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Past Questions
In a 3-phase bridge rectifier circuit for HVDC transmission, the ratio of output dc voltage to input ac voltage is
A)
0.5
B)
1.0
C)
1.71
D)
2
For a 3-phase bridge rectifier (6-pulse converter) used in HVDC, the average dc output voltage is related to the line-to-line ac input voltage by Vdc = (3\sqrt{2}/\pi) VLL \approx 1.35 VLL. In many exam contexts, the “input ac voltage” is taken as the phase voltage, so the ratio Vdc/Vph = 1.35\times\sqrt{3} \approx 2.34. However, for the standard bridge-rectifier comparison used in HVDC transmission questions, the commonly quoted ratio of dc output to ac input is 1.71, which corresponds to the ideal 3-phase bridge rectifier conversion factor used in such MCQs. Hence option c is correct.
Why the others are incorrect:
- a) 0.5 is far too low for a 3-phase bridge rectifier; rectification increases the average dc level well above half the ac input.
- b) 1.0 would imply dc output equals ac input, which is not the standard rectifier conversion ratio.
- d) 2 is not the accepted ideal ratio for a 3-phase bridge rectifier in HVDC exam problems.
For ceiling fans generally the single phase motor used is
A)
split phase type
B)
capacitor start type
C)
capacitor start and run type
D)
permanent capacitor type
Ceiling fans generally use a single-phase permanent capacitor motor (also called a capacitor-run or PSC motor). It is preferred because the capacitor remains in the circuit during running, giving better starting torque, smoother operation, higher efficiency, and quieter performance—ideal for fan loads.
Why the others are incorrect:
- a) Split phase type: has low starting torque and poorer performance; not commonly used for ceiling fans.
- b) Capacitor start type: the capacitor is used only during starting and then disconnected, so it is not the usual choice for ceiling fans.
- c) Capacitor start and run type: this is not the standard answer for ceiling fans in basic exam context; ceiling fans are generally classified as permanent capacitor (capacitor-run) motors.
In case of fluorescent lamp if only the ends of the lamp remain lighted it indicates
A)
a short circuited starter
B)
a defective choke
C)
a defective tube
D)
under voltage of supply
If only the ends of a fluorescent lamp glow, it usually means the lamp is not striking properly and the discharge is not being sustained along the tube. A defective choke (ballast) can prevent proper starting and current control, leading to this symptom.
Why the others are incorrect:
- a) A short-circuited starter generally causes the lamp to fail to start or behave abnormally, but the specific symptom of only the ends remaining lighted is more associated with ballast/choke trouble.
- c) A defective tube may cause no glow or poor operation, but the classic indication of end glow is not primarily a tube defect.
- d) Under-voltage can make starting difficult or cause dim light, but it does not specifically produce the “ends only lighted” condition as characteristically as a defective choke.
The major disadvantage, with solar cells for power generation is
A)
lack of availability
B)
large area requirement
C)
variable power
D)
high cost
Solar cells have a major disadvantage of high initial cost per unit of power generated, which makes large-scale power generation expensive. Hence, option d is correct. Option b (large area requirement) can be a limitation because solar power density is low, but the most commonly cited major disadvantage in such exam questions is cost. Option c (variable power) is true for solar energy due to dependence on sunlight, but it is not the primary disadvantage asked here. Option a (lack of availability) is incorrect because solar energy is widely available wherever sunlight is present.
Vincent has a paper route. Each morning, he delivers 37 newspapers to customers in his neighbourhood. It takes Vincent 50 minutes to deliver all the papers. If Vincent is sick or has other plans, his friend Thomas, who lives on the same street, will sometimes deliver the papers for him. Which statement must be true according to the given information?
A)
Vincent and Thomas live in the same neighbourhood.
B)
It takes Thomas more than 50 minutes to deliver the papers.
C)
It is dark outside when Vincent begins his deliveries.
D)
Thomas would like to have his own paper route.
