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

Explanation:

In a DC Series Motor, at Low or Light Load, Why is the Torque Low?

Definition: A DC series motor is a type of direct current (DC) motor where the field winding is connected in series with the armature winding. This configuration ensures that the same current flows through both the armature and the field winding. The torque produced in a DC series motor is proportional to the product of the armature current and the magnetic flux generated by the field winding. The relationship between the load, current, flux, and torque is a critical factor in understanding the behavior of the motor under different load conditions.

Correct Option Analysis:

The correct option is:

Option 1: Due to the low armature current and low field flux.

This is the correct explanation for the low torque in a DC series motor at light or low load. The torque in a DC series motor is given by the formula:

T ∝ Φ × I_a

Where:

• T is the torque.
• Φ is the magnetic flux produced by the field winding.
• I_a is the armature current.

At light or low load, the motor requires less torque to overcome the load. As a result, the armature current (I_a) is low. Since the field winding in a series motor is directly connected in series with the armature, the field flux (Φ) is also reduced because it depends on the armature current. The reduced flux and low armature current together result in a significantly lower torque output. Thus, the torque decreases due to the combined effect of low armature current and low field flux.

Detailed Explanation:

1. **Torque Equation in a DC Series Motor**:

The torque in a DC motor is generated by the interaction of the magnetic field and the armature current. The relationship can be expressed as:

T = k × Φ × I_a

Here, k is a constant that depends on the design of the motor. In a series motor, since the field winding is connected in series with the armature, the flux (Φ) is directly proportional to the armature current (I_a) up to the point of core saturation.

2. **Effect of Light Load**:

When the load on the motor is light, the required torque is minimal. As a result, the motor draws less current from the supply. The reduced current leads to a decrease in the field flux (Φ), since the field winding is in series with the armature. With both the armature current (I_a) and the field flux (Φ) being low, the torque produced by the motor is also low.

3. **Performance at Low Load**:

At low load, the motor operates at higher speeds because the back EMF (E_b) approaches the supply voltage (V). The reduced current (I_a) in the armature and field winding results in lower magnetic flux (Φ), which, combined with the low armature current, causes the torque to be minimal. This behavior is a characteristic of DC series motors and highlights their dependency on load conditions for efficient operation.

Conclusion:

The low torque at light load in a DC series motor is primarily due to the low armature current and the corresponding low field flux. This explanation aligns with the correct option (Option 1).

Additional Information

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

Option 2: Due to the saturation of the field cores.

This option is incorrect. While core saturation can affect the performance of a DC motor, it is not the primary reason for low torque at light load. Saturation occurs when the magnetic field in the core reaches its maximum level and cannot increase further, even with an increase in current. At light load, the current (and hence the flux) is low, so saturation does not occur.

Option 3: Due to the high field flux.

This option is also incorrect. At light load, the armature current is low, resulting in low field flux, not high field flux. High field flux would require a higher armature current, which is not the case under light load conditions.

Option 4: Due to the high armature current.

This option is incorrect because, at light load, the armature current is low. High armature current occurs only under heavy load conditions, where the motor needs to produce more torque to overcome the load.

Conclusion:

Understanding the behavior of DC series motors under different load conditions is essential for their proper application and operation. At light load, the low armature current and corresponding low field flux result in low torque, as explained in the correct option. The other options misinterpret the underlying principles and do not accurately describe the motor's behavior in this scenario.

Torque in a DC Series Motor at Saturation

Definition: In a DC series motor, torque is initially proportional to the square of the armature current under unsaturated conditions. However, as the motor approaches magnetic saturation, the relationship between torque and armature current changes due to the limitation in further increase of flux.

Working Principle: The torque (T) produced in a DC series motor is generally given by the expression:

T ∝ Φ × I_a

Here:

• T = Torque
• Φ = Magnetic flux produced by the field winding
• I_a = Armature current

Under normal unsaturated conditions, the magnetic flux (Φ) is directly proportional to the armature current (I_a). As a result, the torque becomes proportional to the square of the armature current:

T ∝ Φ × I_a ∝ I_a²

However, as the motor operates at higher currents, the magnetic circuit of the motor approaches saturation. Once saturation occurs, further increases in armature current do not lead to a proportional increase in flux (Φ). Instead, the flux becomes nearly constant, and the torque becomes proportional to the armature current only:

T ∝ I_a

Correct Option Analysis:

The correct option is:

Option 1: Armature current only

At the point of saturation, the magnetic flux (Φ) produced by the field winding becomes constant due to the limitations in further magnetizing the core material. Therefore, the torque is no longer dependent on the flux and becomes directly proportional to the armature current (I_a). This is why, under saturation, the torque in a DC series motor is proportional to the armature current only.

Additional Information

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

Option 2: Both armature current and field flux

Under normal (unsaturated) conditions, torque in a DC series motor is proportional to both the armature current and the magnetic flux (Φ × I_a). However, once saturation occurs, the flux becomes constant and no longer varies with armature current. Hence, this option is incorrect under saturation conditions.

Option 3: Field flux only

This option is incorrect because the torque in a DC series motor is never solely dependent on the field flux (Φ). Torque is always a product of flux and armature current (Φ × I_a). Therefore, even under saturation, armature current is essential in determining the torque.

Option 4: The square of the armature current

This option is correct only under unsaturated conditions, where flux is proportional to armature current (Φ ∝ I_a) and torque becomes proportional to the square of the armature current (T ∝ I_a²). However, under saturation conditions, flux becomes constant, and torque is no longer proportional to the square of the armature current.

Conclusion:

At the point of saturation in a DC series motor, the torque becomes proportional to the armature current (I_a) only. This is due to the magnetic saturation of the core material, which prevents further increases in flux despite increases in armature current. Understanding this behavior is crucial for analyzing the performance characteristics of DC series motors in various operating conditions.

Explanation:

Shunt Motor Current Calculation

Definition: A shunt motor is a type of DC motor where the field windings are connected in parallel (or shunt) with the armature windings. It is widely used for applications requiring constant speed under varying load conditions.

Given Data:

• Motor speed = 1500 r/min
• Source voltage (V_s) = 120 V
• Line current (I_L) = 51 A
• Shunt field resistance (R_sh) = 120 Ω
• Armature resistance (R_a) = 0.1 Ω

Objective: To calculate the armature current (I_a).

Solution:

In a shunt motor, the line current (I_L) is the sum of the armature current (I_a) and the shunt field current (I_sh).

Step 1: Calculate the shunt field current (I_sh)

The shunt field current is given by Ohm's law:

I_sh = V_s / R_sh

Substitute the values:

I_sh = 120 V / 120 Ω

I_sh = 1 A

Step 2: Calculate the armature current (I_a)

The armature current is given by:

I_a = I_L - I_sh

Substitute the values:

I_a = 51 A - 1 A

I_a = 50 A

Conclusion:

The armature current in the motor is 50 A.

Additional Information

To further analyze the options, let’s evaluate why the other choices are incorrect:

Option Analysis:

Option 1: 1 A

This option is incorrect because 1 A is the shunt field current (I_sh), not the armature current (I_a). The armature current is calculated by subtracting the shunt field current from the line current, resulting in 50 A.

Option 2: 51 A

This option is incorrect because 51 A is the line current (I_L), which is the sum of the armature current and the shunt field current. The armature current alone is 50 A.

Option 3: 50 A

This is the correct answer. The armature current is calculated as the difference between the line current and the shunt field current, which is 50 A.

Option 4: 12 A

This option is incorrect and does not correspond to any calculation or parameter in this problem. It might be a distractor or an arbitrary value.

Conclusion:

In a shunt motor, the line current is the sum of the armature current and the shunt field current. By understanding this relationship and applying Ohm's law, the armature current is determined to be 50 A, making Option 3 the correct choice.

The most suitable method to measure the resistance of the series field winding of a DC series motor is the Kelvin double bridge method.

Kelvin Double Bridge

• The series field winding of a DC motor has low resistance, typically in the milliohm range.
• The Kelvin double bridge is specifically designed to measure low resistances accurately.

Measurement of resistance, inductance, and capacitance

Concept

The induced EMF in a DC shunt motor is given by:

$E_b=V_t-I_a{R}_a$

where, E_b = Induced EMF

V_t = Terminal voltage

I_a = Armature current

R_a = Armature resistance

Calculation

Given, V_t = 220 V

Ia = 20 A

Ra = 0.5 Ω

$E_b=220-(20\times 0.5)$

E_b = 210 V

Explanation:-

• A traction motor is an electric motor used for propulsion of a vehicle, such as locomotives (trains).
• DC series motor is used as traction motor.

