DC Generator EMF Equation MCQ || DC Generator EMF Equation Questions and Answers

1. The emf equation of DC generator is E

  1. Øp/nza
  2. Øp60/nza
  3. Øpa/60nz
  4. Øpnz/60a

Answer: 4. Øpnz/60a

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

 

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

  1. 1000 rpm
  2. 2000/π  rpm
  3. 3000/π  rpm
  4. 4000/π rpm

Answer: 4. 4000/π rpm

Explanation: 

Induced emf Ef = Kϕω

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?

  1. Motoring
  2. Generating
  3. Can’t be determined using directions
  4. 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?

  1. 0.188 Wb
  2. 18.88 Wb
  3. 0.0188 Wb
  4. 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?

  1. Ea = φωmNc/P
  2. Ea = φωmNcP/60
  3. Ea = ϕωmNcP
  4. 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

${E_a} = \dfrac{{\phi ZNP}}{{60}}$

 

ωm = speed in rad/sec

ωm = 2πN/60

N = 30ω/ π ……(2)

Z = 2 × Nc …….(3)

Nc = number of coil turns

From equations (1), (2) and (3)

${E_a} = \frac{{\phi \times 2{N_c} \times 30{w_m} \times P}}{{60 \times \pi }}$

 

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

 

6. Nature of the flux density wave in the air gap is __________ (for armature current equal to 0).

  1. Flat-topped with quarter-wave symmetry
  2. Point topped with quarter-wave symmetry
  3. Flat-topped with half-wave symmetry
  4. 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?

  1. 188 V
  2. 276 V
  3. 94 V
  4. 188 mV

Answer: 1. 188 V

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

E.M.F. generated by a generator is

${E_a} = \dfrac{{\phi ZNP}}{{60A}}$

 

Where,

ϕ = flux

Z = number of conductors = 400

N = speed in RPM = 1500

P = number of poles = 4

A = number of parallel paths = 4

${E_a} = \dfrac{{0.0188 \times 1500 \times 400 \times 4}}{{60 \times 4}}$

 

Ea = 188 V

 

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.

  1. 154 V
  2. 196 V
  3. 120 V
  4. 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.

  1. Equal
  2. Dependent
  3. Independent
  4. 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. 1/2
  2. 2/1
  3. 1/4
  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.

Thus, ratio

T1/T2 = 20/40

T1/T2 = 1/2

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