1. The efficiency of the DC motor at maximum power will be ________
100%
Around 90%
Anywhere between 75% and 90%
Less than 50%
Answer: 4. Less than 50%
Explanation:
The DC motor develops maximum power when the Back EMF is half the applied voltage is Eb = V/2.
Practically it is not possible to develop an exact 50% of maximum power because in that case, the current would be much beyond the rated current of the motor.
Some of the energy is wasted in the form of heat and other losses. Therefore motor efficiency is below 50%.
Under the condition of the maximum power output of d.c. motor, half of the input power is wasted in the armature circuit. In fact, if we take into account other losses (iron and mechanical), the efficiency will be well below 50%
2. Constant losses of DC motor can be measured by:
Swinburne’s Test
Retardation Test
Hopkinson’s Test
Load Test
Answer: 1. Swinburne’s Test
Explanation:
Swinburne’s test is performed to determine the constant losses in a DC shunt machine. In this test, the machine is operated as a motor on no-load. This no-load test is also known as Swinburne’s test. This test is very convenient and economical as it is required very less power from the supply to perform the test.
The following are the losses at no-load:
(i) Iron losses in the core
(ii) Windage and friction losses at bearing and commutator.
(iii) Shunt field copper losses.
(iv) Armature copper losses at no-load (very small)
3. The No-Load Test in DC Motor is known as ______
Swinburne’s Test
Retardation Test
Hopkinson’s Test
Load Test
Answer: 1. Swinburne’s Test
Explanation:
Swinburne’s test is performed to determine the constant losses in a DC shunt machine. In this test, the machine is operated as a motor on no-load. This no-load test is also known as Swinburne’s test. This test is very convenient and economical as it is required very less power from the supply to perform the test.
The following are the losses at no-load:
(i) Iron losses in the core
(ii) Windage and friction losses at bearing and commutator.
(iii) Shunt field copper losses.
(iv) Armature copper losses at no-load (very small)
4. The efficiency of a DC motor is defined as _______.
η = input power / (input power – losses)
η = output power / (input power – losses)
η = (input power – losses) / input power
η = output power / (input power + losses)
Answer:3. η = (input power – losses) / input power
Explanation:
Motor efficiency is defined that the ratio of mechanical power developed to the total electrical power input.
η = ( Motor input power – losses) / Motor input power
5. The hysteresis loss in a DC machine least depends on ______
Frequency of magnetic reversals
Maximum value of flux density
Volume and grade of iron
Rate of flow of ventilating air
Answer: 4. Rate of flow of ventilating air
Explanation:
Hysteresis loss: Whenever a magnetic material is subjected to reversal of magnetic flux, this loss occurs. It is due to the retentivity (a property) of the magnetic material. It is expressed with reasonable accuracy by the following expression:
Ph = Kh × V × f × Bm1.6
where,
Kh = hysteresis constant in J/m3 i.e., energy loss per unit volume of magnetic material during one magnetic reversal, its value depends upon the nature of the material;
V = Volume of magnetic material in m3.
f = frequency of magnetic reversal in cycle/second and
Bm = maximum flux density in the magnetic material in Tesla.
Hysteresis loss occurs in the rotating armature. To minimize this loss, the armature core is made of silicon steel which has low hysteresis constant.
Hence Rate of flow of ventilating air does not contribute to hysteresis losses.
6. Find the current required by a 400 V, 10 H.P. DC motor at 73.55% efficiency.
35 A
20 A
30 A
25 A
Answer:4. 25 A
Explanation:
The efficiency of the motor is = Output Power(Pm)/Input power(P)
Pm = 10 HP
= 10 × 746 = 7460 W
V = 400
η = 0.7355
η = 7460 / (400 × I)
⇒ 0.7355 = 7460 / (400 × I)
⇒ I = 25.35 ≈ 25 A
7. Which of the following test is usually conducted to determine the efficiency of traction motors?
Field’s test
Swinburne’s test
Hopkinson’s test
Retardation test
Answer:1. Field’s test
Explanation:
Field’s test is for obtaining the efficiency of two similar series motors.
Traction motors are usually available in pairs.
Two series machines coupled together are used for the test, one as a motor and the other as a generator.
