = 5amp
    Current in branch GC = 5 + 3 = 8 amp
    Now, Vs = VGC + VCD
    = 1 × 8 + 1 × 5
    = 13 V


    Correct Option: D

    Current in branch EF = 1 1 amp
    1

    Current in Branch AB =
    VAB
    =
    2
    = 2 amp
    1Ω 1

    Current in branch CA = 1 + 2 = 3 amp.
    Current in CD =
    VCD
    =
    ACA + VAB
    1 1

    = 3 × 1 + 2 × 1
    1

    = 5amp
    Current in branch GC = 5 + 3 = 8 amp
    Now, Vs = VGC + VCD
    = 1 × 8 + 1 × 5
    = 13 V



  1. The network shown, if iR = (4 e— 3t + 5 e— 4t) amp and i L (0) = 15 amp, then at t = 0 φ would be equal to—











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    From given figure
    IL = IR + I2 cos (t – φ)
    15 = (4e–3t + 5e–4t) + 12 cos (t – φ)
    at t = 0
    15 = 4 + 5 + 12 cos (–φ)
    or 15 – 9 = 12 cos φ
    or 6 = 12 cos φ

    or cos φ = 1
    2

    or φ = π
    3


    Correct Option: B

    From given figure
    IL = IR + I2 cos (t – φ)
    15 = (4e–3t + 5e–4t) + 12 cos (t – φ)
    at t = 0
    15 = 4 + 5 + 12 cos (–φ)
    or 15 – 9 = 12 cos φ
    or 6 = 12 cos φ

    or cos φ = 1
    2

    or φ = π
    3




  1. Which of the following networks in the equivalent of the circuit shown in figure?











  1. View Hint View Answer Discuss in Forum

    This can be solved by using Star to Delta transformation.

    R12 =
    R1R2 + R2R3 + R3R1
    R3

    = J1 × -J1 + (-J1) + (-J1)× (J1)
    -J1

    = J2 + J2- J2 = +J
    -J

    R23 =
    R1R2 + R2R3 + R3R1
    =
    -J2
    = -J
    R1 J

    R31 =
    R1R2 + R2R3 + R3R1
    =
    -J2
    = J
    R2 J


    Correct Option: A

    This can be solved by using Star to Delta transformation.

    R12 =
    R1R2 + R2R3 + R3R1
    R3

    = J1 × -J1 + (-J1) + (-J1)× (J1)
    -J1

    = J2 + J2- J2 = +J
    -J

    R23 =
    R1R2 + R2R3 + R3R1
    =
    -J2
    = -J
    R1 J

    R31 =
    R1R2 + R2R3 + R3R1
    =
    -J2
    = J
    R2 J