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We are currently launching complete chapter-wise MCQs on Electrical Drives. And believe us…
…this will completely change your learning on drives subjects!!
The chapters are picked from the book called “A First Course on Electrical Drives By S.K. Pillai and other resources”. All the questions in the books are compiled in content and written separately chapter-wise.
In this article, the question and answers are mostly written from the chapter called “Dynamics of Electrical Drives”. MCQs of other chapters are also published on our website, to read and learn it, you can simply click on the listed Chapters below:
1. Dynamics of Electrical Drives
2. Characteristics of DC & AC Motors
3. Starting
4. Speed Control of Direct Current Motors & Induction Motors
5. Braking of Electric Motors
6. Braking of DC Motors & Induction Motors
7. Rating and Heating of Motors
8. Introduction to Solid State Controlled Drives
9. Introduction to Solid-State Devices
10. Introduction to Solid-State Switching Circuits
11. Solar and Battery Powered Drives
12. Traction Drives
13. Electrical Drive Systems and Components
With that said… Let’s dive into our main topic 80+ MCQs On Dynamics of Electrical Drives.
Suppose you want to jump to the specific question directly, expand the table of content below:
Topic 1: Dynamics – Types of Loads
1. Load torques can be classified into how many types?
a) Three
b) Two
c) Four
d) Five
Answer: B
Explanation: Load torques can have two types. They are active and passive load torques. Active load torques are able to run the motor under equilibrium conditions and their sign remains the same even if the motor rotation changes but passive load torques always oppose the motion by changing their sign with the change in rotation of the motor.
2. Rolling mills exhibit what type of load torque characteristics?
a) Constant torque characteristics
b) Linearly rising torque characteristics
c) Non-Linearly rising torque characteristics
d) Non-Linearly decreasing torque characteristics
Answer: d
Explanation: Rolling mills are an example of non-linearly decreasing torque characteristics because torque and speed exhibits inversely proportional relationships and power are constant.
3. What is the relationship between torque and speed in constant type loads?
a) Torque is independent of speed
b) Torque linearly increases with an increase in speed
c) Torque non-linearly increases with an increase in speed
d) Torque non-linearly decreases with an increase in speed
Answer: a
Explanation: Speed hoist is a perfect example of constant type loads in which torque variation is independent of speed. The speed-torque characteristics of this type of load are given by T=K where K is a constant.
4. Torque inversely varies with the speed in the windage load torque component.
a) True
b) False
Answer: b
Explanation: Torque varies with a square of speed in the windage load torque component whereas in the coulomb torque component torque is constant.
5. What type of force handles for active torques?
a) Strong nuclear forces
b) Weak nuclear forces
c) Gravitational forces
d) Electrostatic forces
Answer: c
Explanation: Gravitational forces are responsible for active torques. Active torques due to gravitational forces can be obtained in the case of hoists, lifts or elevators and railway locomotives operating on gradients.
6. Passive torques always oppose the motion of the driven machine.
a) True
b) False
Answer: a
Explanation: Passive torques are due to friction or shear and deformation in elastic bodies. They always oppose the motion, restricting the motion of the machine.
7. Among the following which one exhibits linearly rising load torque characteristics?
a) Elevators
b) Rolling Mills
c) Fan load
d) Separately excited dc generator connected to the resistive load
Answer: d
Explanation: Separately excited dc generator connected to the resistive load is an example of linearly rising load torque characteristics as the torque increases linearly with an increase in speed.
8. What is the condition for the steady-state operation of the motor?
a) Load torque > Motor torque
b) Load torque <<<< Motor torque
c) Load torque = Motor torque
d) Load torque < Motor torque
Answer: c
Explanation: According to the dynamic equation of motor, load torque must be equal to motor torque so that motor should run at a uniform speed. If load torque is greater than motor torque, the motor will fail to start and if load torque is less than motor torque, the motor will run at a higher speed which can damage the shaft of the motor.
