Dynamic braking of an induction motor is commonly arranged by
Answer: A
A stationary field produced by dc in the stator makes the rotating rotor generate into its own resistance, and the kinetic energy becomes heat in the rotor. It works whatever the supply is doing, which suits an emergency stop.
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Plugging of an induction motor is achieved by
Answer: B
Reversing the phase sequence sends the field the other way, so the slip becomes nearly two and a heavy braking torque appears. Both the supply energy and the stored kinetic energy are dissipated in the machine, so it heats severely.
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Regenerative braking of an inverter-fed induction motor requires the drive to
Answer: C
Lowering the commanded frequency below the rotor speed makes the machine generate into the link, and the link voltage then rises unless something takes the energy. A braking chopper and resistor or an active front end are the two solutions.
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A four-pole induction motor on a 50 Hz supply runs at 1,440 rpm. Its slip is
Answer: D
Correct answer: 4 per cent
The synchronous speed is
\[ N_s = \frac{120 \times 50}{4} = 1500 \text{ rpm} \]so the slip is
\[ s = \frac{1500 - 1440}{1500} = \frac{60}{1500} = 0.04 = 4\% \]The 96 per cent figure is \( N/N_s \), the per-unit speed, and 6 per cent comes from dividing 60 by 1000 rather than 1500.
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For that machine, the rotor current frequency is
Answer: A
Correct answer: 2 Hz
The rotor frequency is the slip times the supply frequency:
\[ f_r = s f = 0.04 \times 50 = 2 \text{ Hz} \]The low value is typical of normal running and is why the rotor reactance \( sX_2 \) is small compared with \( R_2 \) at full load. It rises to the full 50 Hz only at standstill, where \( s = 1 \).
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An induction motor takes 20 kW across its air gap while running at 5 per cent slip. Its rotor copper loss is
Answer: B
Correct answer: 1 kW
Rotor copper loss is the slip times the air gap power:
\[ P_{cu,r} = sP_g = 0.05 \times 20 = 1 \text{ kW} \]The 19 kW figure is the mechanical power, which is what remains, and 0.05 kW would follow from multiplying the slip by 1 kW rather than by the air gap power.
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That machine's mechanical output power is
Answer: C
Correct answer: 19 kW
The mechanical power is what is left of the air gap power after the rotor copper loss:
\[ P_{mech} = (1-s)P_g = 0.95 \times 20 = 19 \text{ kW} \]The shaft output is a little less again, by the friction and windage. Note that the rotor efficiency here is 95 per cent, equal to \( 1-s \), which is the ceiling any scheme controlling speed by slip must live under.
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A six-pole induction motor is to run at 900 rpm from a variable-frequency drive. The frequency required is nearest to
Answer: D
Correct answer: 45 Hz
Under a variable-frequency drive the machine runs just below the synchronous speed set by the frequency. Taking 900 rpm as the synchronous speed for a six-pole machine,
\[ f = \frac{N_s P}{120} = \frac{900 \times 6}{120} = 45 \text{ Hz} \]The 30 Hz answer follows from using four poles and 90 Hz from inverting the ratio.
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A 400 V 50 Hz machine is operated at 25 Hz under constant V/f control. The stator voltage should be about
Answer: A
Correct answer: 200 V
Constant V/f means the ratio is held at its rated value:
\[ \frac{V}{f} = \frac{400}{50} = 8 \text{ V/Hz}, \qquad V = 8 \times 25 = 200 \text{ V} \]In practice a small boost is added above this figure at low frequency to make up the stator resistance drop, but 200 V is the value the constant-ratio rule gives.
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An induction motor develops 60 N·m at rated voltage. At 80 per cent of rated voltage, at the same slip, it develops
Answer: B
Correct answer: 38.4 N·m
Torque at a fixed slip varies as the square of the applied voltage:
\[ T = 60 \times (0.8)^2 = 60 \times 0.64 = 38.4 \text{ N·m} \]The 48 N·m answer treats the relation as linear in the voltage, which understates the loss of torque. This square law is why undervoltage is so damaging to starting performance.
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