DC Generator EMF Equation MCQ || DC Generator EMF Equation Questions and Answers - Page 3 of 3

21. Emf and torque produced in a DC machine are proportional to ________ and _________ respectively.

Armature speed and armature emf
Armature emf and armature speed
Armature current and armature emf
Armature speed and armature current

Answer: 4. Armature speed and armature current

Explanation:

The electromagnetic torque produced in d.c. machines is expressed in terms of interaction between main field flux φ_f and armature MMF F_a.

T = KφI_a —— (1)

Where

K = PZ/2πA

$E_{a} = \frac{ϕ ω_{m} N_{c} P}{π}$ ——– (2)

N_c = number of coil turns

Thus, the average coil emf generated can also be represented as

Eg = K.φ.ω

So, average coil emf is directly proportional to ω (armature spee Eg ∝ ω

and average torque is directly proportional to Ia (armature current). T ∝ Ia

22. In a DC machine the torque induced beyond the pole shoes is ________

Zero
2/π∅i
π∅i/2
Can’t be calculated

Answer: 1. Zero

Explanation:

In DC machine pole-pair with brushes is placed in a geometrical neutral axis (GN, which is also a magnetic neutral axis (MN when the armature is not carrying current. Thus the magnetic field at the pole terminals in a DC machine will be equal to 0.

The magnetic neutral regions are located in the interpolar gaps. Without any significant loss of accuracy, the reluctance of the iron path will be neglected. Assuming that the air gap over the pole shoes is uniform, the flux density in the gap over the pole shoes is constant (as MMFacting along any flux path is constant for the concentrated field winding); the flux density in the air layer along the stator periphery gradually falls off to zero in the interpolar region.

23. The generated emf per parallel path in the armature of a DC Generator is:

Directly proportional to flux
Inversely proportional to number of poles
Directly proportional to rotational speed of armature
All of the above

Answer: 4. All of the above

Explanation:

The generated emf per parallel path in the armature of a DC Generator is given by

$E_{g} = \frac{ϕ P N}{60} \times \frac{Z}{A}$

ϕ = flux per pole

P = number of poles

N = Rotational speed in rpm

Z = number of conductors

A = number of parallel paths

Therefore, Generated EMF(E) is directly proportional to flux(ϕ ),

rotational speed(N), and Number of poles(P).

24. What is the value of Np in an average coil emf equation, for 10 armature conductors with 2 parallel paths?

Answer: 3. 2.5

Explanation:

In an emf equation of DC generator number of coil turns Nc is given as

Nc = Cp × Np.

Where

Cp = coils/ parallel path (which is twice the total armature turns)

Np = number of turns per parallel paths

Np = Nc/Cp

Np= 10/(2 × 2)= 10/4= 2.5.

25. Calculate the EMF generated by a 6 pole, 1200 rpm lap wound DC generator. The magnetic flux produced per pole is 7 Wb and there are 150 Conductors.

21 V
21 kV
42 kV
42 V

Answer: 2. 21 kV

Explanation:

Given

P = 6

Z = 150

N = 1200 rpm

ϕ = 7 Wb/pole

Lap winding

A = P = 6

The generated emf per parallel path in the armature of a DC Generator is given by

$E_{g} = \frac{ϕ P N}{60} \times \frac{Z}{A}$

$E_{g} = \frac{6 \times 1200 \times 150 \times 7}{60 \times 6}$

26. What is the torque equation in terms of B, Ic, l, Zr (r= mean air gap radius)?

Bav × Ic × l × Zr
Bav × Ic × l/Zr
Bav × Ic × Zr/l
Can’t be expressed

Answer: 1. Bav × Ic × l × Zr

Explanation:

Average force on an armature conductor is

f_av = B_av × l × I_c × R

where

l = active conductor length

Ic = conductor current.

B_av= Average flux density over pole

r = mean air-gap radius

Average torque on armature caused by one conductor is

T = B_av × l × I_c × R

Total torque developed by armature conductors is

T = B_av × l × I_c × Z × R

27. For a constant emf, if field current is reduced then the speed of the DC machine will _____

Remains same
Increases
Decreases
Can’t say

Answer: 3. Decreases

Explanation:

When the field current is decreased, the field flux is reduced, and the counter emf decreases. This permits more armature current. Therefore, the DC machine speed increased. For all other parameters kept the constant speed of the DC machine is inversely proportional to the field.

28. For an ideal DC machine, which phenomenon will reduce the terminal voltage?

Armature reaction
Commutation
Armature ohmic losses
All will contribute in reducing the terminal voltage

Answer: 1. Armature reaction

Explanation:

Armature reaction: It is the effect of the flux produced due to the armature current on the main field. It results in MNA shifting which causes sparking at the commutator bars and brushes. 3 The second effect of armature reaction is flux weakening which results in a reduction in terminal voltage.

29. What is the value of pole pitch (in SI unit) for mean air gap radius= 0.5mm and P=4?

0.785 × 10^-6
0.785 × 10^-3
0.785 × 10^-2
0.785 × 10^-4

Answer: 2. 0.785 × 10^-3

Explanation:

The pole pitch is related to the machine stator inner radius r and the number of poles by

τ_p = 2πr/P

τ_p = 2 × 3.14 × 0.5 × 10^-3 / 4

τ_p = 0.785 × 10^-3

30. Consider a 6 pole DC generator having 300 conductors and is rotating at a speed of 500 rpm. The flux per pole is 20 mWb. What will be the induced voltage, If the winding type is wave connected?

50 V
100 V
150 V
250 V

Answer: 3. 150 V

Explanation:

Given that,

Number poles P = 6

Conductors Z = 3000

Speed N = 500 rpm

The flux per pole (ϕ) = 20 × 10-3 Wb

As winding is wave type

Number of parallel paths A = 2

The generated emf per parallel path in the armature of a DC Generator is given by

$E_{g} = \frac{ϕ P N}{60} \times \frac{Z}{A}$

⇒ Induced or generated emf = (20 × 10-3 × 500 × 300 × 6) / (60× 2) = 150 V

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