The fields of the motor and the generator are connected together in series and thus the field of the generator is supplied from the motor current.
The motor is supplied by dc supply.
To maintain the voltage across the motor terminals at its rated voltage, it is necessary to have the total supply voltage higher than the rated voltage of the motor by an amount to overcome the resistance drop in the generator field winding connected in series with the motor field.
Iron and friction losses of two machines are made equal by joining the series field winding of the generator in the motor armature circuit so that both machines are equally exciting and by running them at equal speed.
8. Which of the following is not the effect of iron loss?
Loss of efficiency
Excessive heating of core
Increase in terminal voltage
Rise in temperature of ventilating air
Answer: 3. Increase in terminal voltage
Explanation:
Iron losses are sometimes described as ‘core losses’. The two-loss mechanisms are hysteresis and eddy current losses. Both of these increase with increasing flux density in the teeth and back iron.
The heat generated by these losses has two major serious effects:
(i) It increases the temperature inside the machine which affects the performance and life of the material of the machine, particularly insulation. Hence, the machine rating is directly affected by the losses.
(ii) Ultimately, losses are a waste of energy. In other words, it is a waste of money because these losses increase the operating cost of the machine.
Iron loss causes excessive heat production in the core of a machine, which will rise the temperature of ventilating air, as it acts as a heat exchanger. Iron loss does not affect terminal voltage.
9. A 500 V shunt motor takes 2 A on no-load. The armature resistance is 0.1 ohms and field current is 1 A. Brush resistance is negligible. If the line input current is 50 A then the efficiency will be
100%
50%
95%
90%
Answer:3. 95%
Explanation:
At no-load all the energy supplied is utilized in losses
At no-load, power input = 500 × 2 = 1000 W
Power input = no load losses
= constant loss + no load copper loss (I2R)
⇒ 1000 = constant loss + (2 – 1)2 × 0.1
⇒ Constant losses = 999.9 W
Ia = 50 A
Ia = 50 – 1 = 49 A
Copper loss = I2aR = 492 × 0.1 = 240.1 W
Total losses = constant losses + copper losses
= 999.9 + 240.1 = 1240 W
η = (Pin − Losses)/Pin
= (500 × 50) -1240/(500 × 50)
= 95.04%
10. The gross mechanical power developed by the motor is maximum when the back EMF is equal to _______ of the supply voltage.
Twice
2/3
1/2
1/3
Answer:3. 1/2
Explanation:
The gross mechanical power developed by the motor is maximum when the back EMF is equal to 50% (1/2) of the supply voltage.
Eb = V/2
11. Which of the following loss is likely to have the highest proportion at the rated load of the DC Motor?
Hysteresis loss
Field copper loss
Armature copper loss
Eddy current loss
Answer:3. Armature copper loss
Explanation:
Electric losses are also known as winding losses, copper losses, losses or ohmic losses.
These losses occur in the winding due to the flow of current in them. These are:
(i) Armature copper losses
(ii) Shunt field winding losses
(iii) Series field winding losses
(iv) Interpole winding losses
(v) Compensating winding losses
The copper loss in any winding depends on the resistance and square of the current flowing in that winding. If the current is doubled, the copper loss increases four times. Armature copper loss and series field copper loss depend on the armature current and hence they also depend on the load on the machine.
When a motor is loaded, its armature current increases. This increases the losses due to the armature resistance, so the copper losses in a motor increase with load. Armature copper losses the highest proportion at the rated load of the DC Motor.
12. A 400 V shunt motor has armature resistance 0.5 Ω and field resistance of 100 Ω. If the motor is delivering 10 kW at 90% Efficiency. Then the value of induced emf will be
388.11 V
367.54 V
351.23 V
391.91 V
Answer:1. 388.11 V
Explanation:
Efficiency η = Pout/Pin
Pin = (10 × 103)/0.9
Pin = 11.11 kW
Motor Line current IL
IL = V/Rsh = 400/100 = 4A
Armature current,
Ia = IL – Ish = 27.78 – 4 = 23.78A
Back emf,
Eb = V – IaRa = 400 – (23.78) (0.5) = 388.11 V
14. ________ test in DC Machine is carried on Full load
Swinburne’s Test
Retardation Test
Hopkinson’s Test
Load Test
Answer:3. Hopkinson’s Test
Explanation:
Hopkinson’s test is basically a regenerative test. It is also known as a back-to-back test.
