SSC JE 3-Phase Induction Motor Solved Question (2018-2009)

Answer.2. Short-circuited via end ring

Explanation:

• When 3 phase supply is given to the stator of a 3 phase induction motor, rotating stator flux will be produced, which will induce emf in rotor windings, according to Faraday’s law of electromagnetic induction.
• Now if the rotor windings are kept open-circuited, no current will flow in these windings.
• Copper bars are short-circuited at both ends via end rings.
• Due to short-circuited rotor windings, short circuit rotor currents will flow in them, giving rise to their own magnetic field linking with the stator flux hence the motor start running.

Answer.3. Pole-changing method

Explanation:

• The pole changing method is used in squirrel cage IM because like slip ring IM we can’t add external resistance in the rotor to reduce speed.
• In this method, it is possible to have one or two speeds by changing the number of poles. This is possible by changing the connection of the stator winding with the help of simple switching.

Answer.4. 1/√3 times of normal voltage

Explanation:-

• Star Delta Starter (Y – Δ) is a common type of three-phase (3 phase) induction motor starters generally used in low starting torque motors.
• At the time of starting the motor, the rotor is at a standstill and the slip ( between the stator’s magnetic field and rotor) is large which causes the large inrush of armature current (which is 6–7 times the rated value). This large current can damage the stator windings and burn the motor.
• To avoid this situation, we use the star-delta starter. At the time of start, motor connections (stator connections) are made in star mode so the impressed voltage is reduced by 1/√3 (Phase voltage = Line voltage/√3) which reduces the starting current. Once the rotor is at 80–90% speed, the centrifugal switch operates changing the connections from star to delta mode (full line voltage is impressed).

Answer.1. Tst/3

Explanation:-

A DOL starter connects the motor terminals directly to the power supply. Hence, the motor is subjected to the full the voltage of the power supply. Consequently, high starting current flows through the motor.

In the above figure, the motor is designed for delta running and is started in star.

Zsc = Short circuit phase Impedance

V = Line to line voltage

For DOL starting in delta

Hence with star-delta starting, the starting torque is also reduced to one-third of starting torque as compared to DOL starting.

Answer.3. Rotor copper loss/rotor power input

Explanation:-

Power is the energy available at the shaft to drive machine and it is given as

P =2πnT/60

where,

P=Power available at the shaft (W or J/s);

N=Speed of rotation of the shaft (rpm);

T=Torque available at the shaft (N-m)

Hence T = P/602πn

If P2 is the power input to the rotor and Pm is the mechanical output power(including friction losses)

P2 – Pm is the copper losses in the rotor i.e P2 – Pm = I_r²R₂

Answer.2. 2.36 kW

Explanation:-

Air gap power Pg

Pg = P_Input – stator Loss

Pg = 60 – 1 = 59 kW

Rotor copper loss = Spg

= 0.04 x 59 = 2.36 kW

Answer.3. Interturn short circuit on the rotor

Explanation:-

The interturn short circuit of the windings inside the rotor is a common electrical fault for the 3-phase induction motor. Bad manufacturing and assembling, overheating due to the jam in the windings, al also the long-term stresses and vibrations may result in this fault. It may intensify the vibrations, burn the windings, and even lead to earth fault.

In interturn short circuit fault mainly causes the rotor to vibrate at fr (the rotating frequency of the rotor), while the static air-gap eccentricity fault primarily produces vibrations at 2fr and the combined fault generates vibrations at fr, 2fr and 3fr at the same time which causes humming in an induction motor.

Answer.2. R₂ = X₂

Explanation:-

The equation of torque is

When slip s = R₂ / X₂, the torque will be maximum and this slip is called maximum slip Sm and it is defined as the ratio of rotor resistance to that of rotor reactance.

If supply voltage V is kept constant, then flux ɸ and E₂ both remain constant. Hence,

At starting S = 1, so the maximum starting torque occurs when rotor resistance is equal to rotor reactance. So it can be proved that maximum starting torque is obtained when rotor resistance is equal to standstill rotor reactance. i.e. R₂² + X₂² =2R₂² .

Answer.3. 1500 RPM

Explanation:-

Speed of Synchronous Motor

Ns = 120f₁/P

f₁ = 50 Hz

P₁ = 4

Now the frequency and no of poles are halved

f₁ = 25 Hz

P₁ = 2

= 120 x 25/2 = 1500 RPM

Answer.1. V²

Starting torque if 3 phase induction motor is given as

Where ns = synchronous speed in r.p.s i.e Ns/60

V = Rotor voltage

R₂ = Rotor resistance

X₂ = Rotor Reactance

∴ T_st ∝ V²

So starting torque of a 3-phase induction motor is directly proportional to V²

Answer.1. (1 – s)Pg

Explanation:-

Let Pg = Air gap power

P_cu = Rotor copper loss = sP_g

P_m = Mechanical power or Gross power output

P_m = Air gap power – Rotor copper loss

= Pg – sP_g

P_m = (1 – s)P_g