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

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:

\(E_b=V_t-I_aR_a\)

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:

To determine the efficiency of a traction motor, various testing methods can be employed. The correct method, in this case, is the Field's test. Let’s delve into why this method is appropriate and why the other options are less suitable for this purpose.

Correct Option Analysis:

The correct option is:

Option 1: Field's test

Field's Test: The Field's test is a recognized method for determining the efficiency of traction motors, particularly for series motors used in electric traction. This test is also known as the regenerative method of testing.

Working Principle: The Field's test involves using two identical motors, where one acts as a generator and the other as a motor. The motor under test (MUT) is mechanically coupled to the generator. The generator is driven by the MUT, and the electrical output of the generator is fed back to the motor. This setup allows for the energy to circulate within the system, with minimal external power input. The losses in the system are measured to determine the efficiency of the traction motor.

Advantages:

• Allows for the accurate measurement of the efficiency of the motor under actual load conditions.
• Energy-efficient testing method as it recirculates energy between the motor and generator.
• Reduces the need for large external power sources, making it cost-effective.

Disadvantages:

• Requires two identical motors, which may not always be available.
• Complex setup and instrumentation are needed for accurate measurements.

Applications: Field's test is widely used for testing traction motors in electric locomotives, electric vehicles, and other applications where series motors are used for traction purposes.

Additional Information

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

Option 2: Acceleration test

The acceleration test is used to determine the performance characteristics of traction motors, such as their acceleration capabilities, torque, and speed. While this test provides valuable information about the motor's dynamic performance, it is not primarily used to determine efficiency.

Option 3: Swinburne's test

Swinburne's test is a method used to determine the efficiency of shunt and compound motors. It is an indirect method where the motor is run at no-load conditions, and the losses are measured. This test is not suitable for series motors used in traction applications as it does not account for the varying load conditions experienced by traction motors.

Option 4: Retardation test

The retardation test, also known as the retardation method or running down test, is used to determine the moment of inertia and frictional losses of rotating machines. It is not primarily used to determine the efficiency of traction motors.

Conclusion:

Understanding the appropriate testing methods for different types of motors is crucial for accurate efficiency determination. The Field's test is the most suitable method for determining the efficiency of traction motors, particularly series motors, due to its ability to simulate actual load conditions and its energy-efficient testing approach. The other methods mentioned are valuable for specific purposes but do not provide the same level of accuracy and relevance for traction motor efficiency testing.

Concept

The torque produced by the motor is given by:

T = W × r × g

where, W = Load on pulley

r = Radius of pulley

g = acceleration due to gravity = 9.81 m/s

The mechanical power in watts is given by:

P = 2π × N × T

where, N is in revolutions per second and T is in Nm

∵ 1 BHP (Metric) = 735.5 watts

∴ \(BHP \space (metric)={P\over 735.5}BHP\space(metric)\)

Calculation

Given, Current (I) = 20 A

Voltage (V) = 200 V

Radius of brake pulley (r) = 10 cm = 0.1 m

Effective load (W) = 20 kg

Speed (N) = 10 rps = 10 × 60 = 600 rpm

T = 20 × 0.1 × 9.8 

T = 19.62 Nm

P = 2π × 10 × 19.62 = 1233.67 watts

\(BHP \space (metric)={1233.67\over 735.5}BHP\space(metric)\)

BHP (in metric) = 0.5 π

We have,

V = 200 volts

R_a = 0.5 Ω

I_a = 20 A

From Equation of DC Motor:

V = E + I_aR_a

200 = E + 20 × 0.5

⇒ E = 200 - 10 = 190 volts

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 = \(P×ϕ × N× Z\over 60\times A\)

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 = \(P×ϕ × N× Z\over A\)

 200= \( 4\times 20 \times 10 ^{-3}\times 200 \times N \over 2\times 60\)

N = \(400 \times 60 \over 800 \times 20 \times 10 ^{-3}\)

= 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,

\(N \propto \frac{1}{\phi} \propto \frac{1}{I_a} \propto \frac{1}{\sqrt T}\)

∴ \(N\propto \frac{1}{\sqrt T}\)

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 = \( \frac{{{\tau _2} - {\tau _1}}}{{{\tau _1}}} \times 100\;\% \) 

 \( \Delta \tau \) =\( \;\;\frac{{\frac{9}{{16}}{\tau _1} - {\tau _1}}}{{{\tau _1}}} \times 100\% \;\)

\(\Delta \tau \) = \(\;\; - \frac{7}{{16}} \times 100\% \)

\(\Delta \tau \) = \( - 43.75\% \)

(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

\({V} = E_b + {I_a}{R_a}=I_fR_{f}\)   .... (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  \(I\; = \;70\;A\)

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

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

\({E_f} ∝ \omega\)  or \(\dfrac{{{E_{f1}}}}{{{E_{f2}}}} = \dfrac{{{\omega _1}}}{{{\omega _2}}}\)  or \(\dfrac{{228.4}}{{202.4}} = \dfrac{{1440}}{{{\omega _2}}}\)

Thus, \({\omega _2} = \dfrac{{202.4}}{{228.4}} \times 1440 = 1276\;rpm\)

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

\({E_1 \over E_2}= {N_1 \over N_2}\)

\({210 \over 200}= {630 \over N_2}\)

N₂ = 600 rpm