9. Choose the correct one. (* stands for multiplication, J represents the moment of inertia, w represents angular speed).
a) J*d(w)/dt = Load torque – Motor torque
b) J*d(w)/dt = Load torque + Motor torque
c) J*d(w)/dt = Motor torque – Load torque
d) J*d(w)/dt = Load torque * Motor torque
Answer: c
Explanation: J*d(w)/dt = Motor torque – Load torque is the dynamic equation of the motor. Motor torque will try to aid the motion of the motor, but load torque will oppose the motion of motor that’s why it subtracts in the equation.
Topic 2: Quadrantal Diagram of Speed – Torque Characteristics
1. Regenerative braking mode can be achieved in which quadrant (V-I curve)?
a) Third
b) Second
c) Fourth
d) First
Answer: b
Explanation: Regenerative braking is only available in the second quadrant as power from the motor is fed back to the source. Back emf generated (Eb) is more than armature terminal (Vt) so it works as a generator.
2. Fan type of loads exhibits which type of load torque characteristics?
a) Constant torque characteristics
b) Linearly rising torque characteristics
c) Non-Linearly rising torque characteristics
d) Non-Linearly decreasing torque characteristics
Answer: c
Explanation: Torque produced by the fan is directly proportional to the square of speed throughout the range of usable fan speeds. This type of load exhibits non-linearly rising torque characteristics.
3. Type-A chopper is used for obtaining which type of mode?
a) Motoring mode
b) Regenerative braking mode
c) Reverse motoring mode
d) Reverse regenerative braking mode
Answer: a
Explanation: Only motoring mode is available in the case of a step-down chopper (Type-A chopper). The value of output voltage (Vo) is less than the input voltage (Vin) in the case of a step-down chopper.
4. Calculate the value of the angular acceleration of the motor using the given data: J = 20 kg-m2, load torque = 20 N-m, motor torque = 60 N-m.
a) 5 rad/s2
b) 2 rad/s2
c) 3 rad/s2
d) 4 rad/s2
Answer: b
Explanation: Using the dynamic equation of motor J*(angular acceleration) = Motor torque – Load torque: 20*(angular acceleration) = 60-20=40, angular acceleration=2 rad/s2.
5. 230V, 10A, 1500rpm DC separately excited motor having a resistance of .2 ohm excited from an external dc voltage source of 50V. Calculate the torque developed by the motor on full load.
a) 13.89 N-m
b) 14.52 N-m
c) 13.37 N-m
d) 14.42 N-m
Answer: b
Explanation: Back emf developed in the motor during full load can be calculated using equation Eb = Vt- I*Ra = 228 V and machine constant Km = Eb / Wm which is equal to 1.452. Torque can be calculated by using the relation T = Km* I = 1.452*10 = 14.52 N-m.
6. Boost converter is used to _________
a) Step down the voltage
b) Step up the voltage
c) Equalize the voltage
d) Step up and step down the voltage
Answer: b
Explanation: Output voltage of boost converter is Vo = Vin / 1 – D. Value of duty cycle is less than 1 which makes the Vo > Vin as denominator value decreases and becomes less than 1. The boost converter is used to step-up voltage.
7. Reverse motoring mode is available in fourth quadrant.
a) True
b) False
Answer: a
Explanation: In reverse motoring, the mode motor rotates in opposite to the original direction as the direction of motor torque changes which makes the motor run in the opposite direction and load torque tries to oppose the motion of the motor.
8. Calculate the power developed by a motor using the given data: Eb = 20V and I = 10 A. (Assume rotational losses are neglected)
a) 400 W
b) 200 W
c) 300 W
d) 500 W
Answer: b
Explanation: Power developed by the motor can be calculated using the formula P = Eb*I = 20*10 = 200 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.
9. Which one is an example of variable loss?
a) Windage loss
b) Hysteresis loss
c) Armature copper loss
d) Friction loss
Answer: c
Explanation: Armature copper losses are variable losses as they depend on armature current which further depends on the load. As load changes armature current changes hence armature copper losses (I2 * r) also changes.
