This set of DC Machines Multiple Choice Questions & Answers (MCQs) focuses on “EMF and Torque Production – 1”.
1. In which mode machine is operating, given that conductor current is in the same direction of conductor emf?
c) Can’t be determined using directions
d) In both modes for different cycles
Explanation: If the conductor current is in the same direction of conductor emf then machine outputs electrical power and absorbs mechanical power. So, when mechanical power is absorbed machine is said to be in a generating mode. When conductor emf and conductor current are in opposite directions then machine is said to be in a motoring mode.
2. Nature of the flux density wave in the air gap is__________ (for armature current equal to 0)
a) Flat topped with quarter wave symmetry
b) Point topped with quarter wave symmetry
c) Flat topped with half wave symmetry
d) Point topped with half wave symmetry
Explanation: In a DC machine magnetic structure is such that the flux density wave in the air gap is flat topped with quarter wave symmetry as long as armature current is equal to 0. For non-zero value of armature current, this quarter wave symmetry is disturbed because of armature reaction.
3. In a DC machine, average energy stored in the magnetic field remains constant independent of the armature rotation.
Explanation: In a DC machine, barring the irrecoverable losses of both electric and magnetic origin, there is balance between electrical and mechanical powers of the machine; the average energy stored in the magnetic field remains constant irrespective of armature rotation.
4. Emf produced by DC machine, for zero armature current (E1) and non-zero armature current (E2) can be related as__________
a) E1 = E2
b) E1 > E2
c) E1 < E2
d) Can’t be determined
Explanation: In a DC machine flux density wave in the air gap is flat topped with quarter wave symmetry as long as armature current is equal to 0. For non-zero value of armature current, this quarter wave symmetry is disturbed because of armature reaction. Emf produced is independent of B-wave shape, thus we will get same value for both cases.
5. Average coil emf for 20 coil turns (E1) and 40 coil turns (E2), will have ratio E1/E2=____ (assuming all other parameters same for both machines).
Explanation: Emf generated in a DC machine is directly proportional to number of coil turns, Flux per pole, number of poles and armature speed in rad/s. Thus, ratio E1/E2= 20/40 (assuming all other parameters same for both machines).
6. What is the average coil emf generated in a 4-pole DC machine having flux/pole equal to 0.1 wb rotating at 1500 rpm? (No. of coil sides = 100)
a) 19 kV
b) 1.9 kV
c) 190 V
d) 19 V
Explanation: Average coil emf generated= ∅ωNP/π.
E≅ 19 Kv.
7. Emf and torque produced in a DC machine are proportional to ________ and _________ respectively.
a) Armature speed and armature emf
b) Armature emf and armature speed
c) Armature current and armature emf
d) Armature speed and armature current
Explanation: Average coil emf generated= ∅ωNP/π. Machine torque = ka*∅*Ia. Thus, average coil emf generated can also be represented as ka*∅*ω. So, average coil emf is directly proportional to ω (armature speed) and average torque is directly proportional to Ia (armature current).
8. What is the value of Np in an average coil emf equation, for 10 armature conductors with 2 parallel paths?
Explanation: In an emf equation Nc= Cp * Np. Here, Cp= coils/ parallel path. Np is defined as number of turns per parallel paths which is also called as ratio of total armature conductors to the twice of number of parallel paths. Np= 10/(2*2)= 10/4= 2.5.
9.What is the torque equation in terms of B, Ic, l, Zr (r= mean air gap radius)?
d) Can’t be expressed
Explanation: Avg. conductor force f= Bav*l*Ic. Here, Bav= Average flux density over pole, l= acyive conductor length. Thus, torque T= Z*f = Bav*l*Ia*Z. This torque is constant because both the flux density wave and current distribution is fixed in space at all times.
T developed= Bav*Ic*l*Zr (Here, r= mean air gap radius).
10. What is the value of pole pitch (in SI unit) for mean air gap radius= 0.5mm and P=4?
a) 0.785* 10-6
b) 0.785* 10-3
c) 0.785* 10-2
d) 0.785* 10-4
Explanation: Pole pitch is called as center to center distance between two adjacent poles. When measured in electrical degrees one pole itch is equal to 1800. Pole pitch can be calculated as ratio of 2πr/P.
Pole pitch= 2*3.14* 0.5* 10-3 / 4= 0.785* 10-3 m.
Sanfoundry Global Education & Learning Series – DC Machines.
To practice all areas of DC Machines, here is complete set of 1000+ Multiple Choice Questions and Answers.