The torque of an induction motor is proportional to
Answer: C
Torque is the air gap power referred to synchronous speed, which is why air gap power is sometimes called torque in synchronous watts. Using the mechanical power and the rotor speed gives the same torque, since both are reduced by the same factor.
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The starting torque of an induction motor corresponds to a slip of
Answer: D
Correct answer: unity
Starting means standstill, so \( N = 0 \) and
\[ s = \frac{N_s - 0}{N_s} = 1 \]Slip zero is synchronous speed, where no torque is developed at all. The slip of maximum torque and the rated full-load slip are both particular operating points on the characteristic and are neither of them the starting condition.
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The slip at which an induction motor develops its maximum torque is approximately
Answer: A
Correct answer: the ratio of rotor resistance to rotor reactance
Maximum torque occurs where the rotor circuit transfers most power, which is where the effective rotor resistance \( R_2/s \) equals the standstill rotor reactance:
\[ \frac{R_2}{s} = X_2 \quad \Rightarrow \quad s_{max} = \frac{R_2}{X_2} \]Because it depends on the ratio, adding external rotor resistance moves the peak to a higher slip without changing its height. A machine with \( R_2 = 0.1\,\Omega \) and \( X_2 = 0.5\,\Omega \) peaks at 20 per cent slip.
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Adding resistance to the rotor circuit of a wound rotor induction motor
Answer: B
Correct answer: increases the starting torque without changing the maximum torque
The maximum torque depends on \( V \) and \( X_2 \) but not on \( R_2 \), so the height of the peak is fixed. What the added resistance does is raise \( s_{max} = R_2/X_2 \), sliding the peak towards standstill. If the resistance is chosen so that \( R_2 = X_2 \), the peak lands exactly at \( s = 1 \) and the machine starts with its maximum torque. This is the classic reason for using a wound rotor machine on a hard-starting load.
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The maximum torque of an induction motor is independent of
Answer: C
Correct answer: the rotor resistance
The maximum torque is
\[ T_{max} = \frac{3}{2\omega_s}\cdot\frac{V^2}{2X_2} \]in which the rotor resistance does not appear. Adding rotor resistance therefore moves the peak but does not raise or lower it. The applied voltage, the rotor reactance and the supply frequency all do affect it, the voltage as its square.
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The torque of an induction motor at a given slip varies with the applied voltage as
Answer: D
Correct answer: the square of the voltage
At a fixed slip the rotor current is proportional to the applied voltage, and the torque is proportional to the product of flux and rotor current, both of which scale with \( V \). Hence
\[ T \propto V^2 \]A ten per cent drop in supply voltage costs about nineteen per cent of the torque, which is why an induction motor on a weak supply may fail to start a load it handles easily at rated voltage.
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An induction motor operating between standstill and the slip of maximum torque is
Answer: A
Correct answer: in an unstable region with most load characteristics
Between standstill and \( s_{max} \) the torque rises as the speed falls. If the load torque exceeds the motor torque at some point in that range, the machine slows, its torque falls further, and it stalls; there is no stable intersection with a constant-torque load. Stable running is on the other side of the peak, between \( s_{max} \) and zero slip, where the torque falls steeply as the speed rises and any speed disturbance is self-correcting.
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An induction motor driven above synchronous speed by its load
Answer: B
Correct answer: generates and returns power to the supply
Driven above \( N_s \) the slip becomes negative, the rotor emf reverses, and the torque opposes the direction of rotation. The machine acts as an induction generator, delivering real power to the supply while still drawing its magnetising reactive power from it. This is the regenerative braking region, used on hoists and on wind turbines with cage generators.
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Star-delta starting of an induction motor reduces the starting current to
Answer: C
Correct answer: one third of its direct-on-line value
In star each winding sees \( V_L/\sqrt{3} \), so the phase current falls by \( \sqrt{3} \); and the line current in star equals the phase current, whereas in delta it is \( \sqrt{3} \) times the phase current. The two factors combine to give
\[ \frac{I_{star}}{I_{delta}} = \frac{1}{3} \]The torque, proportional to the square of the voltage, falls to one third as well. The one-ninth figure is a common trap: it is the ratio of the phase currents multiplied out incorrectly.
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Star-delta starting reduces the starting torque to
Answer: D
Torque follows the square of the phase voltage, and the phase voltage falls by \( \sqrt{3} \), so the torque falls by three. A load needing more than a third of full-load torque to break away cannot be started this way.
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