Network Elements and the Concept of Circuit


Network Elements and the Concept of Circuit


  1. For the network shown above, match List I (yparameter) with List II (Value) and select the correct answer using the code given below the Lists:
    List I (y-parameter)List II (Value)
    A. Y111. s + 1
    B. Y122. –1
    C. Y213. 1 + 1/s
    D. Y224. s









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    The given network.

    From above figure
    V1 = (I1 – Ia) s
    or V1 = I1 s – Ia s ....(i)
    Again V1 = Ia 1 + V2 ....(ii)
    and V1 = Ia 1 + (I2 + Ia) 1/s

    or V1 = 1 +
    1
    Ia +
    I2
    ....(iii)
    ss

    From equation (ii)
    Ia = V1 – V2
    On putting the value of Ia in equation (i)
    V1 = I1 s – (V1 – V2) s

    or V1 =
    1 + s
    V2 s ....(iv)
    s

    On comparing equation (iv) with standard equation
    I1 = V1 Y11 + V2 Y12, we get
    Y11 =
    1 + s
    ; Y12 = s
    s


    and V1=1 +
    1
    V1 – V2 +
    I2
    ss

    I2 = s 1 – 1 –
    1
    V1 + V21 +
    1
    s
    ss

    or I2 = – V1 + V2 (s + 1) ....(v)
    On comparing equation (v) with standard equation
    I2 = – V1 Y21 + V2 Y22, we get
    Y21 = – 1; Y22 = s + 1
    Hence alternative (C) is the correct choice.

    Correct Option: C

    The given network.

    From above figure
    V1 = (I1 – Ia) s
    or V1 = I1 s – Ia s ....(i)
    Again V1 = Ia 1 + V2 ....(ii)
    and V1 = Ia 1 + (I2 + Ia) 1/s

    or V1 = 1 +
    1
    Ia +
    I2
    ....(iii)
    ss

    From equation (ii)
    Ia = V1 – V2
    On putting the value of Ia in equation (i)
    V1 = I1 s – (V1 – V2) s

    or V1 =
    1 + s
    V2 s ....(iv)
    s

    On comparing equation (iv) with standard equation
    I1 = V1 Y11 + V2 Y12, we get
    Y11 =
    1 + s
    ; Y12 = s
    s


    and V1=1 +
    1
    V1 – V2 +
    I2
    ss

    I2 = s 1 – 1 –
    1
    V1 + V21 +
    1
    s
    ss

    or I2 = – V1 + V2 (s + 1) ....(v)
    On comparing equation (v) with standard equation
    I2 = – V1 Y21 + V2 Y22, we get
    Y21 = – 1; Y22 = s + 1
    Hence alternative (C) is the correct choice.


  1. Consider the following statements: Superposition theorem is applicable to a linear network in determining
    1. The current responses
    2. The voltage responses
    3. The power responses.
    Which of the statements given above are correct?









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    Superposition theorem is applicable to a linear network in determining the current responses and voltage responses and not power responses because power is a square of either voltage or current.
    Hence alternative (A) is the most appropriate choice.

    Correct Option: A

    Superposition theorem is applicable to a linear network in determining the current responses and voltage responses and not power responses because power is a square of either voltage or current.
    Hence alternative (A) is the most appropriate choice.



  1. If two identical 3 A, 4Ω Norton equivalent circuits are connected in parallel with like polarity, the combined Norton equivalent circuit will be—









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    Given,
    IN = 3A
    RN = 4Ω
    When the two identical Norton equivalent circuits are connected in parallel with like polarity then I'N = 3 + 3 = 6A
    and R'N = 4||4 = 2Ω.

    Correct Option: D

    Given,
    IN = 3A
    RN = 4Ω
    When the two identical Norton equivalent circuits are connected in parallel with like polarity then I'N = 3 + 3 = 6A
    and R'N = 4||4 = 2Ω.


  1. Consider the following statements: Network NA in figure (A) can be replaced by the network NB shown in figure (B) when Ic and Rc, respectively, are—









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    Using source transformation

    IC =
    10V
    = 10A
    1 Ω

    RC = 1 Ω

    Correct Option: B

    Using source transformation

    IC =
    10V
    = 10A
    1 Ω

    RC = 1 Ω



  1. For the circuit given above, what is the expression for the voltage V?









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    From given circuit

    iC = C
    dVC
    dt

    VC = iCR
    or    iCR − VC = 0
    or     CR
    dVC
    −VC = 0
    dt

    Correct Option: C

    From given circuit

    iC = C
    dVC
    dt

    VC = iCR
    or    iCR − VC = 0
    or     CR
    dVC
    −VC = 0
    dt