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Past Questions
Megger is operated by
A)
an external voltage source
B)
an internal hand driven generator
C)
internal battery
D)
all of above
A Megger (Mega-Ohmmeter) is a specialized instrument used to measure very high electrical resistances, typically the insulation resistance of cables and motor windings. To measure high resistance, a high DC voltage is required. Traditionally, the Megger is designed with an internal hand-driven DC generator (a magneto) that produces the necessary high voltage when the operator turns a handle. While modern digital insulation testers use batteries, the classic 'Megger' referred to in standard electrical engineering exam contexts is defined by its hand-cranked generator. Option (a) is incorrect because the instrument is designed to be self-contained for safety and portability. Option (c) refers to modern insulation testers, but the fundamental definition of a Megger in technical textbooks emphasizes the hand-driven generator.
E.m.f. equation of a generator is (Where is flax per pole, N is speed in r.p.m., Z is number of conductor, P is number of pole and A is number of parallel paths)
A)
e.m.f = NA P / 60 z volts
B)
e.m.f. = AZ P / 60 N volts
C)
e.m.f. = NZ P / 60 A volts
D)
none of the above
The electromotive force (e.m.f.) generated in a DC generator is derived from Faraday's Law of Electromagnetic Induction. The standard formula for the generated e.m.f. (E) is: E = (ΦZN P) / (60A).
Breaking down the variables:
- Φ (Phi): Flux per pole (Webers)
- Z: Total number of armature conductors
- N: Speed of armature in revolutions per minute (r.p.m.)
- P: Number of poles
- A: Number of parallel paths (A = P for lap winding, A = 2 for wave winding)
- 60: Conversion factor to change r.p.m. to revolutions per second.
Comparing this to the options:
- Option (a) incorrectly places P in the numerator and Z in the denominator.
- Option (b) incorrectly places A in the numerator and N in the denominator.
- Option (c) correctly follows the formula: e.m.f. = (Φ * N * Z * P) / (60 * A).
Transformer breaths in when
A)
load on it reduces
B)
load remains constant
C)
load on it increases
D)
none of the above
The 'breathing' of a transformer refers to the expansion and contraction of the insulating oil due to temperature changes.
1. When the load on a transformer increases, the internal temperature of the oil rises due to copper losses (I²R) and iron losses. This causes the oil to expand, pushing air out of the conservator tank (the transformer 'breathes out').
2. Conversely, when the load on the transformer reduces or is removed, the temperature of the insulating oil decreases.
3. As the oil cools, it contracts, leading to a decrease in volume.
4. To fill this void and maintain pressure, air from the atmosphere is drawn into the conservator tank. This process is known as 'breathing in'.
Therefore, option (a) is correct. Option (b) is incorrect because constant load results in constant temperature and no breathing. Option (c) is incorrect because increasing load causes the transformer to breathe out.
The maximum permissible voltage drop from supply terminals to any point in the installation for light wiring is
A)
3%+1V
B)
2%+1V
C)
4%+2V
D)
none of above
According to standard electrical installation guidelines (such as those derived from the Indian Electricity Rules and the National Electrical Code), the permissible voltage drop is limited to ensure that electrical appliances operate efficiently and safely. For light wiring (lighting circuits), the maximum permissible voltage drop from the supply terminals (the point of entry/distribution board) to any point in the installation is generally specified as 3% of the nominal voltage plus a constant of 1V. This allows for a small margin of error in low-voltage scenarios while maintaining a strict percentage limit for higher nominal voltages to prevent flickering of lights or malfunction of electronic ballasts. Options b, c, and d are incorrect as they do not align with the standard regulatory specifications for lighting installations.
The minimum size of single core copper conductor used for power socket wiring is
A)
1.5mm2
B)
2.5mm2
C)
6.0mm2
D)
4.0mm2
In standard electrical installation practices and according to the Indian Standard (IS) and National Electrical Code (NEC) guidelines for residential house wiring:
1. Lighting circuits typically use a minimum conductor size of 1.5 mm² because the current load for LED or CFL bulbs is relatively low.
