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. For the networks shown in the figures (a) and (b) to be duals, it is necessary that R′, L′ and C′ are respectively equal to—










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    In order to make the dual network. simply replace voltage by current, current by voltage, inductance by capacitance resistance by conductance and capacitance by inductance.
    From fig. (a)

    V = iR + L
    di
    		+
    1
    		∫ i dt ......(i)
    dtC

    from fig. (b)
    I = V′ G′ + C′
    dV
    		+
    1
    		∫ V´ dt ......(ii)
    dtL

    On comparing equation (i) and (ii), we get
    R = G′
    L′ = C
    C′ = L

    Correct Option: A

    In order to make the dual network. simply replace voltage by current, current by voltage, inductance by capacitance resistance by conductance and capacitance by inductance.
    From fig. (a)

    V = iR + L
    di
    		+
    1
    		∫ i dt ......(i)
    dtC

    from fig. (b)
    I = V′ G′ + C′
    dV
    		+
    1
    		∫ V´ dt ......(ii)
    dtL

    On comparing equation (i) and (ii), we get
    R = G′
    L′ = C
    C′ = L

  1. The value of R required for maximum power transfer in the network shown below is—










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    We know for maximum power transfer to the network Rth = RL
    Calculation for Rth:

    Rth = 20 || 20 + 10 = 10 + 10 = 20 Ω

    Correct Option: B

    We know for maximum power transfer to the network Rth = RL
    Calculation for Rth:

    Rth = 20 || 20 + 10 = 10 + 10 = 20 Ω

  1. The rms voltage measured across on admittance (G + jB) is V. The reactive power for the element is—








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    NA

    Correct Option: A

    NA

  1. A series R-L circuit is initially released. A step voltage is applied to the circuit. If is the time constant of the circuit, the voltage across R and L will be same at time t equal to—








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    VL = e – R / L x (t)

    VR = 1 – e – R / L x (t)
    (VR + VL = 1)
    According to the given condition
    VL = VR
    e – R / L x (t) = 1 – e – R / L x (t)
    or 2 e e – R / L x (t) = 1
    or e e – R / L x (t) = 1 / 2
    or – R / L x (t) = loge 1 / 2

    or – t =
    L
    		(loge 1 – loge 2)
    R

    or t =
    L
    		loge 2
    R

    or t = loge 2

    Correct Option: A

    VL = e – R / L x (t)

    VR = 1 – e – R / L x (t)
    (VR + VL = 1)
    According to the given condition
    VL = VR
    e – R / L x (t) = 1 – e – R / L x (t)
    or 2 e e – R / L x (t) = 1
    or e e – R / L x (t) = 1 / 2
    or – R / L x (t) = loge 1 / 2

    or – t =
    L
    		(loge 1 – loge 2)
    R

    or t =
    L
    		loge 2
    R

    or t = loge 2

  1. Switch S in position (a) for a long time and moves to be at t = 0. The value of VC and dVC / dt at t = 0+—










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    or
    IC
    		=
    - 4
    dVC
    		=
    Ic
    		=
    - 4
    		= - 8 V
    C1 / 2dtC1 / 2

    Correct Option: A

    or
    IC
    		=
    - 4
    dVC
    		=
    Ic
    		=
    - 4
    		= - 8 V
    C1 / 2dtC1 / 2