300+ DC Motor MCQ [Free PDF] – Objective Question Answer for DC Motor Quiz

Answer.4 All of the above

Explanation:-

Commutators and brush assemblies are a source of trouble in dc rotating machine. The continuous sliding of the brushes against the commutator wears the brushes down and tends to push them out of alignment causing bad contact between the commutator and brushes. When something does go wrong in commutation, it may be accompanied by excessive sparking, which aggravates the original trouble at once. When corm commutation is taking place, there will be very little sparking.

For satisfactory commutation of dc machines, a continuous contact must be maintained between commutator and brushes. The commutator must be mechanically true, the unit in good balance and the brushes should be in good shape and well adjusted. 

Excessive sparking can be caused by a number of external factors, such as shorted or open armature windings, overload, and improper field strength.

There can be many reasons For sparking at the brushes in a DC motor.

1. Brush holder is not equally spaced:-  this condition appears as the unequal sparking on different holders.
2. Holder is too far from commutator surface– If the holder is too far from the commutator surface it may fail to support the brush properly
3. Overload:- Overload. Overloading of the starter will cause high currents in the brushes, resulting in overheating and burning. Overloading is usually encountered when the starter is run for Periods long than 30 seconds
4. Dirty commutator: Oils, greases, and water will foul the commutator and cause excessive sparking between the commutator and the brushes, thus causing high temperatures

When there is excessive sparking at the commutator than good commutation cannot be obtained, the commutator and brush assembly must be checked, and any defect corrected as soon as possible. The inspection procedure and the steps taken to eliminate troubles are as follows:

1. Watch the machine operating to see if you can spot any arcing or excessive sparking elsewhere that might indicate a loose connection.
2. Check the brush positions to be sure that the brushes are commutating at the proper point (neutral plane).
3. Inspect all connections and make sure that none are loose.
4. Check the relative position of the brushes on the commutator. If they are unequally spaced, look for a bent brush holder.
5. Check the condition of the brushes. If they are badly worn, they should be replaced. When removing a brush, first lift the spring lever to release the pressure, then remove brush. Insert a new brush, making sure that it can move freely in the holder.
6. Check the commutator for dirt, pitting, irregularities, etc. Dirt can be removed with a piece of light canvas. Fine sandpaper will remove slight roughness. Never use emery cloth on a commutator.
7. Badly worn commutator should be skimmed on a lathe. This refers to machining away the irregularities of the surface.

Answer 2. Worn Bearing

Explanation:-

Machine Vibration Causes

Almost all machine vibration is due to one or more of these causes 

1. Repeating forces
2. Looseness
3. Resonance

Repeating forces

Repeating forces in machines are mostly due to the rotation of imbalanced, misaligned, worn, or improperly driven machine components.

1. Imbalance machine components:- Imbalance machine components contain “heavy spots,” which, when rotating, exert a repeating force on the machine. Imbalance is often caused by machining errors, nonuniform material density, variations in bolt sizes, air cavities in cast parts, missing balance weights, incorrect balancing, uneven electric motor windings, and broken, deformed, corroded or dirty fan blades or dirt that has dropped from a fan blade, suddenly creating a big imbalance.
2. Misaligned machine components:- It creates “bending moments,” which, when rotated, exert c repeating force on the machine. Misalignment is often caused by inaccurate assembly uneven floors, thermal expansion, distortions due to fastening torque, and improper mounting of couplings.
3. Worn machine:- Worn components exert a repeating force on the machine because of the rubbing of uneven worn surfaces. The wear in roller bearings, gears, and belts is often due to improper mounting, poor lubrication, manufacturing defects, and overloading.
4. Improperly driven machine components exert a repeating force on the machine because of intermittent power use. Examples include pumps receiving air in pulses, internal combustion engines with misfiring cylinders, and intermittent brush-commutator contact in DC motors.

Looseness

• The looseness of machine parts causes a machine to vibrate. If parts become loose, the vibration that is normally at tolerable levels may become unrestrained and excessive. Looseness can cause vibration in both rotating and non rotating machinery. Looseness is often due to excessive bearing clearances, loose mounting bolts, mismatched parts, corrosion, and cracked structure.

Resonance

• Resonance is a state of operation wherein an excitation frequency is close to a natural frequency of the machine structure, at this frequency, the damping of the vibration is very low and the result is resonance. When resonance occurs, the resulting vibration levels can be very high and can cause damage very quickly.

Answer.2. Their speeds and excitations are identical

Explanation:-

Hopkinson’s Test

Hopkinson’s test is also referred as the regenerative or Back-to-Back test.

When large machines are to be tested, the direct loading method involves a lot of loss of power. To avoid this, the regenerative method of testing is used where possible. This type of test can easily be performed on two identically shunt machines mechanically coupled to each other. One of the machines is run as a motor driving, the other machine as a generator. This generator supplies the motor. i.e. connected back to back so that most of the power required by the motor is fed from the generator, while the losses of both the machines — generator and motor — are supplied from the mains. This reduces the power taken from the mains. This type of test is known as Hopkinson test. Figure. shows the connection diagram for the Hopkinson test on two similar shunt machines.

One of the machines is started as the motor through the starter taking supply from the mains. The field current is adjusted to run the machine at the rated speed. The second machine is mechanically coupled and runs at the same speed. The excitation of the machine is so adjusted that the voltage across the armature is slightly higher than the supply voltage. This can be checked by the voltmeter V. The polarities of the machines should be suitable for parallel operation. When the voltmeter reads zero or slightly higher by 1 or 2 V, indicating that it would be generator action, the switch is closed so that both the machines are now in parallel across the supply. The excitation of the generator is so adjusted that the emf in the generator armature increases allowing more current through the motor armature and, thus, loading the machines. The losses are supplied by the mains. As two machines are similar and of equal size, and rating the constant losses — friction, windage.iron are assumed to be equal.

The field current of the generator is supplied from the mains. As an alternative, it can be supplied from the terminals of the generator. The location of the ammeter is such that it reads current I excluding the field currents of both the machines. The machine with smaller excitation current acts as a motor because back emf is less than that of induced emf of the other machine. Thus, it admits more armature current and develops greater torque. The load on the machines can be adjusted as desired and readings are taken.

Advantages of Hopkinson’s Test

The merits of this test are…

1. This test requires very small power compared to the full-load power of the motor-generator coupled system. That is why it is economical. Large machines can be tested at rated load without much power consumption.
2. Temperature rise and commutation can be observed and maintained in the limit because this test is done under full load condition.
3. Change in iron loss due to flux distortion can be taken into account due to the advantage of its full load condition.
4. Efficiency at different loads can be determined.

Disadvantages of Hopkinson’s Test

The demerits of this test are

1. It is difficult to find two identical machines needed for Hopkinson’s test.
2. Both machines cannot be loaded equally all the time.
3. Since the two machines are not loaded equally, therefore, the effect of armature reaction is not the same.
4. Due to the difference in the excitation, It is not possible to get separate iron losses for the two machines.
5. It is difficult to operate the machines at rated speed because field currents vary widely.

Answer.4. 1200 RPM

Explanation:-

Extracting the data from given problem: Speed of DC shunt motor is 1200 rpm, the excitation voltage is 220V. The new excitation voltage is given as 175 V DC. In a DC shunt motor, the field current is given as

The speed equation of DC motor

$N \propto \dfrac{{{E_b}}}{\Phi } \propto \frac{{V – IaRa}}{\Phi }$

The flux of DC motor is φ = K’V.
Where K is the  constant based on armature design
Now the losses and voltage drop is neglected therefore
Voltage is also equal to back emf

V = Eb = KφN

N = V/Kφ = V/KK’V = 1/KK’

Now, it is observed that speed is constant and independent of voltage. Therefore, the value of voltage is 1200 rpm.

Answer.1. 44 A

Explanation:-

As we know that back EMF of DC motor is directly proportional to the flux and speed.

E_b ∝ Nφ
E_b ∝ N (φ is constant)

We can stall the motor (i.e., reduce its speed to zero) by putting more and more load on the shaft. When the motor is stalled since N = 0 therefore E_b = 0

From the voltage equation, the back EMF of DC motor at no-load

E_b = V − I_aR_a

Since external resistance, R_ext is added in series with the motor. Hence

E_b = V − I_a(R_a + R_ext)

0 = 220 − I_a(4.5 + 0.5)

I_a = 220 ⁄ 5 

I_a = 44 A

Answer C. 4

Explanation:-

Stalling torque is the torque at which speed would fall to zero.

With N = 0 , E_b = 0

Hence the value of armature current

From the voltage equation, the back EMF of DC motor at no-load

E_b = V − I_aR_a

Since external resistance, R_ext is added in series with the motor. Hence

E_b = V − I_a(R_a + R_ext)

0 = 240 − I_a(0.5 + 1)

I_a = 220 ⁄ 1.5

I_a = 160 A

The torque developed by a d.c motor is directly proportional to Flux per pole × Armature Resistance i.e

T ∝ ΦI_a (Now flux is constant hence )
T ∝ I_a

Hence Ratio of stalling torque to full load torque will be

T₂ ⁄ T₁ = I₂ ⁄ I₁

T₂ ⁄ T₁ = 160/40 = 4

T₂ ⁄ T₁ = 4

Answer. 4. 115 V

Explanation:-

For maximum power output, the back emf is equal to half of the applied voltage.

Hence, E_b = V/2 = 230/2 ⇒115 V. where V is the terminal voltage.

or

For maximum power transfer load resistance should be same as series field resistance. So applied voltage 230 V will be divided equally 115 V on each section i.e. series field resistance and load resistance. Hence for maximum power development back emf should be 115 V.

Answer 1. High Resistance, Low Resistance

Explanation:-

In a compound machine, there are two field windings, namely a shunt-field winding and a series-field winding. The shunt-field winding is connected in parallel with the armature and the series-field winding is connected in series with the combinations as shown in Fig.

The series winding will carry a large armature current I_a or I_L, and therefore it is made of wires of large cross-section and has a few turns only. The resistance of a series winding is very small.

The shunt-winding is made up of wires of small cross-sections and has high resistance. Since the resistance of the shunt-field winding is high, the current flowing through it is small in comparison with the machine’s armature current. The main flux is created by the shunt winding but it is modified by the flux of the series winding. A compound machine, therefore, combines the best features of shunt machines and series machines

Answer 4. 0.72

Explanation:- 

For a constant torque load, changing the armature resistance will not change the armature current. So, when the external resistance is added the speed of the motor decreases.

From the voltage equation, the back EMF of DC motor is

E_b1 = V − I_aR_a

= 230 − 0.25 × 60

E_b1 = 215

When the resistance of 1Ω is added in the armature circuit the new back EMF voltage

E_b2 = V − I_a(R_a + R_ext)

 = 240 − 60(0.25 + 1)

 230 − 1.25 × 60

E_b2 = 155 V

Let the on load speed be N. As we know that back EMF of DC motor is directly proportional to the flux and speed.

E_b ∝ Nφ
E_b ∝ N (φ is constant)

The ratio of emf and speed can be equated as

$\begin{array}{l}\dfrac{{{E_{b2}}}}{{{E_{b1}}}} = \dfrac{{{N_2}}}{{{N_1}}}\\\\\dfrac{{{N_2}}}{{{N_1}}} = \dfrac{{155}}{{215}} = 0.72\end{array}$

Answer 2. 31.1Ω

Explanation:-

Given Data:-

Supply Voltage V = 240 V
Line Current I_L = 15 A
Shunt Resistance R_sh = 80Ω
Armature Resistance R_a = 0.5 Ω
Armature Current I_a =?
Back EMF E_b =?

(i) Armature current

The Line current of DC shunt motor is the sum of Armature current and Shunt field current

I_L = I_a + I_sh

And Shunt Field Resistance is given as

I_sh = V/R_sh = 240/80 = 3 A

∴ I_a = I_L − I_sh

I_a = 15 − 3 = 12A

Back EMF

From the voltage equation, the Back EMF of DC Motor is given by

E_b = V − I_aR_a

= 240 − 12 × 0.5
= 240 − 6

E_b = 234 V

External resistance to be added in the armature circuit to limit the armature current upto 125%.

E_b = V − I_a(R_a + R_ext)

 234 = 240 − I_a(0.5 + R_ext)

$\begin{array}{l}12 \times 1.25 = \dfrac{{474}}{{0.5 + {{\mathop{\rm R}\nolimits} _{ext}}}}\\\\15\left( {0.5 + {{\mathop{\rm R}\nolimits} _{ext}}} \right) = 474\\\\7.5 + 15{{\mathop{\rm R}\nolimits} _{ext}} = 474\\\\{{\mathop{\rm R}\nolimits} _{ext}} = 31.1\Omega \end{array}$