DC Generators MCQ Quiz - Objective Question with Answer for DC Generators - Download Free PDF

Last updated on Jun 10, 2025

Latest DC Generators MCQ Objective Questions

DC Generators Question 1:

How are the segments of the commutator insulated from each other?

  1. Using thin layers of mica
  2. Using air gaps
  3. Using thick layers of rubber
  4. Using metal sheets

Answer (Detailed Solution Below)

Option 1 : Using thin layers of mica

DC Generators Question 1 Detailed Solution

Function of the commutator in the DC machine:

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  • A commutator consists of a set of copper segments, fixed around the part of the circumference of the rotating machine, or the rotor, and a set of spring-loaded brushes fixed to the stationary frame of the machine.
  • The main function of the commutator in the DC machine is to collect the current from the armature conductor as well as supply the current to the load using brushes. And also provides uni-directional torque for the DC motor. 
  • The commutator can be built with a huge number of segments in the edge form of hard-drawn copper.
  • The segments in the commutator are protected by the thin mica layer.

DC Generators Question 2:

If the number of parallel paths (A) in a DC generator is increased, the generated EMF will:

  1. decrease
  2. remain the same 
  3. increase
  4. become zero

Answer (Detailed Solution Below)

Option 1 : decrease

DC Generators Question 2 Detailed Solution

Concept

The generated EMF in a DC generator is given by:

\(E={NPϕ Z\over 60A} \)

where, E = EMF generated

N = Speed in RPM

P = Number of poles

ϕ = Flux per pole

Z = No. of conductors

A = No. of parallel paths

Explanation

From the above expression, we observed that the induced EMF is inversely proportional to the no. of parallel paths.

So, if the number of parallel paths (A) in a DC generator is increased, the generated EMF will decrease.

DC Generators Question 3:

Which of the following is an example of a constant load application for a DC shunt generator?

  1. Centrifugal pump
  2. DC arc welding
  3. Induction heating
  4. Electric motor

Answer (Detailed Solution Below)

Option 1 : Centrifugal pump

DC Generators Question 3 Detailed Solution

Explanation:

The application of DC Motors are as follows:

DC shunt motor

  • Shunt DC motors are utilized in Centrifugal Pumps, Lifts, Weaving Machines, Lathe Machines, Blowers, Fans, Conveyors, Spinning Machines, and other applications where a consistent speed is required.
     

DC series motor 

  • DC series motors are used where high starting torque is required. Major applications for DC motors are elevators, steel mills, rolling mills, locomotives, and excavators.
     

Cumulative compound DC motor

  • It has high starting torque and has varying speeds within limits i.e. it has self-adjustable speed with changing load. These are used where we require sudden heavy loads for a short duration.​
  • Hence, it is used for driving constant speed line shafts, lathes, constant speed head centrifugal pumps, fans woodworking machines, reciprocating pumps, etc.
     

Differentially compound DC motor

  • The differential compound generators are used for arc welding purposes where a large voltage drop and constant current are required.

DC Generators Question 4:

Which of the following is the commonly used generator in a DC welding machine? 

  1. Shunt generator
  2. Permanent magnet generator
  3. Compound generator 
  4. Series generator

Answer (Detailed Solution Below)

Option 3 : Compound generator 

DC Generators Question 4 Detailed Solution

Applications of DC generators

Compound generator:

A compound generator is commonly used in DC welding machines because it combines the characteristics of both series and shunt generators:

  • It provides a relatively stable voltage (like a shunt generator).
  • It can handle sudden load changes well (like a series generator).
  • This ensures a consistent arc, which is crucial for welding quality.


Shunt generator:

  • Shunt generators are known for providing a relatively constant output voltage, making them suitable for several practical applications such as battery charging, electroplating and electrolysis.


Series generator:

  • Used in early electric traction, arc lighting, and boosting voltage in long DC lines.


Permanent magnet generator:

  • Used in wind turbines, small hydro generators, bicycles (dynamo), portable generators, engine ignition systems, and renewable energy systems for efficient, maintenance-free power generation.

DC Generators Question 5:

In case of four pole, lap wound machine if the air gap under each pole is the same, then what will be the result?

