s MCQ on Induction Motor | Page – 7 | Electrical Study App by SARU TECH

MCQ on Induction Motor | Page – 7

  1. The stator of a 3 - Φ IM has a 3 slots per pole per phase. If supply frequency is 50 Hz, then the number of stator poles produced

    In 3 φ IM having no of slots are 3.
    We know number of stator poles (P) = 2 × n [ n = no. of slots / phase]
    P = 2 × 3 = 6 poles.

  2. A 3 - Φ, 12 pole alternator is coupled to an engine running at 500 r.p.m the alternator supplies an induction motor which has a full load speed of 1455 r.p.m. The slip will be

    Number of poles of alternator, Pa = 12
    Speed of engine, Ne = 500 r.p.m.
    Full-load speed of the induction motor, Nm = 1455 r.p.m.
    Supply frequency, f = Na × Pa / 120 = 500 × 12 / 120
    i.e. 50 Hz. Now the supply frequency is 50 Hz, the synchronous speed can be 3000, 1500, 1000, 750 rpm etc. Since the full load speed is 1455 rpm. Speed will be 1500 rpm.
    Slip (s), s = Ns -N / Ns
    S = 1500 - 1455 / 1500 = 0.03 or 3 %.

  3. A 4 pole , 3 phase double layer winding is housed in 36 slot stator for an ac machine with 60° phase spread. Coil span is of short pitched. Number of slots in which top and bottom layers belong to different phases is

    Pole pitch = 36 / 4 = 9 slots. Coil span = 7, Slot / pole / phase = 3. So that 3 slots in single phase if it is chorded by 2 slot then, Out of 3 - 2 have different phase. Out of 36 - 24 have different phase.

  4. Rotor ohmic loss a squirrel cage induction motor is 4 KW at slip 0.04 and efficiency is 0.85, air gap power is

    Given by S × Pg = 4 KW or Pg = 4 KW / 0.04 = 100 KW = 0.1 MW.

  5. In a 3 - phase induction motor, the starting electric current is four times the full load current while the rated slip is 3 %. The ratio of starting torque to full load torque

    Tst / Tfl = ( Ist / Ifl )2 × Sfl = (4X / X )2 × 0.03 = 0.48.

  6. A 4 pole, 50 Hz, 3 phase induction motor has blocked rotor reactance, per phase that is four times the rotor resistance per phase. The speed at which maximum torque developed is

    X2 = 4R2
    ⇒ R2 / X2 = 0.25.
    For maximum torque R2 = sX2, S = R2 / X2 = 0.25N = ( 1 - s )Ns = ( 1 - 0.25 ) × 1500 = 1125 rpm.

  7. A 4 pole induction motor by a slightly unbalanced 3 phase 50 Hz source, is rotating at 1440 rpm. The electrical frequency in Hz of the induced negative sequence electric current in the rotor is

    Ns (Synchronous speed) = 120f / P = ( 120 × 50 / 4 ) =1500 rpm, So slip is = Ns + Nr / Ns = ( 1500 + 1440 ) / 1500 = 1.96 (if negative sequence current). Negative sequence electric current in the rotor is = ( 50 × 1.96) = 98 Hz.

  8. If starting electric current of 3 phase induction motor is 5 times the rated current, while rated slip is 4 %, the ratio of starting torque to full load torque is

    Tst / Tfl = ( Ist / Ifl )2 × Sfl = (25) × (0.04) = 1.0.

  9. If the rotor power output of 3 phase induction motor is 12 KW, the rotor copper losses at slip of 5 % will be

    Rotor copper loss = S × Pg = ( 0.05 × 12000 ) = 600 W.

  10. In an induction motor the ratio of starting electric current to full load current is 12.65 and the ratio of starting torque to full load torque is 1.6. Then percentage slip at full load is-

    S = Tst / Tf × ( If / Ist )2 = 1.6 × ( 1 / 12.65)2 = 0.01 × 100 % = 1 %.

  11. A 3 - phase, 6 - pole, 50 Hertz IM developed 3.73 kW at 960 rpm and stator loss is 200 W. Calculated stator input?

    Power developed in rotor / rotor input = N / Ns = 120 × 50 / 6 = 1000 rpm
    3730 / rotor input = 960 / 1000
    ∴ Rotor input = 3885 W
    Stator input = rotor input + stator loss = 3885 + 200 = 4085.