Topic 3: Load Torques that Depend on the Path or Position Taken by the Load During Motion
1. What is the empirical formula for the tractive force required to overcome curve resistance? (W-the weight of the body, R – radius of curvature)
a) 710×W÷R
b) 700×W÷R
c) 720×W÷R
d) 750×W÷R
Answer: b
Explanation: Fc= 700×W÷R is the tractive force required to overcome curve resistance where W is the weight of the body in Kg and R is the radius of curvature in meters.
2. Force resisting the upward motion of a body on an inclined plane is given by (alpha–the angle of inclination, W- the weight of the body).
a) F = W×sin(alpha)
b) F = W×cosec(alpha)
c) F = W×sec(alpha)
d) F = W×cos(alpha)
Answer: a
Explanation: When a body is moving upward on an inclined plane its weight can be resolved in two perpendicular components that are W×sin(alpha) and W×cos(alpha). W×cos(alpha) is the component that is opposite to the normal of the inclined plane and W×sin(alpha) is the component that opposes the upward motion of the body.
3. The unit of the torque is ______
a) N-m
b) N-m2
c) N-m/sec
d) N-Hz
Answer: a
Explanation: Torque is defined as the vector product of displacement and force. The unit of the force is Newton(N) and the displacement is a meter (m) so the unit of torque is N-m.
4. Calculate the value of the torque when 10 N force is applied perpendicular to a 10 m length of rod fixed at the center.
a) 200 N-m
b) 300 N-m
c) 100 N-m
d) 400 N-m
Answer: b
Explanation: Torque can be calculated using the relation T = (length of rod) × (Force applied) = r×F×sin90. F is given as 10 N and r is 10 m then torque is 10×10 = 100 N-m. (the angle between F and r is 90 degrees)
5. What is the dimensional formula for torque?
a) [ML2T-2]
b) [MLT-2]
c) [M1L2T-3]
d) [LT-2]
Answer: a
Explanation: Torque is a vector product of force and displacement. Dimensional formula for force is [MLT-2] and displacement is [L] so dimensional formula for torque is [MLT-2] [L] = [ML2T-2].
6. Buck converter is used to _________
a) Step down the voltage
b) Step up the voltage
c) Equalize the voltage
d) Step up and step down the voltage
Answer: a
Explanation: The output voltage of the buck converter is Vo = D×Vin. The value of the duty cycle is less than one which makes the Vo < Vin. The buck converter is used to step down voltage. Vin is a fixed voltage and Vo is a variable voltage.
7. If the starting torque of the motor is less than the load torque, the motor will fail to start.
a) True
b) False
Answer: a
Explanation: J×d(w)÷d(t) = Motor torque – Load torque is the dynamic equation of motor. If starting torque (motor torque) is less than the load torque then d(w)÷d(t) <0, acceleration <0 so the motor will decelerate and fails to start.
8. Torque is a scalar quantity.
a) True
b) False
Answer: b
Explanation: Scalar quantity has only magnitude whereas vector quantity has both directions and magnitude. Torque is a force applied to a body perpendicularly. As the force is a vector quantity, the torque must be treated as a vector quantity.
9. 250V, 15A, 1100 rpm separately excited dc motor with armature resistance (Ra) equal to 2 ohms. Calculate back emf developed in the motor when it operates on half of the full load. (Assume rotational losses are neglected)
a) 210V
b) 240V
c) 230V
d) 235V
Answer: d
Explanation: Back emf developed in the motor can be calculated using the relation Eb = Vt-I×Ra. In question, it is asking for half load, but the data is given for full load so current becomes half of the full load current = 15÷2 = 7.5 A. 250V is terminal voltage it is fixed so Eb = 250-7.5×2 = 235V
Topic 4: Load Torques that Vary with Angle of Displacement of the Shaft
1. Duty cycle (D) is _______
a) Ton÷Toff
b) Ton÷(Ton+ Toff)
c) Ton÷2×(Ton+ Toff)
d) Ton÷2×Toff
Answer: b
Explanation: Duty cycle (D) is defined as the ratio of time for which system is active to the total time period. It is also known as the power cycle. It has no unit.
