DC Motor Drives Questions and Answers

Answer: A

Correct answer: 230 V

Rearranging the armature circuit equation,

The back emf takes 96 per cent of the applied voltage, which is typical in normal running - and the reason the starting current would be so large, since at standstill \( E \) is zero and the whole 240 V appears across the 0.5 ohm resistance.

Answer: B

The new back emf is \( 120 - 10 = 110 \) V, and speed follows back emf at constant flux, so \( 1{,}500 \times 110/230 = 717 \) rpm. Neglecting the drop gives 750 rpm, which is a useful approximation but not the answer.

Answer: C

Correct answer: 75 N·m

With the flux constant, \( T = k\Phi I_a \) makes torque proportional to armature current:

The relation is linear, not square-law - the 112 N·m answer squares the ratio, which would be right only for a series machine, where the flux itself rises with the current.

Answer: D

Correct answer: 125 per cent of its former value

With the resistive drop neglected, \( V \approx E = k\Phi N \), so at constant voltage the speed varies inversely with the flux:

Weakening the field to 80 per cent therefore raises the no-load speed by 25 per cent. This is field weakening, the standard way of running above base speed - and the reason a shunt machine that loses its field runs away.

Answer: A

With \( V_m = 325 \) V, \( V_{dc} = (2 \times 325/\pi)\cos 60° = 207 \times 0.5 = 104 \) V. Using the rms rather than the peak value of the supply is the usual slip and understates the answer by a factor of about one and a half.

Answer: B

Correct answer: 44 degrees

From \( V_{dc} = (2V_m/\pi)\cos\alpha \) with the 230 V supply of the preceding item, \( 2V_m/\pi \approx 207 \) V, so

The 136 degree answer is the supplement, which would give minus 150 V - the same magnitude but with the bridge inverting rather than rectifying, so the machine would be braking rather than motoring.

Answer: C

With a line peak of 587 V, \( V_{dc} = (3 \times 587/\pi)\cos 30° = 560 \times 0.866 = 485 \) V. Six pulses per cycle give both this high mean and a small ripple, which is why three-phase supply is used for any sizeable drive.

Answer: D

Correct answer: 0.6

From \( V_a = DV_s \),

The duty ratio is a fraction between 0 and 1 by definition, so the answers of 1.67 and 160 can be rejected without calculation - the first inverts the ratio and the second is a percentage with the decimal point lost.

Answer: A

Correct answer: 1.2 ms

The switching period is the reciprocal of the frequency, and the on-time is the duty ratio of that period:

The off-time is the remaining 0.8 ms. Raising the switching frequency shortens both and reduces the current ripple, at the cost of greater switching loss in the device.

Answer: B

Correct answer: 1,050 A

Plugging reverses the supply while the machine still turns, so the applied voltage and the back emf now add instead of opposing:

That is far beyond anything the machine or the converter could survive, which is why plugging is never done without series resistance. Subtracting instead of adding gives 50 A and misses the entire point of the method.