DC Machine EMF and Torque Production MCQ || DC Machine EMF and Torque Production Questions and Answers

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

${E_a} = \dfrac{{\phi {\omega _m}{N_c}P}}{\pi }$ ——– (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

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.

Answer: 4. All of the above

Explanation: 

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

${E_g} = \frac{{\phi {\rm{\;}}P{\rm{\;}}N}}{{60}} \times \frac{Z}{A}$

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).

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.

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

${E_g} = \frac{{\phi {\rm{\;}}P{\rm{\;}}N}}{{60}} \times \frac{Z}{A}$

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

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

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.

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.

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

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

${E_g} = \frac{{\phi {\rm{\;}}P{\rm{\;}}N}}{{60}} \times \frac{Z}{A}$

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