Braking of Electrical Drives Questions and Answers

Answer: C

Because stored energy follows the square of the speed, a speed uprating is far more demanding on the braking system than on the motoring side. Resistors, choppers and friction linings all need reassessment before such a change is made.

Answer: D

In a DC drive supplied from a full converter, regenerative braking returns the energy of the motor to the supply. The converter must then work as an inverter, which for a full converter means a firing angle above \(90^\circ\), where the average output voltage is negative and so the converter can accept power from the DC side. For firing angles below \(90^\circ\) the converter works as a rectifier and the power flows from the supply to the motor.

The key gives more than \(90^\circ\), option D. The item speaks of the converter in the field circuit, which is loosely worded, and I took the key as the usual condition for inverting operation of a full converter. Zero, below 30 degrees and below 60 degrees are all rectifier conditions.

Answer: B

The slip of an induction motor is \(s = (N_s - N)/N_s\). In plugging (reverse rotation of the field) the supply phase sequence is reversed, so the field rotates at \(-N_s\) while the rotor is still turning forwards at \(N\).

\[ s = \frac{-N_s - N}{-N_s} = 1 + \frac{N}{N_s} \]

With \(N\) between 0 and \(N_s\), the slip lies between 1 (at standstill) and 2 (at synchronous speed). A slip of zero is synchronous speed, a slip between 0 and 1 is normal motoring (and up to 0.1 the usual operating range), so the answer is between 1 and 2, option B.