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

Last updated on Jun 13, 2025

Latest DC Generators MCQ Objective Questions

DC Generators Question 1:

How can eddy current loss be reduced in the armature core?

  1. By increasing the motor's speed
  2. By increasing the core resistance
  3. By using a high resistance core material
  4. By laminating the core to reduce the flow of eddy currents

Answer (Detailed Solution Below)

Option 4 : By laminating the core to reduce the flow of eddy currents

DC Generators Question 1 Detailed Solution

Explanation:

Reducing Eddy Current Loss in Armature Core

Definition: Eddy current loss is a type of power loss that occurs in the core of electrical machines such as motors, transformers, and generators. It is caused by circulating currents induced within the conductive material of the core due to the alternating magnetic flux. These currents flow in loops within the material, producing heat and resulting in energy loss.

Working Principle of Eddy Currents:

Eddy currents are induced in a conductor when it is exposed to a changing magnetic field, as per Faraday's law of electromagnetic induction. The magnitude of these currents depends on the rate of change of the magnetic flux, the material's electrical conductivity, and the geometry of the conductor. The circulating currents create their own magnetic field, which opposes the original magnetic field (as stated by Lenz's law), leading to energy dissipation in the form of heat.

Correct Option Analysis:

The correct option is:

Option 4: By laminating the core to reduce the flow of eddy currents.

This is the most effective method for minimizing eddy current loss. The lamination process involves dividing the core into thin layers or sheets of insulated material. These laminations are stacked together, and each layer is electrically insulated from the others, typically using a thin coating of varnish or oxide. The purpose of laminating the core is to restrict the flow of eddy currents by reducing the area available for their circulation. As a result, the eddy current paths are interrupted, and their magnitude is significantly decreased.

Why Laminating the Core Works:

  • The induced voltage in the core due to the alternating magnetic flux is proportional to the rate of change of flux and the area of the loop (as per Faraday's law). By reducing the cross-sectional area of the loops using laminations, the induced voltage and hence the eddy current magnitude are minimized.
  • The heat generated by eddy currents is directly proportional to the square of the current. Therefore, reducing the magnitude of eddy currents through lamination significantly reduces energy losses.
  • Laminations are typically made of high-resistance materials like silicon steel, which further limits the flow of eddy currents.

Advantages of Lamination:

  • Significant reduction in eddy current losses, improving the efficiency of electrical machines.
  • Cost-effective and straightforward technique for core design.
  • Improved thermal performance due to reduced heat generation.

Applications:

Laminated cores are widely used in transformers, electric motors, generators, and other electrical machines where alternating magnetic fields are present. This technique is crucial for ensuring the efficient operation of these devices.

Additional Information:

Eddy current loss is one of the two primary core losses in electrical machines, the other being hysteresis loss. While lamination is effective for reducing eddy current loss, hysteresis loss is minimized by using magnetic materials with low hysteresis, such as silicon steel.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: By increasing the motor's speed.

This option is incorrect as increasing the motor's speed does not reduce eddy current loss. In fact, higher speeds result in a higher rate of change of magnetic flux, which increases the induced voltage and, consequently, the eddy currents. This leads to greater energy loss and heat generation.

Option 2: By increasing the core resistance.

While increasing the core resistance can theoretically reduce eddy current losses, it is not a practical solution. Core resistance is primarily determined by the material properties and geometry of the core. Lamination is a more effective and feasible method for increasing the core's effective resistance to eddy currents without compromising the machine's performance.

Option 3: By using a high resistance core material.

Using high-resistance materials like silicon steel can help in reducing eddy current losses. However, this alone is not sufficient to completely mitigate the problem. Lamination remains the most effective method for minimizing eddy currents, even when high-resistance materials are used.

Option 5: (No option provided in this case).

This option is not applicable in the given context.

Conclusion:

Among the given options, laminating the core to reduce the flow of eddy currents is the most effective and widely used method for minimizing eddy current losses. This technique significantly improves the efficiency and performance of electrical machines by reducing energy dissipation and heat generation. Understanding and implementing effective strategies to mitigate core losses is essential for the optimal design and operation of electrical devices.

DC Generators Question 2:

The material used in the armature core of a DC generator is typically: 

  1. silicon steel 
  2. copper
  3. aluminium
  4. plastic

Answer (Detailed Solution Below)

Option 1 : silicon steel 

DC Generators Question 2 Detailed Solution

Explanation:

Material Used in Armature Core of a DC Generator

Definition: The armature core of a DC generator is a crucial component designed to support the windings and facilitate the electromagnetic induction process, which generates electricity. The material used for the armature core plays a vital role in ensuring efficient operation and minimizing energy losses due to hysteresis and eddy currents.

Correct Option: Option 1: Silicon Steel

Silicon steel is the material commonly used for the armature core of a DC generator. This is because silicon steel possesses excellent magnetic properties, which are essential for efficient electromagnetic induction. Additionally, silicon steel reduces both hysteresis loss and eddy current loss, which are critical factors in the operation of electrical machines.