  1. There will be reduced eddy currents
  2. There will be reduced hysteresis loss
  3. Current in each path will not be the same
  4. It will result in higher terminal voltage 

Answer (Detailed Solution Below)

Option 3 : Current in each path will not be the same

DC Generators Question 5 Detailed Solution

Explanation:

Four Pole Lap Wound Machine with Equal Air Gap Under Each Pole

Definition: A lap wound machine refers to the type of winding configuration in electrical machines where each coil's end is connected to the adjacent commutator segment, resulting in multiple parallel paths. In a four-pole machine, there are four magnetic poles, and the air gap under each pole is typically kept uniform to ensure consistent magnetic flux distribution.

Working Principle: In a lap wound machine, the armature winding is divided into as many parallel paths as the number of poles in the machine. For a four-pole machine, there will be four parallel paths. When the air gap under each pole is uniform, the magnetic flux density across the armature is more evenly distributed. This affects the current and voltage in the winding paths.

Analysis of the Correct Option:

Correct Option: Option 3: Current in each path will not be the same

In a lap wound machine with a uniform air gap under each pole, the magnetic flux distribution should ideally be symmetrical. However, in practical scenarios, even with an equal air gap, other factors such as slight manufacturing imperfections, uneven saturation of the magnetic core, or minor misalignments can cause variations in the flux linkage across different parts of the armature. Since the current in each path of a lap wound machine depends on the induced electromotive force (EMF) in the respective coils, these variations in flux linkage lead to unequal induced EMFs in the parallel paths. Consequently, the currents in the parallel paths will also differ slightly.

For example, if one path experiences a slightly higher flux density due to unavoidable non-uniformities, the induced EMF in that path will be higher, leading to a higher current in that path compared to the others. This imbalance in current distribution can cause uneven heating and may affect the machine's overall efficiency and performance. Therefore, even with an equal air gap under each pole, the current in each path of a four-pole lap wound machine will not be the same.

Important Information:

To further understand the implications of this condition, let us evaluate the other options:

Option 1: There will be reduced eddy currents

This option is incorrect. Eddy currents are induced circulating currents in the core material of the machine caused by changing magnetic fields. The uniformity of the air gap under each pole does not directly influence the eddy currents. Instead, eddy currents depend on factors such as the rate of change of magnetic flux, core material properties, and lamination thickness. Therefore, maintaining an equal air gap under each pole will not necessarily reduce eddy currents.

Option 2: There will be reduced hysteresis loss

This option is also incorrect. Hysteresis loss occurs in the magnetic core material due to the repeated magnetization and demagnetization cycles as the machine operates. Hysteresis loss depends on the properties of the core material, such as its hysteresis loop area, and the frequency of operation. While a uniform air gap can help ensure consistent magnetic flux distribution, it does not directly reduce hysteresis loss, which is determined by the material and operating conditions rather than the air gap uniformity.

Option 4: It will result in higher terminal voltage

This option is incorrect. The terminal voltage of a machine depends on the total induced EMF in the armature windings and the voltage drop across the internal resistances and brushes. While a uniform air gap can contribute to a more consistent flux distribution, it does not inherently result in a higher terminal voltage. The terminal voltage is primarily influenced by the machine design, operating speed, and load conditions.

Option 5: Not provided in the question

Since there is no information or specific context provided for Option 5, it is not considered relevant to the analysis. The correct answer remains Option 3.

Conclusion:

In a four-pole lap wound machine, even if the air gap under each pole is uniform, the practical imperfections and variations in flux linkage can lead to unequal induced EMFs in the parallel paths. This results in unequal currents in the paths, making Option 3 the correct answer. Understanding this phenomenon is crucial for the design and analysis of electrical machines to ensure their efficient and reliable operation. The other options are either unrelated to the air gap uniformity or are not directly impacted by it, as explained above.

Top DC Generators MCQ Objective Questions

How can a load be shifted from one DC shunt generator to another running in parallel?