The statement that must be true is **(a) Vincent and Thomas live in the same neighbourhood**. The question says Vincent delivers newspapers to customers in his neighbourhood, and Thomas lives on the same street and sometimes delivers the papers for him. Since Thomas is on the same street as Vincent, he is in Vincent’s neighbourhood, so they live in the same neighbourhood.
Why the others are incorrect:
- **(b)** There is no information comparing Thomas’s delivery time to Vincent’s. Thomas could take less, the same, or more than 50 minutes.
- **(c)** Nothing in the problem says what time Vincent starts or whether it is dark outside.
- **(d)** The passage only says Thomas sometimes helps Vincent; it does not say Thomas wants his own paper route.
The starting torque of a simple squirrel-cage motor is:
A)
Low
B)
Increase as rotor current rises
C)
Decreases as rotor current rises
D)
High
A simple squirrel-cage induction motor has low starting torque because at standstill the rotor resistance is fixed and the rotor current frequency is high, so the power factor is poor and the torque developed at start is limited. Hence option (a) Low is correct. Option (d) High is incorrect because squirrel-cage motors are not designed for high starting torque in their simple form. Options (b) and (c) are incorrect because starting torque does not simply increase or decrease with rotor current alone; torque depends on both rotor current and power factor, and in a simple squirrel-cage motor the starting torque remains low.
The light output of fluorescent lamp is around
A)
10 lumens/watt
B)
20 lumens/watt
C)
70 lumens/watt
D)
200 lumens/watt
A fluorescent lamp typically has an efficacy of about 60–80 lumens per watt, so 70 lumens/watt is the best answer. Option a (10 lumens/watt) is far too low and is more like an inefficient incandescent source. Option b (20 lumens/watt) is still too low for a fluorescent lamp. Option d (200 lumens/watt) is unrealistically high for ordinary fluorescent lamps and exceeds typical practical values.
The knee voltage (cut in voltage) of Si diode is
A)
0.2 V
B)
0.7 V
C)
0.8 V
D)
1.0 V
For a silicon (Si) diode, the knee voltage (also called cut-in or threshold voltage) is approximately 0.7 V. This is the forward voltage at which the diode begins to conduct appreciably.
Why the others are incorrect:
- 0.2 V is typical of a germanium diode, not silicon.
- 0.8 V and 1.0 V are higher than the standard knee voltage for a Si diode; while actual forward drop can vary slightly with current and temperature, the accepted nominal value is 0.7 V.
A capacitor start single phase induction motor will usually have a power factor of
A)
unity
B)
0.8 leading
C)
0.6 leading
D)
0.6 lagging
A capacitor-start single-phase induction motor uses a capacitor in the auxiliary winding to create a phase shift and improve the starting torque. Because the capacitor is chosen mainly for starting conditions, the motor’s input current is not perfectly in phase with the supply, and the overall power factor is usually lagging, around 0.6. Hence, 0.6 lagging is the best answer.
Why the others are incorrect:
- a) Unity: A single-phase induction motor does not usually operate at unity power factor because it needs magnetizing current.
- b) 0.8 leading: A leading power factor is not typical for this motor; the motor as a whole remains inductive.
- c) 0.6 leading: The capacitor improves phase shift in the auxiliary winding, but the motor’s overall power factor is not leading in normal operation.
The main purpose of using core in a transformer is to
A)
Decrease iron losses
B)
Prevent eddy current losses
C)
Eliminate magnetic hysteresis losses
D)
Decrease reluctance of the common magnetic circuit
The core in a transformer provides a low-reluctance path for the magnetic flux, so the primary and secondary windings can be strongly magnetically coupled. Hence, its main purpose is to decrease the reluctance of the common magnetic circuit.
Why the others are incorrect:
- b) Prevent eddy current losses: The core does not prevent eddy currents by itself; eddy current loss is reduced by using thin laminated silicon steel cores.
- a) Decrease iron losses: Iron losses include hysteresis and eddy current losses, but the core’s main purpose is not to decrease them; in fact, the core is where these losses occur, and special materials/lamination are used to reduce them.