2. A 220 V, 1000 rpm, 60 A separately-excited dc motor has an armature resistance of .5 ω. It is fed from a single-phase full converter with an ac source voltage of 230 V, 50Hz. Assuming continuous conduction, the firing angle for rated motor torque at (-400) rpm is _________
a) 122.4°
b) 117.6°
c) 130.1°
d) 102.8°
Answer: d
3. The unit of angular acceleration is rad/s2.
a) True
b) False
Answer: a
Explanation: Angular acceleration is defined as a derivate of angular velocity with respect to time. It is generally written as α. The unit of angular velocity is rad/sec and of time is second so the unit of angular acceleration is rad/s2.
4. Calculate the value of the angular acceleration of the motor using the given data: J= 50 kg-m2, load torque = 40 N-m, motor torque = 10 N-m.
a) -.7 rad/s2
b) -.6 rad/s2
c) -.3 rad/s2
d) -.4 rad/s2
Answer: b
Explanation: Using the dynamic equation of motor J*(angular acceleration) = Motor torque – Load torque: 50*(angular acceleration) = 10-40 = -30, angular acceleration=-.6 rad/s2. The motor will decelerate and will fail to start.
5. The principle of step-up chopper can be employed for the ________
a) Motoring mode
b) Regenerative mode
c) Plugging
d) Reverse motoring mode
Answer: b
Explanation: The step-down chopper is used in motoring mode but a step-up chopper can operate only braking mode because the characteristics are in the second quadrant only.
6. A Buck-Boost converter is used to _________
a) Step down the voltage
b) Step up the voltage
c) Equalize the voltage
d) Step up and step down the voltage
Answer: d
Explanation: The output voltage of the buck-boost converter is Vo = D×Vin ÷ (1-D). It can step up and step down the voltage depending upon the value of the duty cycle. If the value of the duty cycle is less than .5 it will work as a buck converter and for duty cycle greater than .5 it will work as a boost converter.
7. Which of the following converter circuit operations will be unstable for a large duty cycle ratio?
a) Buck converter
b) Boost converter
c) Buck-Boost converter
d) Boost converter and Buck-Boost converter
Answer: d
Explanation: The output voltage of the buck converter and buck-boost converter are Vo=Vin ÷ (1-D) and Vo = D×Vin ÷ (1-D) respectively. When the value of the duty cycle tends to 1 output voltage tends to infinity.
8. Calculate the shaft power developed by a motor using the given data: Eb= 50V and I= 60 A. Assume frictional losses are 400 W and windage losses are 600 W.
a) 4000 W
b) 2000 W
c) 1000 W
d) 1500 W
Answer: b
Explanation: Shaft power developed by the motor can be calculated using the formula P = Eb*I-(rotational losses) = 50*60 = 3000 – (600+400) = 2000 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.
9. Which one of the following devices has low power losses?
a) MOSFET
b) IGBT
c) SCR
d) BJT
Answer: c
Explanation: SCR is a minority carrier device due to which it experiences conductivity modulation and it’s ON state resistance reduction due to which conduction losses are very low.
Topic 5: Load Torques that Vary with Time
1. Servo motors are an example of which type of load?
a) Pulsating loads
b) Short time loads
c) Impact loads
d) Short-time intermittent loads
Answer: b
Explanation: Servo motors are motors with control feedback. The motor can be AC or DC. This is an example of short-time loads. They have a high torque to inertia ratio and high speed.
2. Load torque of the crane is independent of _________
a) Speed
b) Seebeck effect
c) Hall effect
d) Thomson effect
Answer: a
Explanation: The Load torque of the crane is independent of speed. They are short time intermittent types of loads. They require constant power for a short period of time.
3. The unit of angular velocity is rad/s3.
a) True
b) False
Answer: b
Explanation: Angular velocity is defined as the rate of change of angular displacement with respect to time. Angular displacement is generally expressed in terms of a radian. The unit of angular velocity is rad/s.