  1. Adjust their field rheostat
  2. Insert a resistance in the armature circuit
  3. Adjust the speed of the prime mover
  4. Use an equaliser connection

Answer (Detailed Solution Below)

Option 1 : Adjust their field rheostat

DC Generators Question 6 Detailed Solution

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  • When two generators are operating in parallel, the load may be shifted from one shunt generator to another merely by adjusting the field excitation
  • If generator 1 is to be shut down, the whole load can be shifted onto generator 2 provided it has the capacity to supply that load
  • In that case, the field current of generator 1 gradually reduces to zero

 

Important Points:

For stable parallel operation, the most suitable type of DC generator is a shunt generator as it has slightly drooping characteristics. If there is any tendency for a generator to supply more or less than its proper share of load it changes system voltage which certainly opposes this tendency. This restores the original division of load. Thus the shunt generators automatically remain in parallel, once they are paralleled.

The characteristics of dc shunt generator are shown below

SSC JE Electrical 36 20Q BSPHCL JE EE Part 1 Hindi - Final images Q1

Flemings right hand rule is used to find the

  1. Direction of rotation
  2. Direction of flux
  3. Direction of emf
  4. Direction of torque

Answer (Detailed Solution Below)

Option 3 : Direction of emf

DC Generators Question 7 Detailed Solution

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Fleming right-hand thumb rule:

When a conductor such as a wire attached to a circuit moves through a magnetic field, an electric current is induced in the wire due to Faraday's law of induction

Fleming's right-hand rule (for generators) shows the direction of induced current when a conductor attached to a circuit moves in a magnetic field.

  • The thumb is pointed in the direction of the motion of the conductor relative to the magnetic field.
  • The first finger is pointed in the direction of the magnetic field. (north to south)
  • Then the second finger represents the direction of the induced or generated current within the conductor (from the terminal with lower electric potential to the terminal with higher electric potential, as in a voltage source)

 

F2 Shubham Madhu 07.08.20 D3

Finger

Right-hand rule

Left-hand rule

Thumb

The direction of motion of conductor (input)

The direction of the conductor (output)

Forefinger

Magnetic field

Magnetic field

Middle finger

The direction of induced emf (output)

The direction of current (input)

 

Conclusion:

The direction of induced emf is known by Flemings right-hand thumb rule

Identify the machine shown in the circuit.

F2 U.B Madhu 5.11.19 D 4

  1. DC short shunt compound generator
  2. DC short shunt compound motor
  3. DC shunt motor
  4. DC long shunt compound generator

Answer (Detailed Solution Below)

Option 4 : DC long shunt compound generator

DC Generators Question 8 Detailed Solution

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Based on the connection of armature and field windings DC generators can be classified as:

Type of DC Machine

Circuit diagram

Separately excited DC generator

F1 U.B Deepak 31.12.2019 D 6

DC shunt generator

F1 U.B Deepak 31.12.2019 D 7

DC series generator

F1 U.B Deepak 31.12.2019 D 8

DC short shunt compound generator

F1 U.B Deepak 31.12.2019 D 9

DC long shunt compound generator

F1 U.B Deepak 31.12.2019 D 10


Therefore, the machine shown in the question represents a DC long shunt compound generator.

Which of the following generators at load condition offers positive poorest voltage regulation?

  1. Cumulative compounded
  2. Series
  3. Differential compounded
  4. Shunt

Answer (Detailed Solution Below)

Option 3 : Differential compounded

DC Generators Question 9 Detailed Solution

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The correct answer is option "3'.

Concept :- For differential compound generator it is positive (Poorest positive regulation among all)

  • The voltage regulation of a generator is defined as the change in the voltage drop from no load to full load to full load voltage.
  • Voltage regulation = (no-load voltage - full load voltage)/full load voltage
  • In the case of a series generator, the field is connected in series with the armature. Any increase in load current causes an increase in the field and hence the terminal voltage rises.
  • Hence it has negative voltage regulation and it has the poorest voltage regulation.

During on-load conditions, the differentially compounded DC generator has the poorest voltage regulation as shown.

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During the no-load condition, the DC series generator has the poorest voltage regulation.

