Electrical machines miscellaneous


Electrical machines miscellaneous

Electrical Machines

  1. A single -winding single-phase motor has









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    NA

    Correct Option: B

    NA


  1. If a single-winding single-phase motor is running in a particular direction









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    NA

    Correct Option: C

    NA



  1. An iron-cored choke with 1 mm air-gap length, draws 1 A when fed from a constant voltage ac source of 220 V. If the length of air-gap is increased to 2 mm, the current drawn by the choke would









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    NA

    Correct Option: C

    NA


  1. In a 3-phase Δ/ Y transformer shown in the figure, the phase displacement of secondary line voltages with corresponding primary line voltages will be









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    Consider any pair of line terminals on primary and secondary sides.
    Consider pair AB.
    Let VA1B1 be reference on primary side
    VA1B1 = V ∠0°
    VB1C1 = V ∠– 120°
    VC1A1 = V ∠120°
    VA2n = k VA1C1 = − k VC1A1 = − k ∠120°
    VnB2 = k VA1B1 = k V ∠0°
    VA2B2 = VA2n − VnB2 = k V( ∠−60° + ∠0°)
    = √3 kV ∠−30°
    Thus line voltage VA2B2 on secondary side which equals √3 kV ∠−30°, lags corres-ponding line voltage VA1B1 on primary side, which is V ∠0°; by 30°.

    Correct Option: B

    Consider any pair of line terminals on primary and secondary sides.
    Consider pair AB.
    Let VA1B1 be reference on primary side
    VA1B1 = V ∠0°
    VB1C1 = V ∠– 120°
    VC1A1 = V ∠120°
    VA2n = k VA1C1 = − k VC1A1 = − k ∠120°
    VnB2 = k VA1B1 = k V ∠0°
    VA2B2 = VA2n − VnB2 = k V( ∠−60° + ∠0°)
    = √3 kV ∠−30°
    Thus line voltage VA2B2 on secondary side which equals √3 kV ∠−30°, lags corres-ponding line voltage VA1B1 on primary side, which is V ∠0°; by 30°.



  1. Two single-phase transformers with turns ratio 1 and 2 respectively are connected to a 3-phase supply on the primary side as shown in the figure. The voltmeter V2, will read









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    Voltmeter V2 read rms magnitude of the sum of two voltage phasors,
    viz., 100 ∠0° + 2 × 100 ∠ – 120° υ2
    = | 100 ∠0° + 200 ∠ – 120° |
    = | 100 – 100 – j 100 √3 |
    = | – j 100 √3 | = 173.2 V

    Correct Option: B

    Voltmeter V2 read rms magnitude of the sum of two voltage phasors,
    viz., 100 ∠0° + 2 × 100 ∠ – 120° υ2
    = | 100 ∠0° + 200 ∠ – 120° |
    = | 100 – 100 – j 100 √3 |
    = | – j 100 √3 | = 173.2 V