Induction Motor Drives Questions and Answers

Answer: C

At standstill the rotor frequency is high and the current crowds into the outer high-resistance path, while at running slip it spreads into the low-resistance one. The machine therefore behaves as if its resistance varied with speed.

Answer: D

Correct answer: a resistance divided by the slip

The rotor is referred to the stator as \( R_2/s \) in series with the rotor leakage reactance. Splitting it as \[ \frac{R_2}{s} = R_2 + R_2\frac{1-s}{s} \] separates the actual copper loss from the mechanical power developed, and it also explains the machine’s behaviour at a glance: at standstill \( s=1 \) and the resistance is small, so the current is large; near synchronous speed the resistance is very large and the current small.

Answer: A

Separating that term from the rotor resistance itself makes the power split explicit, and at small slip it is much the larger of the two. At unity slip it vanishes, which is why all the air gap power becomes heat at standstill.

Answer: B

Correct answer: the magnetising branch parameters

With no load the slip is nearly zero, so \( R_2/s \) is very large and almost no current flows in the rotor branch. What remains is the magnetising branch, so the test gives the magnetising reactance and the core loss resistance. The blocked rotor test is its complement: with the slip at unity the magnetising branch is negligible and the series leakage impedance is measured instead.

Answer: C

Correct answer: reduced voltage

With the rotor held still the machine is a short-circuited transformer, so rated voltage would drive five to seven times rated current and overheat the windings in seconds. The voltage is raised only until rated current flows, typically 15 to 25 per cent of rated, and the impedance measured then applies at full voltage because it is essentially linear.

Answer: D

Correct answer: reducing the flux to match the load

At part load the torque needed is small, but a machine held at full flux still carries the full magnetising current and the full core loss. Reducing the flux - lowering the voltage in step with the reduced torque demand - brings the iron loss and magnetising current down and can improve part-load efficiency appreciably. Raising the flux does the opposite, and a higher switching frequency merely increases the inverter loss.

Answer: A

Correct answer: operated with flux reduction or sized more closely

A machine running for long periods at a small fraction of its rating spends most of its energy on losses that do not depend on the load. The two remedies both address that: reduce the flux so the magnetising and core losses fall, or fit a smaller machine so the load is a larger fraction of its rating. Fitting a larger machine makes the problem worse, which is why oversizing for margin is a false economy.

Answer: B

Correct answer: no brushes to spark or wear

A cage induction motor has no brushes, no commutator and no sliding contacts, so there is nothing to spark in a flammable atmosphere and nothing to wear out or need replacing in a dirty one. A dc machine’s commutator sparks by its nature and its brushes must be inspected and changed. The efficiency advantage is real but modest; the maintenance and safety advantage is decisive.

Answer: C

Correct answer: Varying the armature resistance of the machine

An induction motor has no armature - that is a dc machine term, and the corresponding induction motor method is rotor resistance control, which works only on a wound rotor machine. Stator voltage variation, pole changing and frequency variation are all genuine methods, the last being the one that has displaced the others.