Electric circuits miscellaneous


Electric circuits miscellaneous

  1. With the usual notations, a two-port resistive network satisfies the condition A = D =
    3
    B =
    4
    C
    23

    z11 of the network is









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    Z11 =
    A
    =
    4
    C4

    Correct Option: B

    Z11 =
    A
    =
    4
    C4


  1. A two-port network is defined by the relations I1 = 2V1 + V2, I2 = 2V1 + 3V2 Then z12 is









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    From the equations,

    I1 = 2V1 +
    (I2 - 2V1)
    3

    ⇒ V1 = 3I1 -
    I2
    4

    ∴ Z12 =
    1
    Ω
    4

    Correct Option: D

    From the equations,

    I1 = 2V1 +
    (I2 - 2V1)
    3

    ⇒ V1 = 3I1 -
    I2
    4

    ∴ Z12 =
    1
    Ω
    4



  1. An ideal transformer has turns ratio of 2 : 1. Considering high voltage side as port 1 and low voltage side as port 2, then transmission line parameters of transformer will be









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    V1
    =
    I1
    =
    N1
    =
    2
    V2I2N21

    ⇒ V1 = 2V2
    I 1 = 0.5 I2

    In matrix form,

    V1 = 20V2
    I10-0.5I2

    Correct Option: C

    V1
    =
    I1
    =
    N1
    =
    2
    V2I2N21

    ⇒ V1 = 2V2
    I 1 = 0.5 I2

    In matrix form,

    V1 = 20V2
    I10-0.5I2


  1. In a passive two-port network, the open-circuit impedance matrix is 102
    52

    If input port is interchanged with the output port, then open-circuit impedance matrix will be









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    The port-equations are
    V1 = 10 I 1 + 2I 2 ....(i)
    V2 = 2I 1 + 5 I2 ....(ii)
    Rewriting equations (i) and (ii) as
    V2 = 5 I2 + 2I1 ....(iii)
    V1 = 2 I2 + 10 I1 ....(iv)
    Writing in matrix form


    V2 = 52I2
    V1210I1

    Correct Option: B

    The port-equations are
    V1 = 10 I 1 + 2I 2 ....(i)
    V2 = 2I 1 + 5 I2 ....(ii)
    Rewriting equations (i) and (ii) as
    V2 = 5 I2 + 2I1 ....(iii)
    V1 = 2 I2 + 10 I1 ....(iv)
    Writing in matrix form


    V2 = 52I2
    V1210I1



  1. The two electric sub-networks N1 and N2 are connected through three resistors as shown in the figure below. The voltage across 5 Ω resistor and 1 Ω resistor are given to be 10V and 5V respectively. Then voltage across 15 Ω resistor is









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    By KCL, if we take a cutset along all three branches, then total current at the junction is zero.
    i.e. I 1 + I 2 + I 3 = 0

    10
    + I2 +
    5
    = 0
    51

    ⇒ I2 = 7A
    and V = – 105 V.

    Correct Option: A

    By KCL, if we take a cutset along all three branches, then total current at the junction is zero.
    i.e. I 1 + I 2 + I 3 = 0

    10
    + I2 +
    5
    = 0
    51

    ⇒ I2 = 7A
    and V = – 105 V.