Field test for series motor:

• This test is applicable to two similar series motor.
• One of the machines is run as a generator while the other as a motor.
• Therefore input power is only losses in the machines.
• Gives efficiency of DC series machine.

Hopkinson’s or Regenerative or Back To Back Test:

• Two identical DC shunt machines are coupled mechanically and tested.
• One of the machines is run as a generator while the other as a motor.
• Therefore input power is only losses in the machines.
• Performed at rated speed.
• Temperature rise and commutation qualities can be observed.

Retardation or running down test:

• Applicable to shunt motors and generators.
• Used for finding the stray losses.
• Machine is speeded up slightly beyond its rated speed and then supply is cut off from the armature while keeping the field excited.
• Armature will slow down and its kinetic energy is needed to meet rotational losses. i.e., friction and windage losses.

Swinburne’s Test:

• This test is performed no load.
• The rated speed is adjusted by the shunt field resistance.
• As it is a no load test, it cannot be done on a dc series motor.
• Give efficiency of DC machine.

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

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

The correct answer is option 2):(1500 rpm)

Concept:

The back EMF of the DCSeries motor

E_b = $\frac{P\times \varphi \times N\times Z}{60\times A}$

A = 2

where

E_b is the back emf  in volt

ϕ is flux per pole in weber

Z is the number of the conductors

A is the parallel path

N is the speed in RPM

Calculation:

Given 

P = 4

Z = 200

ϕ = 20 × 10^-3 Wb

I = 20 A

E_b = 200 V

Eb = $\frac{P\times \varphi \times N\times Z}{A}$

 200= $\frac{4\times 20\times {10}^{-3}\times 200\times N}{2\times 60}$

N = $\frac{400\times 60}{800\times 20\times {10}^{-3}}$

= 1500 rpm

Concept of  Characteristics of DC Series Motor:

For a DC series motor, the torque is inversely proportional to the speed. The torque sped curve is, therefore, a rectangular parabola. The speed-torque relation in a DC shunt and the series motor is as shown:

• DC series motor has high starting torque; This is one of the most required characteristics of DC series motor.
• Speed control of DC series motor is easy and various methods are available for this.
• The size of the DC series motor is small. (i.e. small size and high-power rating).
• This motor can handle overload easily.
• Regenerative braking can be applied to DC series motor, But with some modifications.
• The maintenance cost required is less.

The motor that is different considering speed and load torque is the differentially compound motor.

• Differentially compound motor: This motor has a shunt field winding and a series field winding, but the two windings are connected in such a way that they oppose each other. As the load torque increases, the current in the series field winding increases, which tends to decrease the total magnetic flux. This causes the motor speed to increase. Therefore, the differentially compound motor has a constant speed characteristic, even under varying load conditions.
• Cumulatively compound motor: This motor has a shunt field winding and a series field winding that are connected in parallel. As the load torque increases, the current in the series field winding increases, which tends to increase the total magnetic flux. This causes the motor speed to decrease. Therefore, the cumulatively compound motor has a variable speed characteristic, with the speed decreasing as the load torque increases.
• Series motor: This motor has a series field winding only. The speed of a series motor is inversely proportional to the load torque. This means that as the load torque increases, the speed of the motor decreases. Therefore, the series motor has a high starting torque but a variable speed characteristic under load.
• Permanent magnet motor: This motor uses permanent magnets to create the magnetic field. Permanent magnet motors have a variety of speed-torque characteristics, depending on the design of the motor. However, the most common characteristic is a constant speed characteristic, even under varying load conditions.

Therefore, the answer is Differentially compound motor.

Important PointsThe speed-torque characteristics of different DC motors are as shown in the below figure.

Concept:

In a DC series machine, both field and armature winding is connected in series. 

Therefore, field current will be equal to armature current.

Torque produced ∝ I_f × I_a

i.e., T ∝ ϕ . I_a

∵ ϕ ∝ I_a (in DC series motor)

∴ T ∝ I²_a

or, √T ∝ I_a

we also know that,

$N\propto \frac{1}{\varphi }\propto \frac{1}{I_a}\propto \frac{1}{\sqrt{T}}$

∴ $N\propto \frac{1}{\sqrt{T}}$

Therefore, curve will be rectangular hyperpolar.

Therefore, correct option will be (2)

In series motor flux is directly proportional to line current flowing to the motor.

