Long motor cables on a PWM drive can cause
Answer: A
Fast switching edges travelling down a cable reflect at the machine's higher impedance and can nearly double the voltage there, stressing the insulation. Output filters, slower edges and inverter-duty windings are the answers.
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Field-oriented control of an induction motor resolves the stator current into
Answer: B
Correct answer: flux-producing and torque-producing components
Field-oriented control transforms the stator current into a frame aligned with the rotor flux, giving \( i_d \) along the flux and \( i_q \) in quadrature with it. The two can then be commanded separately, exactly as field and armature current are in a separately excited dc machine, so torque responds without disturbing the flux. This decoupling is what gives a vector drive its fast torque response.
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Field-oriented control requires knowledge of
Answer: C
The transformation into flux and torque axes is meaningless without knowing where the flux is, and estimating it accurately at low speed is the hard part. Direct schemes measure it and indirect schemes compute it from the currents.
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Direct torque control differs from field-oriented control in that it
Answer: D
A switching table chosen from the errors and the flux sector replaces the coordinate transformations and current loops, giving very fast torque response. The penalties are a variable switching frequency and greater torque ripple.
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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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