Network Elements and the Concept of Circuit


Network Elements and the Concept of Circuit

  1. Consider the following network:

    Which one of the following is the differential equation of V in the above network?









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    The given network

    Applying KCL we get,
    VG = iC

    or VG = C
    dv
    dt

    or C
    dv
    – GV = 0
    dt

    Hence alternative (C) is the correct choice.

    Correct Option: C

    The given network

    Applying KCL we get,
    VG = iC

    or VG = C
    dv
    dt

    or C
    dv
    – GV = 0
    dt

    Hence alternative (C) is the correct choice.


  1. Consider the following circuit:

    What is the current I in the above circuit?









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    The given circuit

    Applying superposition theorem.
    Case 1: When 60V source is taken, the equivalent circuit of the given circuit becomes.

    Req = 10 + {20 || (10 + 20 || 10)}

    or Req = 10 + 20 ||
    50
    3

    or Req = 10 + 20 ×
    50
    3
    20 +
    50
    3

    =
    210
    = Ω
    11

    Ieq =
    V
    =
    60
    =
    60 × 11
    =
    22
    A
    VReq210/112107

    Let the current in the branch AB be I1, then
    I1 = Iq ×
    20
    20 +
    50
    3


    or I1 =
    22
    ×
    3 × 20
    =
    12
    = = 1.71 A (A to B)
    71107

    Case 2: When 15 A source is taken, the equivalent circuit of the given circuit becomes.

    Let the current in the branch BA be I2.

    Where, R1 = (10||20+10)||20
    or R1 = 100/7 Ω
    IR1 = 15 ×
    4
    =
    15 × 4
    4 + 6 + R14 + 6 + 100/7


    or IR1 =
    15 × 4 × 7
    = 2.47 A.
    170

    and
    I2 = IR1 ×
    20
    20 +
    50
    3


    or I2 =
    2.47 × 20 × 3
    =1.34A (B→A)
    110

    Now, I = I1 – I2 = 1.71 – 1.34 = 0.37 A.

    Correct Option: C

    The given circuit

    Applying superposition theorem.
    Case 1: When 60V source is taken, the equivalent circuit of the given circuit becomes.

    Req = 10 + {20 || (10 + 20 || 10)}

    or Req = 10 + 20 ||
    50
    3

    or Req = 10 + 20 ×
    50
    3
    20 +
    50
    3

    =
    210
    = Ω
    11

    Ieq =
    V
    =
    60
    =
    60 × 11
    =
    22
    A
    VReq210/112107

    Let the current in the branch AB be I1, then
    I1 = Iq ×
    20
    20 +
    50
    3


    or I1 =
    22
    ×
    3 × 20
    =
    12
    = = 1.71 A (A to B)
    71107

    Case 2: When 15 A source is taken, the equivalent circuit of the given circuit becomes.

    Let the current in the branch BA be I2.

    Where, R1 = (10||20+10)||20
    or R1 = 100/7 Ω
    IR1 = 15 ×
    4
    =
    15 × 4
    4 + 6 + R14 + 6 + 100/7


    or IR1 =
    15 × 4 × 7
    = 2.47 A.
    170

    and
    I2 = IR1 ×
    20
    20 +
    50
    3


    or I2 =
    2.47 × 20 × 3
    =1.34A (B→A)
    110

    Now, I = I1 – I2 = 1.71 – 1.34 = 0.37 A.



  1. Which one of the following statements is correct? In a four-branch parallel circuit, 50 mA current flows in each branch. If one of the branches opens, the currents in the other branches—









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    Since voltage across the branches remains unaltered, therefore current in other branches are unaffected after one of the branch opens.

    Correct Option: C

    Since voltage across the branches remains unaltered, therefore current in other branches are unaffected after one of the branch opens.


  1. Consider the following waveform diagram:

    Which one of the following gives the correct description of the waveform shown in the above diagram?









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    The given waveform

    f(t) = f1(t) + f2(t) + f3(t)
    Where,

    On combining the above three waveform, we get the required waveform i.e.
    f(t) = u(t) + u(t – 1) + (t – 2) u (t – 2)

    Correct Option: C

    The given waveform

    f(t) = f1(t) + f2(t) + f3(t)
    Where,

    On combining the above three waveform, we get the required waveform i.e.
    f(t) = u(t) + u(t – 1) + (t – 2) u (t – 2)