In the equivalent circuit of an induction machine, the rotor resistance appears as
Answer: C
Dividing by the slip accounts for the rotor being at slip frequency while the circuit is drawn at supply frequency, and the element then represents the copper loss and the mechanical power together. Splitting it separates the two.
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The element \( R_2(1-s)/s \) in the equivalent circuit represents
Answer: D
Separating the total \( R_2/s \) into \( R_2 \) and this term splits the air gap power into the copper loss and the mechanical output explicitly. At unity slip it vanishes, which is why all the air gap power becomes heat at standstill.
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The rotor reactance in the equivalent circuit is drawn as its standstill value because
Answer: A
The actual rotor reactance at slip frequency is \( sX_2 \), and referring the circuit to supply frequency is what turns that into the standstill value with the resistance divided by slip. The two adjustments go together.
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The magnetising branch of the equivalent circuit carries
Answer: B
Correct answer: the current that establishes the air gap flux and supplies the core loss
The shunt branch of the equivalent circuit carries the magnetising current, largely reactive, which sets up the air gap flux, together with the small in-phase component representing hysteresis and eddy current loss in the iron. It is drawn whether or not the machine is loaded, which is why an induction machine has a poor power factor at light load.
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The approximate equivalent circuit of an induction machine moves the magnetising branch to
Answer: C
Shifting it across the stator impedance simplifies the algebra considerably at the cost of a small error, which is acceptable for a machine of appreciable rating. The exact circuit is used where the stator impedance is a large fraction of the total.
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Which of these expressions gives the torque developed at a slip \( s \), for a rotor emf \( E_2 \)?
Answer: D
Correct answer: \( sE_2^2R_2/(R_2^2 + s^2X_2^2) \)
At slip \( s \) the rotor emf is \( sE_2 \) and the rotor reactance \( sX_2 \), so the current is \( sE_2/\sqrt{R_2^2 + s^2X_2^2} \). Torque is proportional to \( I_2^2 R_2/s \), and substituting gives the expression above. It has the familiar shape: rising nearly linearly from zero slip, peaking, then falling away.
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At small slip the torque of an induction machine is approximately proportional to
Answer: A
Correct answer: the slip itself
At small slip \( s^2X_2^2 \) is negligible beside \( R_2^2 \), and the torque expression reduces to \( sE_2^2/R_2 \). This is why the speed of a loaded induction machine falls almost linearly with torque over the working range, and why the machine behaves as a nearly constant speed drive.
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Between the breakdown point and standstill, the torque a machine develops varies roughly with
Answer: B
With \( s^2X_2^2 \) dominating the denominator, the expression reduces to a term inversely proportional to slip, which is why the torque falls away between the peak and standstill. That falling region is the unstable part of the characteristic.
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The ratio of maximum torque to full-load torque of a cage induction machine is typically
Answer: C
Correct answer: between two and three
The breakdown torque of a general purpose cage machine is two to three times its full-load value, and the margin matters: it is what allows the machine to ride through a momentary overload or a voltage dip without stalling. Since torque falls as the square of voltage, a dip to seventy per cent already halves the available torque.
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The ratio of starting torque to full-load torque of an ordinary cage machine is typically
Answer: D
It is enough for a fan or a pump but marginal for a fully laden conveyor, which is why high-torque designs and wound rotor machines exist. Reduced-voltage starting cuts it by the square of the voltage ratio, which is the usual difficulty.
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