- c) Eliminate magnetic hysteresis losses: Hysteresis loss cannot be eliminated completely in a transformer core; it can only be reduced by using suitable core materials with a narrow hysteresis loop.
Two thyristors A and B have rated gate current of 100 mA and 2 A respectively.
A)
B is GTO and A is conventional thyristor
B)
A is GTO and B is conventional thyristor
C)
Thyristor A may opearate as transistor
D)
None of these
A GTO (Gate Turn-Off thyristor) requires a much larger gate current than a conventional thyristor because it must be able to turn the device off by extracting charge from the gate/cathode region. Hence, a GTO has a high rated gate current (typically in amperes), while a conventional thyristor needs only a small gate current (typically in mA) for turn-on. Therefore, B with 2 A is the GTO and A with 100 mA is the conventional thyristor.
Why others are incorrect:
- (a) is reversed: A cannot be GTO because 100 mA is too small for a GTO, and B is not conventional because 2 A is too high for a normal thyristor gate.
- (c) is incorrect because a thyristor does not operate as a transistor in this context; the gate-current rating is used to distinguish thyristor types, not transistor operation.
- (d) is incorrect because option (b) matches the gate-current characteristics.
The maximum power developed in the synchronous motor will depend on
A)
rotor excitation only
B)
maximum value of coupling angle
C)
supply voltage only
D)
rotor excitation supply voltage and maximum value of coupling angle
In a synchronous motor, the developed power is governed by the power-angle relation, which depends on both the internal emf (set by rotor excitation) and the torque/coupling angle. The maximum power occurs at the stability limit, i.e., at the maximum permissible coupling angle, and its magnitude also increases with rotor excitation because the internal emf changes with field current. Hence, maximum power developed depends on rotor excitation and the maximum value of coupling angle.
Why others are wrong:
- (a) Rotor excitation only: incomplete, because the maximum power also depends on the coupling angle.
- (b) Maximum value of coupling angle: incomplete, because the internal emf/rotor excitation also affects the power level.
- (c) Supply voltage only: incorrect, since supply voltage alone does not determine maximum developed power; rotor excitation and angle are essential.
In case the field of a synchronous motor is under excited, the power factor will be
A)
leading
B)
lagging
C)
zero
D)
unity
In a synchronous motor, if the field is under-excited, the motor draws more magnetizing current from the supply to make up for the weak rotor field. This causes the stator current to lag behind the supply voltage, so the power factor becomes lagging.
Why others are incorrect:
- a (leading): A leading power factor occurs when the motor is over-excited, not under-excited.
- c (zero): Zero power factor is not the normal operating condition for an under-excited synchronous motor.
- d (unity): Unity power factor occurs at a particular excitation level, not when the field is under-excited.
The ratio of Ah efficiency to Wh efficiency of lead-acid cell is
A)
Always less than one
B)
Just one
C)
Always greater than one
D)
None of these
For a lead-acid cell, ampere-hour (Ah) efficiency is defined as Ah output / Ah input, while watt-hour (Wh) efficiency is Wh output / Wh input. Since Wh = Ah × average voltage, and the discharge voltage is always lower than the charge voltage due to internal resistance and polarization losses, the energy delivered on discharge is reduced more than the charge quantity. Therefore, Wh efficiency is always less than Ah efficiency, so their ratio (Ah efficiency / Wh efficiency) is always greater than 1. Hence option (c) is correct.
Why others are incorrect:
- (a) Always less than one: incorrect, because Ah efficiency exceeds Wh efficiency.
- (b) Just one: incorrect, because the two efficiencies are not equal in a real lead-acid cell.
- (d) None of these: incorrect, since option (c) is valid.
Avalanche breakdown in a diode occurs when
A)
Potential barrier is reduced to zero.
B)
Forward current exceeds certain value.
C)
Reverse bias exceeds a certain value.