5e26d331f60d5d6c153be4e8 16328251006432

Key Points

  •  Voltage regulation of shunt generator is positive
  • For series generator, it  is negative (Poorest negative voltage regulation among all)
  • For over compound generator it is negative
  • For under compound it is positive
  • For, flat compound generator it is zero (lowest Voltage regulation among all)
  • For differential compound generator it is positive (Poorest positive regulation among all)

In a 4 pole, 20 kW, 200 V wave wound DC shunt generator, the current in each parallel path will be _______.

  1. 100 A
  2. 25 A
  3. 10 A
  4. 50 A

Answer (Detailed Solution Below)

Option 4 : 50 A

DC Generators Question 10 Detailed Solution

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Concept:

The current in each parallel path \(= \frac{I}{A}\)

A is the number of parallel paths

In a wave wound generator, number of parallel paths = 2

In a lap wound generator, number of parallel paths = number of poles

Calculation:

Power (P) = 20 kW

Voltage (V) = 200 V

Total current \(I = \frac{P}{V} = \frac{{20 \times {{10}^3}}}{{200}} = 100\;A\)

Number of parallel paths = 2

The current in each parallel path \(= \frac{{100}}{2} = 50\;A\)

Armature reaction of an unsaturated DC machine results in _______ effect.

  1. Cross-Magnetising
  2. Demagnetising
  3. Axial Magnetising
  4. Two-Pole Magnetising

Answer (Detailed Solution Below)

Option 1 : Cross-Magnetising

DC Generators Question 11 Detailed Solution

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Armature reaction:

The effect of armature flux (ϕa) on main field flux (ϕm) is called armature reaction.

The armature reaction mmf produces two undesirable effects on the main field flux and these are

(1) Net reduction in the main field flux per pole

(2) Distortion of the main field flux wave along the air gap periphery.

The armature mmf in a dc machine is stationary with respect to field poles but rotating with respect to armature.
F4 Vinanti Engineering 22.05.23 D1 V2
F4 Vinanti Engineering 22.05.23 D2 V2

  • Armature reaction of unsaturated DC machine results in cross – magnetizing effect.
  • Armature reaction distorts the main flux, hence the position of M.N.A. gets shifted (M.N.A. is perpendicular to the flux lines of main field flux).
  • Brushes should be placed on the M.N.A., otherwise, it will lead to sparking at the surface of brushes. So, due to the armature reaction, it is hard to determine the exact position of the MNA.
  • For a loaded dc generator, MNA will be shifted in the direction of the rotation.
  • While for a loaded dc motor, MNA will be shifted in the direction opposite to that of the rotation.

F1 Shraddha Uday 22.12.2020 D5

The yoke in a large DC machine is made of:

  1. Grain oriented steel
  2. Cast iron
  3. Cast steel
  4. Mild steel

Answer (Detailed Solution Below)

Option 3 : Cast steel

DC Generators Question 12 Detailed Solution

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F1 J.P Madhu 29.05.20 D2

The above figure shows the constructional details of a simple 4-pole DC machine.

Yoke:

  • The outer frame of a dc machine is called a yoke.
  • It is made up of cast iron or steel. In small DC machines, it is made up of cast iron and in the case of large DC machines, it is made up of cast steel.
  • It not only provides mechanical strength to the whole assembly but also carries the magnetic flux produced by the field winding.

Key Points

Type of material used for the construction of the DC machine:

Yoke

Cast steel

Pole core and pole shoe

Cast steel

Armature core

Laminated steel

Commutator

Hard drawn copper

Brushes (small machine)

Copper or Carbon

Brushes (normal and large machine)

Electro - graphite

A 220V dc machine supplies 20A at 200 V as a generator. The armature resistance is 0.2 . If the machine is now operated as a motor as a same terminal voltage and current but with flux increased by 10%, the ratio of motor speed to generator speed is

  1. 0.87
  2. 0.95
  3. 0.96
  4. 1.06

Answer (Detailed Solution Below)

Option 1 : 0.87

DC Generators Question 13 Detailed Solution

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Concept:

The EMF equation of a DC Machine is

\({E_b} = \frac{{NPϕ Z}}{{60A}}\)

