Network Elements and the Concept of Circuit

Network Elements and the Concept of Circuit

Easy Questions

Network Theory

Network Elements and the Concept of Circuit
Network theory miscellaneous
  1. Two Incandescent light bulbs of 40 W and 60 W rating are connected in series across the supply voltage, V then—








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    As the bulbs are connected in series, and power,
    P = I2 R
    Current in the series connection will remain the same
    i.e. P ∝ R
    It means higher rating bulb will glow bright.


    Correct Option: C

    As the bulbs are connected in series, and power,
    P = I2 R
    Current in the series connection will remain the same
    i.e. P ∝ R
    It means higher rating bulb will glow bright.


  1. Given two coupled inductors L1 and L2, their mutual Inductance M satisfies—








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    NA

    Correct Option: D

    NA

  1. The resonant frequency of the given series is—










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    f1 =
    1
    2 π√Leq C

    here L eq = L1 + L2 – 2M
    = 2 + 2 – 2 × 1 = 2
    so C = 1 F
    fr =
    1
    		=
    1
    2 π√2 × 12 π√2


    Correct Option: D

    f1 =
    1
    2 π√Leq C

    here L eq = L1 + L2 – 2M
    = 2 + 2 – 2 × 1 = 2
    so C = 1 F
    fr =
    1
    		=
    1
    2 π√2 × 12 π√2


  1. In the network shown in the figure, the effective resistance forced by the voltage source is—










  1. View Hint View Answer Discuss in Forum

    Redrawn the given figure.
    In order to calculate the effective resistance faced by the voltage source.
    First off all we must calculate the resistance of the branch AB.
    Calculation of resistance in AB branch:

    i
    		= i ×
    4
    44 + RAB

    or 16 = 4 + RAB
    ⇒ RAB = 16 – 4 = 12 Ω.
    The resistance offered by the voltage source is equal to the parallel combination of the resistance AB branch and CD branch.
    Reflective 12 || 4 =
    12 × 4
    		=
    48
    		3Ω
    12 + 416


    Correct Option: B

    Redrawn the given figure.
    In order to calculate the effective resistance faced by the voltage source.
    First off all we must calculate the resistance of the branch AB.
    Calculation of resistance in AB branch:

    i
    		= i ×
    4
    44 + RAB

    or 16 = 4 + RAB
    ⇒ RAB = 16 – 4 = 12 Ω.
    The resistance offered by the voltage source is equal to the parallel combination of the resistance AB branch and CD branch.
    Reflective 12 || 4 =
    12 × 4
    		=
    48
    		3Ω
    12 + 416


  1. When a unit impulse voltage is applied to an inductor of 4 H, the energy supplied by the source is—










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    Current flow is given by
    V (s)= LsI(s) – L i(0+)
    I=LsI(s) {∵ initial conditions is not given}

    or I(s) =
    1
    		×
    1
    		=
    1
    		u(t) or I
    		=
    1
    LS44

    Low energy supplied =
    1
    		LI2
    2

    =
    1
    		× 4 ×
    1
    		=
    1
    		J
    2428


    Correct Option: C

    Current flow is given by
    V (s)= LsI(s) – L i(0+)
    I=LsI(s) {∵ initial conditions is not given}

    or I(s) =
    1
    		×
    1
    		=
    1
    		u(t) or I
    		=
    1
    LS44

    Low energy supplied =
    1
    		LI2
    2

    =
    1
    		× 4 ×
    1
    		=
    1
    		J
    2428