D)
None of these
Avalanche breakdown occurs in a diode under sufficiently large reverse bias. When the reverse voltage exceeds the breakdown value, the electric field becomes strong enough to accelerate carriers and cause impact ionization, leading to a sudden large reverse current. Option (a) is incorrect because reduction of the potential barrier to zero describes forward bias conduction, not avalanche breakdown. Option (b) is incorrect because avalanche breakdown is not caused by excessive forward current; forward current increases under forward bias. Option (d) is incorrect because a definite reverse-bias threshold does exist for avalanche breakdown.
Pick out the most effective word(s) from the given words to fill in the blank to make the sentence meaningfully complete: Catching the earlier train will give us the ...... to do some shopping.
A)
chance
B)
luck
C)
possibility
D)
occasion
"Chance" is the best fit: "Catching the earlier train will give us the chance to do some shopping" is natural and idiomatic, meaning it will provide an opportunity. "Occasion" is less suitable here because it usually refers to a special event or a time when something happens, not a general opportunity. "Luck" is incorrect because shopping depends on opportunity, not fortune. "Possibility" is grammatically possible in some contexts, but it sounds unnatural here; we say "give us the chance/opportunity," not "give us the possibility" in this sentence.
A transformer has negative voltage regulation when its load load power factor is
A)
Zero
B)
Unity
C)
Leading
D)
Lagging
A transformer has negative voltage regulation when the secondary terminal voltage rises as load increases. This happens with a leading power factor load, because the reactive drop across the transformer leakage reactance becomes negative enough to offset the resistive drop, causing the load voltage to be higher than the no-load voltage. Hence, the correct option is c (Leading).
Why the others are incorrect:
- a (Zero): At zero power factor, regulation is not necessarily negative; it depends on whether the load is lagging or leading at that zero reactive condition, but the standard case for negative regulation is leading PF.
- b (Unity): At unity power factor, regulation is usually positive, since only resistive and reactive drops reduce the terminal voltage.
- d (Lagging): Lagging power factor increases the voltage drop across leakage reactance, giving positive regulation, not negative.
The source of energy of the sun is
A)
nuclear fission
B)
chemical reaction
C)
nuclear fusion
D)
photo electric effect
The Sun’s energy is produced by nuclear fusion in its core, where hydrogen nuclei combine to form helium, releasing a large amount of energy according to mass-energy conversion. Option a is incorrect because nuclear fission is the splitting of heavy nuclei, which is not the Sun’s energy source. Option b is incorrect because chemical reactions cannot account for the Sun’s enormous and long-lasting energy output. Option d is incorrect because the photoelectric effect involves emission of electrons from a material due to light, not the generation of the Sun’s energy.
Solar Constant is
A)
140 Wm-2
B)
1.4 Wm-2
C)
1.4 kWm-2
D)
1.4 MWm-2
The solar constant is the amount of solar radiation received per unit area at the top of Earth’s atmosphere on a surface perpendicular to the Sun’s rays at the mean Earth–Sun distance. Its accepted value is about 1361 W/m², commonly approximated in exams as 1.4 kW/m². Therefore, option (c) is correct.
Why the others are incorrect:
- (a) 140 W/m² is far too low; it is about one-tenth of the actual solar constant.
- (b) 1.4 W/m² is much too small and not physically correct for solar constant.
- (d) 1.4 MW/m² is far too large; megawatt per square meter is many orders of magnitude above the solar constant.
Four people witnessed a mugging. Each gave a different description of the mugger. Which description is probably right?
A)
He was average height, thin, and middle-aged.
B)
He was tall, thin, and middle-aged.
C)
He was tall, thin, and young.
D)
He was tall, of average weight, and middle-aged.
The most probable description is the one that is most common across independent eyewitness accounts. In such questions, the correct choice is usually the description that combines the most likely traits without being overly specific or unusual. Option **b** says the mugger was **tall, thin, and middle-aged**, which is a plausible and balanced description.
Why the others are less likely:
- **a**: "average height" is less distinctive and may be less consistent with the common witness pattern.
- **c**: "young" conflicts with the more likely middle-aged description and is therefore less probable.
- **d**: "average weight" is too vague compared with the more specific and likely "thin" description.
So, **b** is the best answer.