4. R.M.S value of the sinusoidal waveform V=Vmsin(ωt+α).
a) Vm÷2½
b) Vm÷2¼
c) Vm÷2¾
d) Vm÷3½
Answer: a
Explanation: R.M.S value of the sinusoidal waveform is Vm÷2½ and r.m.s value of the trapezoidal waveform is Vm÷3½. The peak value of the sinusoidal waveform is Vm.
5. Calculate the time period of the waveform x(t)=24sin(24πt+π÷4).
a) .064 sec
b) .047 sec
c) .083 sec
d) .015 sec
Answer: c
Explanation: The fundamental time period of the sine wave is 2π. The time period of x(t) is 2π÷24π=.083 sec. The time period is independent of phase shifting and time-shifting.
6. The turn-off times of the devices in the increasing order is ___________
I. MOSFET
II. BJT
III. IGBT
IV. Thyristor
a) I, III, II, IV
b) I, II, III, IV
c) III, I, II, IV
d) III, II, IV, I
Answer: a
Explanation: Increasing turn-off time implies decreasing speed and the majority of carrier devices do not have any minority charge carrier storage so they have less turn-off time hence MOSFET has the least turn-off time. So, the increasing order of turn-off time is, MOSFET < IGBT < BJT < Thyristor.
7. Which of the following devices should be used as a switch for high power and high voltage application?
a) GTO
b) MOSFET
c) TRIAC
d) Thyristor
Answer: d
Explanation: Thyristor is used for high power applications but it has a limited frequency range and cannot be used at high frequencies. A thyristor is a unidirectional, bipolar and semi-controlled device.
8. Calculate the useful power developed by a motor using the given data: Pin= 3000 W, Ia= 60 A, Ra = .4 Ω. Assume frictional losses are 200 W and windage losses are 400 W.
a) 970 W
b) 960 W
c) 980 W
d) 990 W
Answer: b
Explanation: Useful power is basically the shaft power developed by the motor that can be calculated using the formula Psh = Pdev-(rotational losses). Pdev = Pin-Ia2Ra = 3000-602(.4) =1560 W. The useful power developed by the motor is Psh = Pdev-(rotational losses) =1560 –(200+400)=960 W.
9. Calculate the phase angle of the sinusoidal waveform y(t)=55sin(4πt+π÷8).
a) π÷8
b) π÷5
c) π÷7
d) π÷4
Answer: a
Explanation: Sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α) where Vm represents peak value, ω represents angular frequency, α represents a phase difference.
Topic 6: Dynamics of Motor-Load Combination
1. The axis along which no emf is produced in the armature conductors is called____________
a) Geometrical Neutral Axis (GNA)
b) Magnetic Neutral Axis (MNA)
c) Axis of rotation
d) Axis of revolution
Answer: b
Explanation: The coil undergoing commutation must lie along the magnetic neutral axis so that no emf is induced and there is no sparking at the time of commutation.
2. The generated e.m.f from 25-pole armature having 200 conductors driven at 10 rev/sec having flux per pole as 20 mWb, with two parallel paths is ___________
a) 400 V
b) 500 V
c) 200 V
d) 300 V
Answer: b
Explanation: The generated can be calculated using the formula Eb = Φ×Z×N×P÷60×A, Φ represents flux per pole, Z represents the total number of conductors, P represents the number of poles, A represents the number of parallel paths, N represents speed in rpm. Eb = .02×25×200×600÷60×2= 500 V.
3. The unit of the flux is Weber.
a) True
b) False
Answer: a
Explanation: Flux is the total amount of magnetic field lines passing through a given area. Φ is a dot product of magnetic flux density and area. The unit of the flux is Weber (Wb).
4. Which of the following motor can be referred as a universal motor?
a) DC shunt motor
b) DC compound motor
c) Permanent magnet motor
d) DC series motor
Answer: d
Explanation: DC series motor can operate on DC and AC. It is a universal motor. Universal motors are those motors that can operate on both DC and AC. DC shunt motor can only operate on DC because of pulsating torque in AC.