Reason for Using Silicon Steel:

  • High Permeability: Silicon steel has high magnetic permeability, which allows it to easily magnetize and demagnetize as required during the operation of the DC generator.
  • Low Hysteresis Loss: When the magnetic field changes direction rapidly during operation, hysteresis loss occurs. Silicon steel minimizes this loss due to its favorable magnetic properties.
  • Reduced Eddy Current Loss: Eddy currents are induced in the armature core due to changing magnetic fields. Silicon steel is often laminated to further reduce eddy current losses, ensuring more efficient operation.
  • Mechanical Strength: Silicon steel provides the necessary mechanical strength to support the armature windings and withstand the stresses during operation.
  • Durability: The material is robust and durable, ensuring long-term reliable performance of the DC generator.

Applications: Silicon steel is widely used in armature cores not only in DC generators but also in other electrical machines, such as transformers and AC motors, due to its excellent magnetic properties and efficiency in reducing losses.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: Copper

Copper is an excellent conductor of electricity and is widely used in electrical machines for windings and conductors. However, it is not suitable for the armature core material because it lacks the magnetic properties required for efficient operation. Copper does not offer high magnetic permeability, nor does it minimize hysteresis and eddy current losses, making it an unsuitable choice for the armature core.

Option 3: Aluminium

Aluminium is another material known for its electrical conductivity and lightweight properties. However, like copper, aluminium is not suitable for the armature core because it lacks the necessary magnetic properties. Aluminium's inability to efficiently magnetize and demagnetize makes it a poor choice for this application.

Option 4: Plastic

Plastic is a non-magnetic material and cannot support electromagnetic induction. It is not suitable for use in the armature core of a DC generator. While plastic may be used for insulation or housing components, it is entirely unsuitable for the magnetic core.

Conclusion:

Silicon steel is the ideal material for the armature core of a DC generator due to its excellent magnetic properties, high permeability, and ability to minimize energy losses such as hysteresis and eddy current losses. The other options—copper, aluminium, and plastic—are unsuitable for the armature core because they lack the required magnetic properties and efficiency for this application.

DC Generators Question 3:

A DC shunt generator produces 450A at 230V. The resistance of shunt field and armature are 50 ohms and 0.025 ohm respectively. The armature voltage drop will be

  1. 11.36 V
  2. 22.7 V
  3. 31.6 V
  4. 38.4 V

Answer (Detailed Solution Below)

Option 1 : 11.36 V

DC Generators Question 3 Detailed Solution

Explanation:

Step 1: Understanding the parameters

  • Generated Current (IG): The total current generated by the DC shunt generator is given as 450A.
  • Terminal Voltage (VT): The terminal voltage of the generator is given as 230V.
  • Shunt Field Resistance (Rsh): The resistance of the shunt field is 50 ohms.
  • Armature Resistance (Ra): The resistance of the armature is 0.025 ohms.

Step 2: Calculating the shunt field current (Ish)

The shunt field current can be calculated using Ohm's law:

Ish = VT / Rsh

Substituting the known values:

Ish = 230 / 50 = 4.6 A

Step 3: Calculating the armature current (Ia)

The armature current can be determined as the difference between the total generated current (IG) and the shunt field current (Ish):

Ia = IG - Ish

Substituting the known values:

Ia = 450 - 4.6 = 445.4 A

Step 4: Calculating the armature voltage drop (Va)

The armature voltage drop is given by Ohm's law:

Va = Ia × Ra

Substituting the known values:

Va = 445.4 × 0.025 = 11.36 V

Conclusion:

The armature voltage drop is 11.36V, making Option 1 the correct answer.

DC Generators Question 4:

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 4 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 5:

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 5 Detailed Solution

Concept

The generated EMF in a DC generator is given by:

E=NPϕZ60A

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.

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.

5e26d331f60d5d6c153be4e8 16328251006251

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 =IA

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=PV=20×103200=100A

Number of parallel paths = 2

The current in each parallel path =1002=50A

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

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 12 Detailed Solution

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Concept:

The EMF equation of a DC Machine is

Eb=NPϕZ60A

From the above equation,

NEbϕ

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 

∴ Eg=200+20×0.2=204 V

Eb=20020×0.2=196 VN2N1=EbEg×ϕ1ϕ2=196204×ϕ11.1ϕ1=0.87

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 13 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 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

Ra=RLap=Z/Pm.pρA=ZP2ρA        (Parallel path = m.p and m = 1)

Now, when the generator is connected as wave winding.

Rwave=Z(Ap)ρA

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

No. of conductors in each parallel path (AP) = Z2

(Ra)wave=Z4ρA

(Ra)lap(Ra)wave=(ZP2ρA)(Z4ρA)=4P2=462=19

⇒ (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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