Induction Motor Drives Questions and Answers

Answer: C

Correct answer: 100 A

Star-delta starting reduces the line current to one third of the direct-on-line value:

The 173 A figure would be \( 300/\sqrt{3} \), which accounts for the winding voltage reduction but forgets that the line and phase currents are also related differently in the two connections.

Answer: D

Correct answer: multiplied by five

The slip of maximum torque is \( s_{max} = R_2/X_2 \). At present \( R_2/X_2 = 0.2 \); to place the peak at standstill we need \( s_{max} = 1 \), so the rotor resistance must be raised by the factor

The reactance is unchanged by the external resistor, so the total rotor resistance is five times the original. The height of the peak is unaffected, only its position.

Answer: A

Correct answer: 1 and 2 only

Statement 1 is correct: reducing the stator voltage lowers the speed by raising the slip, and the rotor loss \( sP_g \) rises with it. Statement 2 is correct: holding \( V/f \) constant holds \( E/f \), and therefore the flux, nearly constant. Statement 3 is wrong: rotor resistance control needs access to the rotor circuit through slip rings, so it applies only to wound rotor machines, never to a cage machine whose rotor bars are permanently short-circuited.

Answer: B

Correct answer: 1 and 2 only

Statements 1 and 2 are the two standard slip relations, \( P_{cu,r} = sP_g \) and \( f_r = sf \). Statement 3 has the inequality backwards: the rotor efficiency is \( 1 - s \), so a machine running at 4 per cent slip converts about 96 per cent of the air gap power to mechanical power. Efficiency is bounded above by \( 1-s \), not below by \( s \).

Answer: C

Correct answer: 1-Q, 2-R, 3-P

V/f control varies the frequency and the voltage together so as to keep the flux constant, which is Q. Rotor resistance control leaves the synchronous speed alone and changes the slip at which the machine settles, which is R. Pole changing switches the winding between, say, four and eight poles, so the synchronous speed \( 120f/P \) jumps between two discrete values, which is P.

Answer: D

Correct answer: 1-Q, 2-P, 3-R

Rotor resistance control dissipates the slip power as heat in the external resistors, which is Q. A static Kramer drive uses a diode rectifier, so slip power can only leave the rotor and return to the supply, which is P and confines the machine to sub-synchronous speeds. A static Scherbius drive uses a bidirectional converter and can both take power from the rotor and inject it, which is R and allows super-synchronous running.

Answer: A

Correct answer: 2, 3, 1

Relative to direct on line, star-delta gives one third of the torque, and an autotransformer at the 70 per cent tapping gives \( 0.7^2 = 0.49 \) of it. In increasing order the torques are therefore 0.33, 0.49 and 1.00, that is star-delta, then the autotransformer, then direct on line. Direct on line always gives the largest torque because it applies the full voltage, at the cost of the largest current.

Answer: B

Correct answer: Both A and R are true, and R is the correct explanation of A

The assertion is true: rotor resistance control is inefficient. The reason is true and explains it exactly. Speed is reduced only by increasing the slip, and the rotor loss is \( sP_g \), all of it burnt in the external resistors. Running at 25 per cent slip means a quarter of the air gap power is thrown away, so the efficiency falls roughly as \( 1-s \).

Answer: C

Correct answer: Both A and R are true, and R is the correct explanation of A

Both statements are true and the second explains the first. The strict constant-ratio rule sets the terminal voltage from \( V = (V/f)_{rated} \times f \), but part of that voltage is lost across the stator resistance and never reaches the airgap. At 50 Hz that drop is a negligible fraction; at 2 Hz the commanded voltage is only a few per cent of rated and the resistive drop may be most of it, leaving the flux and hence the torque badly short. A boost added at low frequency makes up the difference.

Answer: D

Correct answer: A is false but R is true

The reason is true: \( s_{max} = R_2/X_2 \), so adding rotor resistance does raise the slip at which the peak occurs. The assertion is false, because

contains no \( R_2 \) at all. The peak moves along the slip axis without changing height. What the added resistance buys is a larger torque at starting, which is a different quantity from the maximum torque.