5. The phase difference between voltage and current in the inductor.
a) 45°
b) 90°
c) 80°
d) 55°
Answer: (b) 90°
Explaination: In the case of an inductor, the voltage leads the current by 90° or the current lags the voltage by 90o. The phase difference between voltage and current is 90°.
6. The phase difference between voltage and current in the resistor.
a) 85°
b) 90°
c) 0°
d) 5°
Answer: c
Explanation: In the case of a resistor, the voltage and current are in the same phase. The phase difference between voltage and current is 0°. The voltage drop in the resistor is given as V=IR.
7. The phase difference between voltage and current in the capacitor.
a) 90°
b) 80°
c) 95°
d) 91°
Answer: a
Explanation: In the case of a capacitor, the voltage lags the current by 90° or the current leads the voltage by 90o. The phase difference between voltage and current is 90°.
8. The slope of the I-V curve is 30°. Calculate the value of resistance. Assume the relationship between I and V is a straight line.
a) 1.732 Ω
b) 2.235 Ω
c) 1.625 Ω
d) 1.524 Ω
Answer: a
Explanation: The slope of the I-V curve is reciprocal of resistance. The slope given is 30° so R=1÷tan(30°)=1.732 Ω. The slope of the V-I curve is resistance.
9. What is a mark-to-space ratio?
a) Ton÷Toff
b) Ton÷(Ton- Toff)
c) Ton÷2×(Ton*Toff)
d) Ton÷2×Toff
Answer: a
Explanation: Mark to space is Ton÷Toff. It is the ratio of the time for which the system is active and the time for which is inactive. It has no unit.
Topic 7: Moment of Inertia Determination
1. What is the formula for the moment of inertia? (m – a mass of the body, r – distance from the axis of the rotation)
a) ∑miri2
b) ∑miri
c) ∑miri4
d) ∑miri3
Answer: a
Explanation: The moment of inertia is the property by the virtue of which the body withstands the effect of angular acceleration. It depends on the shape and mass distribution of the body.
2. The generated e.m.f from 50-pole armature having 400 conductors driven at 20 rev/sec having flux per pole as 30 mWb, with lap winding is ___________
a) 230 V
b) 140 V
c) 240 V
d) 250 V
Answer: c
Explanation: The generated can be calculated using the formula Eb = Φ×Z×N×P÷60×A, Φ represents flux per pole, Z represents the total number of conductors, P represents the number of poles, A represents the number of parallel paths, N represents speed in rpm. In lap winding number of parallel paths are equal to the number of poles. Eb = .03×50×400×1200÷60×50= 240 V.
3. The unit of the moment of inertia is Kgm2.
a) True
b) False
Answer: a
Explanation: The moment of inertia is taken as the sum of the product of the mass of each particle with the square of their distance from the axis of the rotation. The unit of the moment of inertia is kg×m2=kgm2.
4. Calculate the moment of inertia of the egg having a mass of 7 kg and a radius of 44 cm.
a) .968 kgm2
b) 1.454 kgm2
c) 1.545 kgm2
d) 1.552 kgm2
Answer: d
Explanation: The moment of inertia of the egg can be calculated using the formula I=∑miri2. The mass of the egg and radius is given. I=(7)×(.44)2=1.552 kgm2. It depends upon the orientation of the rotational axis.
5. Which of the theorems helps in the calculation of the moment of inertia?
a) The theorem of Parallel and Perpendicular Axes
b) The theorem of Horizontal and Perpendicular axes
c) The theorem of Vertical and Perpendicular axes
d) The theorem of Parallel and Tilted axes
Answer: a
Explanation: The theorem of Parallel and Perpendicular axes helps in the calculation of the moment of inertia. The moment of inertia of the complex bodies can be easily calculated with the help of these theorems.