2. Power socket circuits (which are intended for appliances like irons, mixers, or vacuum cleaners) carry higher current loads. To prevent overheating and voltage drop, a minimum conductor size of 2.5 mm² is mandated for power sockets.
3. 4.0 mm² and 6.0 mm² conductors are generally reserved for high-power appliances like air conditioners, geysers, or the main incoming supply line to the distribution board.
Therefore, for general power socket wiring, 2.5 mm² is the standard minimum requirement.
1(one) Ton of refrigeration is equal to
A)
15,000 BTU/hr
B)
13,000 BTU/hr
C)
12,000 BTU/hr
D)
none of above
A 'Ton of Refrigeration' (TR) is a unit of power used to describe the cooling capacity of air conditioning and refrigeration systems. It is defined as the amount of heat required to melt one short ton (2,000 lbs) of pure ice at 0°C (32°F) in 24 hours.
1. The latent heat of fusion for ice is approximately 144 BTU/lb.
2. For one ton of ice: 2,000 lbs × 144 BTU/lb = 288,000 BTU.
3. To find the hourly rate: 288,000 BTU / 24 hours = 12,000 BTU/hr.
Therefore, 1 TR = 12,000 BTU/hr. Options (a) and (b) are numerically incorrect based on this standard thermodynamic definition.
Which is better earth resistance-
A)
1 ohm
B)
50 ohm
C)
100 ohm
D)
0.5 ohm
In electrical earthing systems, the primary goal is to provide a low-impedance path for fault currents to flow safely into the earth. According to Ohm's Law (V = I * R), for a given fault current (I), a lower resistance (R) results in a lower voltage (V) appearing on the equipment chassis or the ground surface. This minimizes the risk of electric shock to personnel and ensures that protective devices (like circuit breakers or fuses) trip quickly by allowing a high enough fault current to flow. Therefore, the lower the earth resistance, the more effective and 'better' the earthing system is. Comparing the given options: 0.5 ohm < 1 ohm < 50 ohm < 100 ohm. Thus, 0.5 ohm is the best (lowest) resistance value among the choices provided.
Induction type of instruments are used on-
A)
d.c. supply only
B)
a.c. supply only
C)
both a.c. and d.c. supply
D)
battery supply
Induction-type instruments operate on the principle of electromagnetic induction. They require a changing magnetic field to induce eddy currents in a conducting disc (usually aluminum). According to Faraday's Law of Electromagnetic Induction, a changing magnetic flux is necessary to produce these currents.
1. Why AC only: In an Alternating Current (AC) supply, the current changes direction and magnitude periodically, creating the necessary time-varying magnetic flux to induce eddy currents and produce a deflecting torque.
2. Why not DC: In a Direct Current (DC) supply, the magnetic field produced is stationary (constant). A constant magnetic field cannot induce eddy currents in a stationary disc, meaning no torque is produced and the instrument will not deflect.
3. Battery supply: Since batteries provide DC, they are unsuitable for induction-type instruments.
Therefore, induction-type instruments (such as induction wattmeters and energy meters) are exclusively used for AC measurements.
The standard pit depth of elevator is in between
A)
1,200-1,400mm
B)
1,000-1,500mm
C)
both (a) and (b)
D)
none of above
In elevator installation and engineering standards, the 'pit' is the area below the lowest landing floor level. The pit depth is critical for housing the buffers, tensioning devices, and providing safety clearance for technicians. For standard passenger elevators (low to medium speed), the typical standard pit depth ranges between 1,200mm and 1,400mm. While specific depths can vary based on the manufacturer (e.g., Otis, Kone, Schindler) and the speed of the lift, the range of 1,200-1,400mm is the most widely accepted standard for general departmental engineering examinations. Option (b) is too broad, and since (a) represents the specific standard range, (c) is incorrect.
The function of a silica gel is to absorb moisture from the air when transformer
A)
breaths out
B)
breaths in
C)
breaths in and breaths out
D)
none of above
In a transformer, the oil expands when the transformer is loaded (heating up) and contracts when the transformer is unloaded or cooled down. This process is referred to as 'breathing'.
1. When the transformer 'breathes out', oil expands and air is pushed out of the conservator tank into the atmosphere. No moisture is entering the system during this phase.
