11. A 6-pole, wave-wound armature has 600 conductors and is driven at 700 rpm. Determine the generated EMF in the armature if the flux per pole is 10 mWb.
420 V
230 V
500 V
210 V
Answer: 4. 210 V
Explanation:
Given that,
Number poles P = 6
Conductors Z = 600
Speed N = 700 rpm
The flux per pole (ϕ) = 0.01 Wb
As winding is wave type
Number of parallel paths A = 2
E.M.F. generated by a generator is
${E_g} = \dfrac{{\phi ZNP}}{{60A}}$
${E_g} = \frac{{(0.01)(600)(700)(6)}}{{60(2)}}$
Eg = 210 V
12. Emf produced by DC machine, for zero armature current (E1) and non-zero armature current (E2) can be related as __________
E1 = E2
E1 > E2
E1 < E2
Can’t be determined
Answer: 1. E1 = E2
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).
However, this fact does not affect the constancy of emf (between brushes) and torque developed by the machine with magnitudes of each of these being determined by the flux/pole independent of the shape of the B-wave.
Hence Emf produced by DC machine, for zero armature current (E1) and non-zero armature current (E2) can be related as E1 = E2
13. If the no-load speed of DC Machine is 1300 rpm and full load speed is 1100 rpm, then its voltage regulation is ____________
12.56%
18.18 %
17.39%
18.39%
Answer: 2. 18.18 %
Explanation:
The voltage regulation of a generator (independent of the kind of excitation employe is defined as
% Regulation = (Vo − Vf)/Vf
where
Vf = full-load voltage = V rated
Vo = no-load voltage
In a dc generator, the induced emf is directly proportional to the conductor per pole, speed, number of poles, and flux per pole.
% Regulation = (1300 − 1100)/1100
% Regulation = 18.18 %
14. What is the EMF generated per path in a simplex wave-wound DC generator?
φZN/60P
φZNP/60
φZN/120P
φZNP/120
Answer: 4. φZNP/120
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
If multiplexity is not mentioned, then always take simplex lap winding i.e. m = 1
EMF generated per path in a simplex wave-wound DC generator
∴ A = 2, m = 1
${E_g} = \frac{{\phi ZNP}}{{120}}V$
15. Average coil emf for 20 coil turns (E1) and 40 coil turns (E2), will have a ratio E1/E2= ____ (assuming all other parameters same for both machines).
1/2
2/1
1/4
4/1
Answer: 1. 1/2
Explanation:
In a dc generator, the induced emf is directly proportional to the conductor per pole, speed, number of poles, and flux per pole.
Let’s suppose all the parameters are constant then
E1/E2 = 20/40
E1/E2 = 1/2
16. If the average coil emf of a DC generator is doubled and flux is halved (keeping other parameters constant) then its shaft speed will become ___________
Twice of the original speed
Square of the original speed
Four times of the original speed
Half of the original speed
Answer: 3. Four times of the original speed
Explanation:
E.M.F. generated by a generator is
${E_g} = \dfrac{{\phi ZNP}}{{60A}}$
Where,
ϕ = flux
Z = number of conductors
N = speed in RPM
P = number of poles
A = number of parallel paths
Now other parameter Z, P, A is kept constant then
Eg = φN
N = Eg/φ
N = 2Eg/1/2φ
N = 4 times the original speed
17. A 6 pole, wave-connected DC armature has 250 conductors and runs at 1200 rpm. The emf generated is 600V. The useful Flux/pole is:
19. A 4-pole wave wound DC motor drawing an armature current of 20 A has provided with 360 armature conductors. If the flux per pole is 0.015 Wb then the torque developed by the armature of the motor is _______
10.23 N-m
34.39 N-m
17.17 N-m
19.08 N-m
Answer: 2. 34.39 N-m
Explanation:
The electromagnetic torque produced in d.c. machines is expressed as
T = KφIa
Where
K = PZ/2πA
Z= total armature conductors = 360
P= No. of poles = 4
A= No. of parallel paths. For a wave winding A = 2
∴ K = (4 × 360)/(2 × 3.14 × 2)
K = 114.64
φ = Flux = 0.015 Wb
Ia = Armature current = 20 A
T = 114.64 × 20 × 0.015
T = 34.39 N-m
20. A 4-pole generator having wave wound armature winding has 60 slots, each slot containing 20 conductors. What will be voltage generated in the machine when driven at 1000 rpm assuming the flux per pole to be 5 mWb?