From the above equation,

\(N \propto \frac{{{E_b}}}{ϕ }\)

In a dc generator, induced emf is

Eg = V + IaRa

In a dc motor, back emf is

Eb = V – IaRa

Where,

N is the speed in rpm

ϕ is the flux per pole

P is the number of poles

Z is the number of conductors

V is the terminal voltage

Ia is the armature current

Ra is the armature resistance

Calculation:

Given-

V = 200, Ra = 0.2 Ω, Ia = 20 A, ϕ2 = 1.1 ϕ1

Let ϕ1 = Generator flux, ϕ2 = Motor flux, N1 = Generator speed, N2 = Motor speed 

∴ \({E_g} = 200 + 20 \times 0.2 = 204\ V\)

\(\begin{array}{l} {E_b} = 200 - 20 \times 0.2 = 196\ V\\ \frac{{{N_2}}}{{{N_1}}} = \frac{{{E_b}}}{{{E_g}}} \times \frac{{{ϕ _1}}}{{{ϕ _2}}} = \frac{{196}}{{204}} \times \frac{{{ϕ _1}}}{{1.1{ϕ _1}}} = 0.87 \end{array}\)

A 6-pole lap-connected DC generator has 480 conductors and armature circuit resistance is 0.06 ohm. If the conductors are reconnected to form wave winding, other things remaining unchanged, the value of armature circuit resistance will be

  1. 0.01 Ω
  2. 0.08 Ω
  3. 0.36 Ω
  4. 0.54 Ω

Answer (Detailed Solution Below)

Option 4 : 0.54 Ω

DC Generators Question 14 Detailed Solution

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Given that:

DC generator wit Pole (P) = 6

Conductor = 480 (Z)

For lap winding armature resistance Ra = 0.06 Ω

When the generator is connected as lap winding then

\({R_a} = {R_{Lap}} = \frac{{Z/P}}{{m.p}} \cdot \frac{{\rho \ell }}{A} = \frac{Z}{{{P^2}}}\frac{{\rho \ell }}{A}\)        (Parallel path = m.p and m = 1)

Now, when the generator is connected as wave winding.

\({R_{wave}} = \frac{Z}{{\left( {{A_p}} \right)}} \cdot \frac{{\rho \ell }}{A}\)

In wave winding, the no. of parallel paths = A = 2

No. of conductors in each parallel path (AP) = \(\frac{Z}{2}\)

\({\left( {{R_a}} \right)_{wave}} = \frac{Z}{4}\frac{{\rho \ell }}{A}\)

\(\frac{{{{\left( {{R_a}} \right)}_{lap}}}}{{{{\left( {{R_a}} \right)}_{wave}}}} = \frac{{\left( {\frac{Z}{{{P^2}}}\frac{{\rho \ell }}{A}} \right)}}{{\left( {\frac{Z}{4} \cdot \frac{{\rho \ell }}{A}} \right)}} = \frac{4}{{{P^2}}} = \frac{4}{{{6^2}}} = \frac{1}{9}\)

⇒ (Ra)wave = 9 × (Ra)lap

= 9 × 0.06 = 0.54 Ω 

Alternate Method

 A2lap × Rlap = A2wave × Rwave

In lap winding A = P = 6

Rlap = 0.06 Ω

In wave winding A = 2

62 × 0.06 = 22 × Rwave

Rwave = (36 × 0.06) / 4

Rwave = 9 × 0.06 = 0.54 Ω 

What will happen if the back emf of a D.C. motor vanishes suddenly?

  1. The motor will stop
  2. The motor will continue to run
  3. The armature may burn
  4. The motor will run noisy

Answer (Detailed Solution Below)

Option 3 : The armature may burn

DC Generators Question 15 Detailed Solution

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If back emf of a dc motor vanishes suddenly, motor circuit will try to retain back emf by drawing more current from supply.

The voltage equation of dc motor is, Eb = V – IaRa

As the back emf vanishes zero, the whole supply voltage appears across armature and heavy current flows.

If supplying unit didn’t trip down by this time, excess current in armature may heat up the armature and it may cause burning of armature winding.
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