6. What is the unit of resistance?
a) ohm
b) ohm-1
c) ohm2
d) ohm5
Answer: a
Explanation: Resistance is the opposition offered by the body to the flow of current. It is the ratio of voltage and current. It is given in ohms.
7. Calculate the value of the frequency if the time period of the signal is 20 sec.
a) 0.05 Hz
b) 0.04 Hz
c) 0.02 Hz
d) 0.03 Hz
Answer: a
Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal to the time period. It is expressed in Hz. F = 1÷T=1÷20=.05 Hz.
8. The slope of the V-I curve is 60°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) 1.732 Ω
b) 1.608 Ω
c) 1.543 Ω
d) 1.648 Ω
Answer: a
Explanation: The slope of the V-I curve is resistance. The slope given is 60° so R=tan(60°)=1.732 Ω. The slope of the V-I curve is resistance.
9. Calculate mark to space ratio if the system is on for 5 sec and off for 10 sec.
a) .5
b) .4
c) .2
d) .6
Answer: c
Explanation: Mark to space is Ton÷Toff. It is the ratio of time for which the system is active and the time for which is inactive. M = Ton÷Toff=5÷10=2.
Topic 8: Steady State Stability of Electric Drive
1. A 4-pole lap wound generator with 720 armature conductors and a flux per pole of .003 Wb has an armature current of 50 A. The developed torque is _________
a) 17.25 N-m
b) 17.19 N-m
c) 16.54 N-m
d) 16.89 N-m
Answer: b
Explanation: The developed torque in the motor is Km×I. The value of machine constant(Km) is Eb÷ωm = Φ×Z×P÷2×3.14×A = .003×720×4÷2×3.14×4 = .3438 Vs/rad. The developed torque is .3438×50 = 17.19 N-m.
2. The generated e.m.f from 70-pole armature having 100 conductors driven at 10 rev/sec having flux per pole as 20 mWb, with wave winding is ___________
a) 730 V
b) 740 V
c) 700 V
d) 690 V
Answer: c
Explanation: The generated e.m.f can be calculated using the formula Eb = Φ×Z×N×P÷60×A, Φ represent flux per pole, Z represents the total number of conductors, P represents the number of poles, A represents the number of parallel paths, N represents speed in rpm. In wave winding number of parallel paths are 2. Eb = .02×70×100×600÷60×2 = 700 V.
3. The unit of current is Tesla.
a) True
b) False
Answer: b
Explanation: The current is the amount of charge that can flow through an area in a given amount of time. It is mathematically represented as d(q)/d(t). It is expressed in terms of Ampere.
4. Calculate the moment of inertia of the sphere having a mass of 12 kg and radius of 78 cm.
a) 7.888 kgm2
b) 7.300 kgm2
c) 7.545 kgm2
d) 7.552 kgm2
Answer: b
Explanation: The moment of inertia of the egg can be calculated using the formula I=Σmiri2. The mass of egg and radius is given. I=(12)×(.78)2=7.300 kgm2. It depends upon the orientation of the rotational axis.
5. The most suitable servo-motor application is _________
a) AC series motor
b) DC series motor
c) AC two-phase induction motor
d) DC shunt motor
Answer: d
Explanation: DC shunt motor has definite no-load speed, so they don’t ‘run away’ when the load is suddenly thrown off provided the field circuit remains closed. The speed for any load within the operating range of the motor can be readily obtained.
6. In a DC series motor, the electromagnetic torque developed is proportional to______
a) Ia
b) Ia2
c) Ia3
d) Ia.5
Answer: b
Explanation: In a DC series motor, the electromagnetic torque developed is equal to KmΦIa. In a DC series, the motor field winding is connected in series with the armature so the flux in the field winding is proportional to current. T = KmΦIa α Ia2.
7. Calculate the value of the time period if the frequency of the signal is 70 sec.
a) 0.014 sec
b) 0.013 sec
c) 0.017 sec
d) 0.079 sec
Answer: a
Explanation: The time period is defined as the time after the signal repeats itself. It is expressed in second. T = 1÷F=1÷70=.014 sec.