To perform this test, two identical machines are required. These machines are mechanically coupled to each other.
One of them works as a motor which acts as a prime-mover for the other machine which works as a generator.
The electrical power or energy supplied to the motor is converted into mechanical energy which is further converted into electrical energy by the second machine coupled to it and fed back to the motor through the supply system.
In the process, in fact, the two machines draw electrical power or energy to meet the losses of the two machines.
Since the mechanics are identical, the losses in each machine are determined by dividing the input into two equal parts. Usually, this test is performed on large-size machines at full-load for a longer duration.
15. What is M. N. A. in the context of DC motors?
Minimum Natural Alignment
Magnetic Neutral Axis
Magnetized Natural Alignment
None of these
Answer: 2. Magnetic Neutral Axis
Explanation:
Magnetic Neutral Axis (MN:
MNA is the axis at which induced emf in the armature conductor is zero.
The brush axis is also known as MNA.
MNA is perpendicular to resultant flux or MMF axis.
MNA is a variable axis and it depends upon armature reaction
16. Which of the following loss in a DC generator varies significantly with the load current?
Field copper loss
Windage loss
Armature copper loss
Cannot be determined
Answer:3. Armature copper loss
Explanation:
The copper loss in any winding depends on the resistance and square of the current flowing in that winding. If the current is doubled, the copper loss increases four times. Armature copper loss and series field copper loss depend on the armature current and hence they also depend on the load on the machine.
When a motor is loaded, its armature current increases. This increases the losses due to the armature resistance, so the copper losses in a motor increase with load. Armature copper losses the highest proportion at the rated load of the DC Motor.
17. The total losses in a well-designed DC generator of 10 kW will be nearly equal to ______
100 W
500 W
1000 W
1500 W
Answer:2. 500 W
Explanation:
From the experimental observations, the total losses in a DC machine can be approximated to 4-5% of its rating. Thus, 5% of 10 kW is equal to 500 W.
18. Which of the following methods is likely to result in a reduction of hysteresis loss in a DC Machine?
Providing laminations in armature core
Providing laminations in stator
Using non-magnetic material for frame
Using material of low hysteresis coefficient for armature core material
Answer:4. Using material of low hysteresis coefficient for armature core material
Explanation:
Hysteresis loss: Whenever a magnetic material is subjected to reversal of magnetic flux, this loss occurs. It is due to the retentivity (a property) of the magnetic material. It is expressed with reasonable accuracy by the following expression:
Ph = Kh × V × f × Bm1.6
where,
Kh = hysteresis constant in J/m3 i.e., energy loss per unit volume of magnetic material during one magnetic reversal, its value depends upon the nature of the material;
V = Volume of magnetic material in m3.
f = frequency of magnetic reversal in cycle/second and
Bm = maximum flux density in the magnetic material in Tesla.
Hysteresis loss occurs in the rotating armature. To minimize this loss, the material of low hysteresis coefficient for armature core material the armature core is made of silicon steel which has low hysteresis constant is used.
19. Which of the following losses are significantly reduced by laminating the core of a DC machine?
Hysteresis losses
Eddy current losses
Copper losses
Windage losses
Answer: 2. Eddy current losses
Explanation:
Eddy current loss: When flux linking with the magnetic material changes (or flux is cut by the magnetic material) an emf is induced in it which circulates eddy currents through it. These eddy currents produce eddy current loss in the form of heat.
The major part of this loss occurs in the armature core. To minimize this loss, the armature core is laminated into thin sheets (0·3 to 0·5 mm) since this loss is directly proportional to the square of the thickness of the laminations.
20. Which of the following loss/losses in a DC generator is dissipated in the form of heat?
(i) It increases the temperature inside the machine which affects the performance and life of the material of the machine, particularly insulation. Hence, the machine rating is directly affected by the losses.
(ii) Ultimately, losses are a waste of energy. In other words, it is a waste of money because these losses increase the operating cost of the machine.
The various losses occurring in a DC machine can be sub-divided as:
