This set of Advanced Electric Drives Questions and Answers focuses on “Countercurrent Braking of DC Shunt Motors”.
1. Plugging is suitable for __________ type loads.
Explanation: Plugging is suitable for gravitational type loads. It halts the motor operation. It can reverse the direction of rotation of the motor. It can hold the motor at zero speed.
2. In DC shunt machine ∅ is directly proportional to Ia.
Explanation: In shunt DC machine, the field winding is in parallel with the armature winding. The armature current flows through the armature winding. IF∝∅∝Vt.
3. Braking resistance should be selected in order to limit the braking current.
Explanation: During braking, a high current flows in the armature circuit. To limit the braking current a proper value of braking resistance should be selected.
4. Torque developed in the motor is ___________
Explanation: The torque developed in the motor is directly proportional to the flux, machine constant, armature current. It is mathematically represented as T= KmΦIa.
5. Full form of TVR is _________
a) Terminal voltage reversal
b) Total voltage reversal
c) Terminal voltage redirect
d) Total voltage reversal
Explanation: TVR stands for terminal voltage reversal. It is one of the methods of countercurrent braking of the DC shunt motor. It halts the motor rapidly and reverses its direction of rotation.
6. Calculate the current in DC shunt field winding using the following data: Ia=15 A, IL=21, R=22, N=484.
a) 4 A
b) 2 A
c) 7 A
d) 6 A
Explanation: In DC shunt motor, field windings are connected in parallel with the armature circuit. IF=IL-Ia=11-5=6 A. Motor current is the sum of field and armature current.
7. Calculate the flux produced by the DC shunt field winding using the following data: Ia=5 A, IL=11, R=2, N=4.
a) 14 Wb
b) 12 Wb
c) 17 Wb
d) 18 Wb
Explanation: In DC shunt motor, field windings are connected in parallel with the armature circuit. IF=IL-Ia=11-5=6 A. F=NIF=R∅. IF∝∅. ∅=F÷R=12 wb.
8. The shape of the speed-armature current characteristics in DC shunt motor is __________
c) Negative slope linear line
Explanation: The shape of the speed-armature current characteristics in DC shunt motor is negative slope linear line. The motor equation is Eb=Vt-IaRa=Km∅ωm. N=(Vt-IRa)/Km. N vs Ia is a negative slope line with slope = -Ra/Km.
9. The shape of the speed-torque characteristics in DC shunt motor is __________
a) Rectangular Hyperbolic
b) Whole x-y plane
Explanation: The shape of the speed-torque characteristics in the DC shunt motor is a rectangular hyperbola. The motor equation is Eb=Vt-IaRa=Km∅ωm. N=(Vt/KmIa)-Ra/Km. Ia∝√T .N∝1÷√T.
10. DC shunt motors are used where the high _________ is required.
a) Starting torque
b) Maximum torque
c) Minimum torque
d) Breakdown torque
Explanation: DC series motors are used where the high starting torque is required. At starting the torque produced by DC series motor is very high.
11. The shape of the current-torque characteristics in DC shunt motor is __________
a) Rectangular Hyperbola
c) Straight line
Explanation: The shape of the current-torque characteristics in the DC shunt motor is a straight line. The motor equation is Eb=Vt-IaRa=Km∅ωm. T=Km∅Ia. ∅∝If. Torque ∝ Ia.
12. The relationship between the torque(T) and power(P) developed in the DC shunt motor is ___________ (Neglecting all the losses)
a) T ∝ √P
b) T ∝ P2
c) T ∝ ∛P
d) T ∝ P
Explanation: The torque developed in the DC shunt motor is the ratio of power developed and angular
speed of the motor. T=EbIa÷ωm=P÷ωm. T ∝ P.
13. The value of current at time of starting is ___________
c) Very low
Explanation: At the time of starting the motor is at rest. The e.m.f developed in the motor is zero. Ia×Ra = Vt-Eb. Ia=Vt÷Ra. The starting current is very high about 10-15 times of full load current.
15. Calculate the bandwidth in series RLC circuit if the frequency is 20 Hz and the quality factor is 5.
a) 2 Hz
b) 4 Hz
c) 6 Hz
d) 8 Hz
Explanation: Bandwidth is defined as the range of frequencies for which the signal exists. Selectivity is inversely proportional to the bandwidth. B.W(Hz)=f÷Q=20÷5=4 Hz.
Sanfoundry Global Education & Learning Series – Electric Drives.
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