Induction Motor Drives Questions and Answers

Answer: A

Correct answer: a short-circuited transformer

At standstill the slip is unity, the rotor is a shorted secondary at full supply frequency, and the only thing limiting the current is the leakage impedance. Five to seven times rated current is the usual result. As the machine accelerates the effective rotor resistance \( R_2/s \) rises, the current falls, and the machine settles at its running point - which is why the high current lasts only for the run-up.

Answer: B

Correct answer: a converter in the rotor circuit rated for the slip power

That is the whole point of the arrangement. With the stator connected directly to the grid and only the slip power passing through the rotor converter, a machine operating within about plus or minus 30 per cent of synchronous speed needs a converter rated at roughly 30 per cent of the machine output. The saving in converter cost and losses is what made the doubly fed generator the dominant wind turbine topology for many years.

Answer: C

Correct answer: flow into the rotor from the supply

The rotor electrical power is \( P_r \approx -sP_{ag} \), so above synchronous speed the slip is negative and \( P_r \) takes the same sign as the air gap power. For a doubly fed machine motoring above synchronous speed that means power must be injected into the rotor from the supply, while below synchronous speed the slip power flows the other way and is recovered to the supply rather than burnt in a resistance. A doubly fed generator reverses both directions, since its air gap power flows outward.

Answer: D

The machine has no field of its own, so the reactive current establishing the flux must come from somewhere, and on a network it comes from the network. Standing alone, it needs capacitors and the arrangement is difficult to control.

Answer: A

Correct answer: the largest load peak it can ride through

The breakdown or pull-out torque is the maximum on the torque-slip curve. A load demanding more than this stalls the machine, because beyond the peak the torque falls as the slip rises and there is no stable operating point. It is typically two to three times full-load torque, and that margin is what lets a motor survive a momentary overload or a supply dip.

Answer: B

Raising the rotor resistance moves the torque peak towards standstill and increases the full-load slip, which suits punch presses and hoists needing a soft characteristic. The penalty is greater rotor loss at rated load.

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.