  12. An IM rotor input is 4500 watts and rotor cu loss is 300 watts, find slip?

    Slip = rotor Cu loss / rotor input
    Slip = 300 / 4500 = 0.0667 = 6.67 %.

  13. In a 50 Hz, 4 pole, 3 phase induction motor running at 1440 rpm, the frequency of rotor electric current is

    We know slip(s) = Ns - N / Ns.
    i.e. 1500 - 1440 / 1500 = 60 / 1500
    but, Ns = 120f / p = 120 × 50 / 4 = 1500 rpm
    Now frequency induced e.m.f in rotor (fr) = sf
    ∴(fr) = 60 / 1500 × 50 = 2 Hz.

  14. A 3 - Φ IM draws an electric current of 50 A from mains when started by direct switching. If an auto transformer with 60 % tapping is used for starting, the electric current drawn from mains will be

    The 3 - Φ IM draws electric current of 50 Amp from mains. Auto transformers being tapped with 60 % i.e 60 / 100 with starting.
    ∴ K2Ist = ( 60 / 100)2 × 50. i.e. 18 Amps.

  15. A 6 pole IM is excited by a 3 phase 50 Hz source. The motor is set to turn at 500 rpm in direction opposite to that of rotating field. The frequency of rotor induced electric current is

    We know,fr= sf.
    Where s = slip, f = supply frequency in Hz.
    Now, s = ( Ns -N / Ns ) × 100.
    ∴ Ns = 120f / P = 120 × 50 / 6 = 1000 rpm.
    S= ( 1000 - 500 / 1000 ) × 100 = 0.5.
    Sf = 0.5 × 50 = 25 Hz.

  16. A 3 - Φ IM has a starting torque of 320 N-m when started by direct switching. When started by a auto transformer with 50 % tapping, the starting torque will be

    We know that torque with auto transformer = k2 × torque with direct switching.
    K = 50 % i.e. 0.5
    ∴ Torque with auto transformer = 0.52 × 320 = 80 Nm.

  17. The rotor voltage of a slip ring induction motor gives 120 oscillation per minute when the motor is connected to 3 - Φ 50 Hz supply. The percentage slip of rotor is

    The oscillation of rotor fr = sf.
    Where fr = 120 and frequency(f) = 50 Hz.
    ∴ 120 / 60 = s × 50
    ∴ %s = ( 120 / 60 ) × ( 1 / 50 × 100 ) = 4 %.

  18. A power station consists of two synchronous generators A and B of rating 250 MVA and 500 MVA with inertia 1.6 p.u. and 1 p.u. respectively on their own base MVA ratings the equivalent p.u. inertia for the system on 100 MVA common base will be

    H1 × S1 + H2 × S 2 = H × S
    Equivalent p.u. = 250 × 1.6 + 500 × 1 / 100 = 9 p.u.

  19. A 750 kVA, 11 kV, 4 pole, 3 - Φ, star connected alternator has % reactance and reactance of 1 and 15 respectively. In no-load the synchronizing power per mechanical degree of displacement will be

    Full load current I = 75 × 103 / √3 × 11 × X3 = 40 A
    Vph = 11000 / √3 = 6350 V
    IRa = 1% of 6350 = 63.5
    ⇒ 40 Ra = 63.5, Ra = 1.6 Ω.
    40 × Xs = 15 % of 6350 = 952.5 V
    ∴ Xs = 23.8 Ω Zs = √( 1.62 + 23.82 ) = 23.8 Ω
    On no load, α (mech) = 1° and α(elect) = 1 × ( 4 / 2) = 2° = 2 ×π / 180 = π / 90 elect radian.
    PSY = αE2 / Xs = (π/90) × 63502 / 23.8 = 59140 W = 59.14 kW.

  20. A 2200 V, 50 Hz, 3 phase, star connected alternator having effective resistance of 0.5Ω / phase. A field electric current of 30 A produced the full load current of 200 A on short circuit and line to line emf of 1100 V on per circuit. The short circuit ratio of an alternator will be

    Under construction.