Power systems miscellaneous

Power systems miscellaneous

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Power Systems

Power systems miscellaneous
  1. For the network shown in the given below, the zero sequence reactances in p.u. are indicated. The zero sequence driving point reactance of the node 3 is ___________








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    NA

    Correct Option: B

    NA

  1. An IDMT over current relay has a current setting of 15% and has a time multiplier setting of 0.5. The relay is connected in the circuit through a CT having 500: 5. The time of operation of the relay if the circuit carries a fault current of 6000 A is ________ sec








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    Secondary fault current = 6000 ×
    5
    		 = 60 A
    500

    P.S.M. =
    actual arms current in relay
    setting current

    = 60 ×
    11
    		 = 8
    5 × 1.5

    Operating time = 3.15 × 0.5 = 1.575 sec.

    Correct Option: C

    Secondary fault current = 6000 ×
    5
    		 = 60 A
    500

    P.S.M. =
    actual arms current in relay
    setting current

    = 60 ×
    11
    		 = 8
    5 × 1.5

    Operating time = 3.15 × 0.5 = 1.575 sec.

  1. Base load power plants are
    P → wind farms
    Q → run-of-river plants
    R → nuclear power plants
    S → diesel power plants








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    Base Load power plants are
    (i) Wind farms
    (ii) Run of river plants
    (iii) Nuclear power plants

    Correct Option: C

    Base Load power plants are
    (i) Wind farms
    (ii) Run of river plants
    (iii) Nuclear power plants

  1. For a power system network with n nodes, Z33 of its bus impedance matrix is j0.5 per unit. The voltage at node 3 is 1.3 ∠– 10° per unit. If a capacitor having reactance of – j3.5 per unit is now added to the network between node 3 and the reference node, the current drawn by the capacitor per unit is








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    Thevenin’s equivalent impedance between bus (3) and ref. bus is Z33
    Z33 = 10.5 pu
    Thevenin’s equivalent voltage between bus (3) and ref. bus = 1.3 ∠ – 10°

    Correct Option: D


    Thevenin’s equivalent impedance between bus (3) and ref. bus is Z33
    Z33 = 10.5 pu
    Thevenin’s equivalent voltage between bus (3) and ref. bus = 1.3 ∠ – 10°

  1. Consider a three-phase, 50 Hz, 11 kV distribution system. Each of the conductors is suspended by an insulator string having two identical porcelain insulators. The self capacitance of the insulator is 5 times the shunt capacitance between the link and the ground, as shown in the figure given below. The volt age across the two insulators are








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    Using KCL at point ℗, we have
    ⇒ I = I1 + I2
    ⇒ V1 (5Cω) = V2 (Cω) + (5Cω)V2
    ⇒ 5V1 = 6V2 ...(A)
    and V1 + V2 = 11 kV ...(B)
    Equation (A) and Equation (B) gives,
    V1 = 3.46 kV, V2 = 2.89 kV

    Correct Option: B


    Using KCL at point ℗, we have
    ⇒ I = I1 + I2
    ⇒ V1 (5Cω) = V2 (Cω) + (5Cω)V2
    ⇒ 5V1 = 6V2 ...(A)
    and V1 + V2 = 11 kV ...(B)
    Equation (A) and Equation (B) gives,
    V1 = 3.46 kV, V2 = 2.89 kV