Dynamics of Electrical Drives Questions and Answers

Answer: A

Correct answer: take a long time to reach its working speed

From \( t = J\omega/(T - T_l) \), the run-up time is proportional to the inertia and inversely proportional to the torque margin, so a large inertia with a limited starting torque gives a long start. It reaches the same steady speed as a lighter drive - that is set by where the characteristics cross - but takes far longer about it, and the motor must be rated to survive the starting current for that time.

Answer: B

Reversal means decelerating, passing through zero and accelerating the other way, all of which the inertia resists and all of which need torque of both signs. Servo drives are built with deliberately small rotor inertia.

Answer: C

Correct answer: 1 and 2 only

Statements 1 and 2 follow directly from the torque equation. Statement 3 confuses the transient with the steady state: inertia multiplies \( d\omega/dt \), which is zero at steady speed, so it drops out of the equation entirely and cannot affect where the drive settles. It governs how long the drive takes to get there, and nothing else.

Answer: D

Correct answer: 1 and 2 only

Statements 1 and 2 are the two referral rules: torque by the first power of the ratio, inertia by the square. Statement 3 has the efficiency the wrong way round for motoring - the motor must supply the gearbox loss as well as the load, so the referred torque is \( T_l/(i\eta) \) and the efficiency divides, making the referred torque larger. It multiplies only when the power flows the other way, during braking.

Answer: A

Correct answer: 1-R, 2-P, 3-Q

Three of the four standard load classes. A hoist lifting a fixed weight needs the same torque at any speed, so it is constant-torque. A centrifugal pump, like a fan, needs torque proportional to the square of the speed. A lathe working at constant power needs torque inversely proportional to speed. The fourth class, torque proportional to speed, is viscous friction.

Answer: B

Correct answer: 1-R, 2-P, 3-Q

The three methods differ in where the kinetic energy ends up. Regeneration returns it to the supply, less the losses, and is the only one that recovers anything. Dynamic braking dissipates it in an external resistor, which can be sized and cooled for the duty. Plugging dissipates it - together with further energy drawn from the supply throughout - inside the machine, which is why it is so hard on it.

Answer: C

The first quadrant has both quantities positive, the second positive speed with negative torque, the third both negative, and the fourth negative speed with positive torque. Reading the sign of their product tells motoring from braking in each case.

Answer: D

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason is exactly why. In \( J\,d\omega/dt = T - T_l \) the inertia multiplies the rate of change of speed, so at steady speed that term is zero and the inertia disappears from the equation altogether. The steady speed is fixed by where the two torque curves cross. Inertia governs the transient - how long the run-up takes and how the drive responds to a disturbance - and nothing else.

Answer: A

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the mechanism. A flywheel gives up energy only by slowing down, since \( \Delta E = \tfrac{1}{2}J\Delta(\omega^2) \). A synchronous motor is locked to the supply frequency and cannot slow at all, so no energy can be drawn from the flywheel and none returned to it. Load equalisation by flywheel therefore requires a machine with a drooping characteristic - an induction or a dc machine.

Answer: B

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason is the stability criterion applied to this case. With the load torque flat and the motor torque rising, a small increase in speed leaves the motor torque above the load torque, so the net torque accelerates the drive still further and the departure grows. Stability requires the load curve to rise more steeply than the motor curve, or the motor curve to fall.