And also torque is directly proportional to flux and current.

So torque will be directly proportional to the square of the current.

Calculation:

τ ∝ ϕ I ∝ I²

where, ϕ = flux 

I = current

Then $\frac{{\tau }_2}{{\tau }_1}=\frac{I_2^2}{I_1^2}$

$\frac{{\tau }_2}{{\tau }_1}=\frac{{30}^2}{{40}^2}=\frac{900}{1600}=\frac{9}{16}$

${\tau }_2=\frac{9}{16}{\tau }_1$

% change in torque = $\frac{{\tau }_2-{\tau }_1}{{\tau }_1}\times 100\mathrm{\%}$ 

 $\mathrm{\Delta }\tau$ =$\frac{\frac{9}{16}{\tau }_1-{\tau }_1}{{\tau }_1}\times 100\mathrm{\%}$

$\mathrm{\Delta }\tau$ = $-\frac{7}{16}\times 100\mathrm{\%}$

$\mathrm{\Delta }\tau$ = $-43.75\mathrm{\%}$

(Negative sign indicates that the torque is reduced from the original value.)

Answer is 43.75%

DC Shunt Motor:

DC Shunt motor is a self-excited machine because the supply given to the field winding in this motor is fed by the armature itself.

The figure is shown below the configuration of the DC shunt motor,

From the circuit diagram,

I = Ia + If

Where I is supply current

If is field current or exciting current

Ia is armature current

Ra is armature resistance

R_f is field winding resistance.

$I_f=\frac{V}{R_f}$

Applying KVL to the circuit

$V=E_b+I_a{R}_a=I_f{R}_f$   .... (1)

Calculation:

For the given data, we draw the circuit as

So the current through field resistance is $I_f=\frac{V}{R_f}=\frac{230}{230}=1A$

Hence, we have the armature current $I_a=I-I_f$ = 5 – 1 = 4A

From the given circuit, we have $V=E_{f1}+I_a\times R_s$  Or, $230=E_{f1}+4\times 0.4$  Or, $E_{f1}$ = 228.4 V

At full load  $I=70A$

So, I_a = 70 - 1 = 69 A   and $E_{f1}=230-69\times 0.4=202.4V$

as emf ∝ flux $\times$ speed ,  for DC shunt motor flux ($\varphi$) is constant. So,

$E_f\propto \omega$  or ${\displaystyle \frac{E_{f1}}{E_{f2}}}={\displaystyle \frac{{\omega }_1}{{\omega }_2}}$  or ${\displaystyle \frac{228.4}{202.4}}={\displaystyle \frac{1440}{{\omega }_2}}$

Thus, ${\omega }_2={\displaystyle \frac{202.4}{228.4}}\times 1440=1276rpm$

Concept: Torque current relation in DC machines:

T = K_a ϕ Ia

i.e., it is proportional to the product of flux and armature current.

In DC series matrix, field current and armature current are equal. Therefore,

T ∝ Ia² 

$\frac{\delta (T)}{\delta T}=2\frac{\delta (Ia)}{\delta Ia}$

∴ $\frac{\delta T}{\delta Ia}=2$

or $\frac{\delta Ia}{\delta T}=\frac{1}{2}$

In DC shunt motor, field current is constant.

∴ T ∝ Ia

∴ $\frac{\delta T}{\delta Ia}=1$

In compound motor, it will initially behave as series motor and when saturation sets in, it will behave like shut motor. Hence, current to torque ratio will lie b/w 0.5 to 1.

Hence, the corresponding ratio will be minimum in DC series motor.

Therefore, correct option is (1).

Concept:

The EMF equation of the DC shunt motor is:

E = V - I_aR_a

The speed of a DC motor is given by:

E = kϕω 

The torque produced by DC motor is:

T = kϕI_a

where,  E = Back EMF

V = Terminal voltage

I_a = Armature current

R_a = Armature resistance

ω = Speed in rpm

T = Torque

Calculation:

Given, V = 220 volt

Ia = 50 A

Ra = 0.2Ω 

E₁ = 220 - ( 50 × 0.2)

E1 = 210 V

If torque gets doubled, then armature current also doubles.

E₂ = 220 - ( 100 × 0.2)

E₂ = 200 V

$\frac{E_1}{E_2}=\frac{N_1}{N_2}$

$\frac{210}{200}=\frac{630}{N_2}$

N₂ = 600 rpm