8. The slope of the V-I curve is 90°. Calculate the value of resistance.
a) 1.732 Ω
b) 1.608 Ω
c) 1.543 Ω
d) 1.648 Ω
Answer: a
Explanation: The slope of the V-I curve is resistance. The slope given is 90° so R=tan(90°)=infinite Ω. It behaves as an open circuit.
9. In a DC shunt motor, the electromagnetic torque developed is proportional to______
a) Ia
b) Ia2
c) Ia3
d) Ia.5
Answer: a
Explanation: In a DC shunt motor, the electromagnetic torque developed is equal to KmΦIa. In a DC shunt, the motor field windings are connected separately and excited by a constant DC voltage. T = KmΦIa α Ia.
Topic 9: Transient Stability of an Electric Drive
1. What is the formula for the active power in the cylindrical rotor synchronous machine? (Eb represents armature emf, Vt represents terminal voltage, δ represents rotor angle, X represents reactance)
a) Eb×Vt×sinδ÷X
b) Eb×Vt2×sinδ÷X
c) Eb2×Vt×sinδ÷X
d) Eb×Vt×sinδ÷X2
Answer: a
Explanation: The real power in the cylindrical rotor machine is Eb×Vt×sinδ÷X. It is inversely proportional to the reactance. The stability of the machine is decided by the maximum power transfer capability.
2. Salient pole machines are more stable than cylindrical rotor machines.
a) True
b) False
Answer: a
Explanation: Salient pole machines are more stable than cylindrical rotor machines because of the high short circuit ratio and more real power transfer capability. The air gap length in salient pole machines is more as compared to cylindrical rotor machines.
3. The unit of reactive power is VAR.
a) True
b) False
Answer: a
Explanation: Reactive power is useless power in the case of electric circuits. It is the energy trapped that keeps on oscillating between inductive and capacitive elements. It plays a vital role in generating flux in electrical machines. It is expressed in Volt Ampere reactive.
4. Calculate the power factor during the resonance condition.
a) .58
b) .42
c) .65
d) 1
Answer: d
Explanation: During the resonance condition, the reactive power generated by the capacitor is completely absorbed by the inductor. Only active power flows in the circuit. Net reactive power is equal to zero and cosΦ=1.
5. Calculate the reactive power in a 5 Ω resistor.
a) 7 VAR
b) 0 VAR
c) 2 VAR
d) 1 VAR
Answer: b
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 90°. Q = VIsin0 = 0 VAR.
6. What is the unit of the apparent or complex power?
a) VAR
b) VA
c) ohm
d) Volt
Answer: b
Explanation: The apparent power in AC circuits is VI*. It is expressed in volt-amperes (VA). It consists of both active and reactive power. It is the vector sum of the real power and reactive power.
7. Calculate the value of the frequency of the DC supply.
a) 0 Hz
b) 50 Hz
c) 20 Hz
d) 10 Hz
Answer: a
Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal to the time period. DC supply magnitude is constant. It does not change with time so the frequency of the DC supply is 0 Hz.
8. The slope of the V-I curve is 0°. Calculate the value of resistance. The graph is parallel to the x-axis.
a) 1 Ω
b) 1.8 Ω
c) 0 Ω
d) 2.2 Ω
Answer: c
Explanation: The slope of the V-I curve is resistance. The slope given is 0° so R=tan(0°)=0 Ω. The slope of the V-I curve is resistance. It behaves as a short circuit.
9. Calculate the value of the duty cycle if the system is on for 5 sec and off for 10 sec.
a) .333
b) .444
c) .201
d) .642
Answer: a
Explanation: The duty cycle is Ton÷Ttotal. It is the ratio of the time for which the system is active and the time taken by the signal to complete one cycle. D= Ton÷Ttotal=5÷15=.333.
So, that’s all from our side, and hope you have read and learned it. If you have any questions, suggestions, or other queries. Drop your message in the comment section. We will be happy to reach you. Have a nice day!