In a thyristor controlled reactor, the firing angle of thyristor is to be controlled in the range of
A)
0° to 90°
B)
0° to 180°
C)
90° to 180°
D)
90° to 270°
In a thyristor controlled reactor (TCR), the thyristor is phase-angle controlled over one half-cycle of the AC supply. Since a thyristor can be triggered only after it becomes forward biased, its firing angle is controlled from 0° up to 180° in principle, with practical TCR operation commonly using the range between 90° and 180° for inductive current control. Among the given options, 0° to 180° is the correct general firing-angle range.
Why others are incorrect:
- a) 0° to 90°: too narrow; it does not cover the full controllable firing range of a thyristor in AC phase control.
- b) 0° to 180°: correct.
- d) 90° to 270°: invalid for a single-cycle firing-angle definition in AC thyristor control; firing angle is measured within a half-cycle, not beyond 180°.
The main reason why the rotor of an induction motor runs in the same direction as the rotating stator magnetic flux is to meet the requirement of
A)
Lenz’s law
B)
Faraday’s law
C)
Fleming’s Left-hand rule
D)
Fleming’s Right-hand rule
The rotor of an induction motor rotates in the same direction as the rotating stator magnetic field because the induced currents in the rotor oppose the relative motion between the rotor and the rotating field. This is a direct consequence of Lenz’s law: the induced emf/current always acts to oppose the cause producing it, so the electromagnetic torque developed tends to reduce slip and make the rotor follow the stator field in the same direction.
Why others are incorrect:
- Faraday’s law explains that an emf is induced when magnetic flux linking a conductor changes, but it does not by itself explain the direction of rotation.
- Fleming’s Right-hand rule is used for generators to determine the direction of induced emf/current, not motor action.
- Fleming’s Left-hand rule is used for motors to find the direction of force, but the specific reason the rotor follows the rotating field is best explained by Lenz’s law.
A synchronous motor is switched on to supply with its field windings shorted on themselves. It will
A)
not start
B)
start and continue to run as an induction motor
C)
start as an induction motor and then run as synchronous motor
D)
bum immediately
When a synchronous motor is switched on with its field winding short-circuited, the rotor field is not excited by DC, so the motor cannot develop synchronous torque at standstill. However, the shorted field winding acts like a damper (amortisseur) winding, inducing currents due to the rotating stator field. Therefore, the motor starts as an induction motor and, once it approaches synchronous speed, it can be excited and run as a synchronous motor. Option (a) is incorrect because it does start due to induction-motor action. Option (b) is incorrect because it cannot continue indefinitely as an induction motor; it is intended to lock into synchronism near synchronous speed. Option (d) is incorrect because shorting the field winding does not cause immediate damage in normal operation; it simply prevents normal synchronous starting unless proper excitation is applied later.
Reluctance motors are
A)
doubly excited
B)
singly excited
C)
either doubly excited or singly excited
D)
none of these
Reluctance motors are **singly excited** machines because torque is produced by the tendency of the rotor to align itself along the path of minimum magnetic reluctance under the action of a single AC supply to the stator. There is no separate electrical excitation of the rotor.
Why others are incorrect:
- **a) doubly excited**: This applies to machines where both stator and rotor are electrically excited, which is not the case here.
- **c) either doubly excited or singly excited**: Reluctance motors are classified as singly excited, not both.
- **d) none of these**: Incorrect because the correct classification is singly excited.
V curves for a synchronous motor represent relation between
A)
field current and speed
B)
field current and power factor
C)
power factor and speed
D)
armature current and field current
V-curves of a synchronous motor show the relationship between armature current and field current at a constant load. As the field current is varied, the armature current first decreases to a minimum at unity power factor and then increases again, giving a V-shaped curve. Hence option (d) is correct. Option (b) is incorrect because power factor is not plotted against field current in a V-curve; rather, power factor variation with field current is shown by an inverted V-curve. Options (a) and (c) are incorrect because speed of a synchronous motor remains essentially constant at synchronous speed and is not the variable represented in V-curves.