2. A 220 V separately excited dc motor takes 20 A and the armature resistance is 1.0 Ω. If the armature constant KaΦ = 1.5 rad/sec, the speed of the motor is
1000 rpm
2000/π rpm
3000/π rpm
4000/π rpm
Answer: 4. 4000/π rpm
Explanation:
Induced emf Ef = Kaϕω
ω = angular speed (rad/se
ϕ = flux (weber)
Ka = constant
${K_a} = \dfrac{{PZ}}{{2\pi A}}$
Calculation:
Given that terminal voltage Vt = 220 V,
armature current Ia = 20 A
armature resistance Ra = 1 Ω and Kaϕ = 1.5 rad/sec
For separately excited DC motor
Ef = Vt – Ia Ra = 220 – 20 × 1 = 200 V
Now Ef = Kaϕω
ω = 200/1.5
ω = 2πN/60
N = (200 × 60)/(1.5 × 2π)
N = 4000/π RPM
3. In which mode machine is operating, given that conductor current is in the same direction as conductor emf?
Motoring
Generating
Can’t be determined using directions
In both modes for different cycles
Answer: 2. Generating
Explanation:
If the conductor current flows in the same direction as the conductor emf, the machine outputs electrical power i.e. the machine is operating in generating mode. In Generating Mode the machine Absorb Mechanical Power.
When the conductor current and emf oppose each other, the machine absorbs electrical power and outputs mechanical power, i.e. it operates in the motoring mode.
4. A 4-pole Dc wound machine is lap wound with 400 conductors. The pole shore is 20 cm long and the average flux density over the one-pole pitch is 0.4 T, the armature diameter is 30 cm. What is the value of flux/pole?
0.188 Wb
18.88 Wb
0.0188 Wb
1.888 Wb
Answer: 3. 0.0188 Wb
Explanation:
Flux Per pole of DC machine is given as
Flux/pole = (B × 2π × r × l)/P
Where
B = average flux density over pole
r = mean air-gap radius
l = active conductor length
P = Number of Pole
Flux/pole = (0.4 × 2π × 15 × 10−2 × 20 × 10−2)/4
Flux/pole = 0.0188 Wb
5. In a DC machine, let ϕ be the flux per pole, ωm be the armature speed in rad/s, Nc be the number of coil turns in the armature and P be the number of poles. What will be the average coil EMF in that machine?
Ea = φωmNc/P
Ea = φωmNcP/60
Ea = ϕωmNcP
Ea = φωmNcP/π
Answer: 4. Ea = φωmNcP/π
Explanation:
E.M.F. generated by a generator is
${E_a} = \dfrac{{\phi ZNP}}{{60A}}$
Where,
ϕ = flux
Z = number of conductors
N = speed in RPM
P = number of poles
A = number of parallel paths
As the number of parallel paths is not given take A = 1
6. Nature of the flux density wave in the air gap is __________ (for armature current equal to 0).
Flat-topped with quarter-wave symmetry
Point topped with quarter-wave symmetry
Flat-topped with half-wave symmetry
Point topped with half-wave symmetry
Answer: 1. Flat-topped with quarter-wave symmetry
Explanation:
The air gap under the poles in a dc machine is almost uniform and further, the pole-shoes are wider than in a synchronous machine (pole-arc is about 70% of the pole-pitch). As a result, the flux density in the air around the armature periphery is a flat-topped wave. The voltage induced in the coil (full-pitch has a similar wave shape.
In a dc machine, the magnetic structure is such that the flux density wave in the air gap is flat-topped with quarter-wave symmetry so long as the armature is not carrying any current. The flux density wave gets distorted when the armature carries current (the armature reaction effect destroys quarter-wave symmetry).
7. A 4-pole Dc wound machine is lap wound with 400 conductors. The pole shore is 20 cm long and the average flux density over the one-pole pitch is 0.4 T, the armature diameter is 30 cm. What is the value of induced emf?
8. A four-pole, lap-would DC machine has 462 conductors in the armature. The average flux per pole is 0.02 Wb. Determine the induced armature voltage when the armature rotates at 1000 rpm.
154 V
196 V
120 V
180 V
Answer: 1. 154 V
Explanation:
Given that,
Number poles P = 4
Conductors Z = 462
Speed N = 1000 rpm
The flux per pole (ϕ) = 0.02 Wb
As winding is wave type
Number of parallel paths A = 4
E.M.F. generated by a generator is
${E_g} = \dfrac{{\phi ZNP}}{{60A}}$
${E_g} = \dfrac{{(0.02)(462)(1000)(4)}}{{60(4)}}$
Eg = 154 V
9. In a DC machine, the average energy stored in the magnetic field remains _______ of the armature rotation.
Equal
Dependent
Independent
Any of the above
Answer: 3. Independent
Explanation:
In a DC machine, the average energy stored in the magnetic field remains constant independent of the armature rotation.
In a DC machine, barring the irrecoverable losses of both electric and magnetic origin, there is a balance between the electrical and mechanical powers of the machine; the average energy stored in the magnetic field remains constant irrespective of armature rotation.
10. Coil torque for 20 kA armature current (T1) and 40 kA armature current (T2), will have a ratio T1/T2 = ____ (assuming all other parameters same for both machines).
1/2
2/1
1/4
4/1
Answer: 1. 1/2
Explanation:
The electromagnetic torque produced in d.c. machines is expressed in terms of interaction between main field flux φf and armature MMF Fa.
T = KφIa
Torque produced in a DC machine is directly proportional to the number of coil turns, Flux per pole, number of poles, and the product armature current.