2. When the transformer 'breathes in', the oil contracts, creating a vacuum that draws fresh air from the atmosphere into the conservator tank to fill the space.
3. Atmospheric air contains moisture (water vapor). If this moisture enters the transformer, it can degrade the dielectric strength of the insulating oil and the paper insulation, leading to potential electrical failure.
4. The silica gel breather is placed in the air intake path. As air 'breathes in', it passes through the silica gel, which absorbs the moisture before the air reaches the oil. Therefore, the function of silica gel is specifically to protect the oil during the 'breathing in' process.
1(one) horse-power is equal to
A)
546 W
B)
656 W
C)
746 W
D)
none of above
Horsepower (hp) is a unit of power. In the electrical and mechanical engineering context (specifically the mechanical or imperial horsepower), 1 horsepower is defined as the power needed to lift 550 pounds by 1 foot in 1 second. When converted to the International System of Units (SI), this equals approximately 745.7 watts, which is standardly rounded to 746 W for most engineering examinations.
Breakdown of options:
- Option (a) 546 W: Incorrect value.
- Option (b) 656 W: Incorrect value.
- Option (c) 746 W: Correct standard conversion for 1 hp.
- Option (d) none of above: Incorrect because option (c) is correct.
The standard lift door height for a hospital is
A)
2,100mm
B)
2,200mm
C)
2,000mm
D)
2,300mm
In the design of elevators, standard door heights vary based on the application of the building. For residential or general commercial buildings, a standard door height is often 2,000mm. However, for hospitals and healthcare facilities, the doors must accommodate stretchers, beds, and medical equipment, requiring a higher clearance. The industry standard for hospital lift door heights is typically 2,100mm to ensure the seamless movement of patient trolleys and beds. Options b, c, and d are incorrect because 2,000mm is too low for medical equipment, and 2,200mm or 2,300mm exceed the standard architectural specifications for typical hospital lift shafts unless specifically customized for oversized equipment.
25mm conduit size means that
A)
the mean diameter of the conduit is 25mm
B)
the external diameter of the conduit is 25mm
C)
the internal diameter of the conduit is 25mm
D)
none of the above
In electrical installations and house wiring, conduit sizes are standardized based on their internal diameter (ID). This is because the primary concern for an electrician or engineer is the 'clear space' available inside the pipe to accommodate the electrical cables without exceeding the permissible fill capacity (to prevent overheating and allow for easy pulling of wires). Therefore, when a conduit is referred to as '25mm', it signifies that the internal diameter is 25mm, regardless of the thickness of the conduit wall, which would make the external diameter larger.
Refrigeration and air -conditioning involve heat transfer using
A)
water
B)
ice and water
C)
refrigerants
D)
none of above
Refrigeration and air-conditioning systems operate on the principle of heat transfer, where heat is absorbed from a low-temperature region (the space to be cooled) and rejected to a high-temperature region (the surroundings). This process is facilitated by a specialized working fluid known as a 'refrigerant'. Refrigerants are chosen for their specific thermodynamic properties, most notably a low boiling point, which allows them to evaporate at low temperatures and absorb significant amounts of heat (latent heat of vaporization) from the environment.
Why other options are incorrect:
- (a) Water: While water is used as a secondary coolant in chilled water systems or as a heat sink in condensers, it is not the primary medium used to perform the refrigeration cycle itself due to its high freezing point.
- (b) Ice and water: These were used in primitive 'ice-box' refrigeration, but modern refrigeration and air-conditioning systems rely on mechanical cycles using chemical refrigerants.
- (d) None of above: This is incorrect because option (c) is the standard technical answer.
The power in a 3-phase delta connection is
A)
V I Cos
B)
V I Cos / 2 ph 2
C)
3 V I Cos / ph 2
D)
none of above
In a 3-phase system, the total active power (P) is the sum of the power in all three phases. For a balanced load, the formula is P = 3 × V_ph × I_ph × cosΦ, where V_ph is the phase voltage, I_ph is the phase current, and cosΦ is the power factor.
In a Delta (Δ) connection:
1. Line Voltage (V) is equal to Phase Voltage (V_ph): V = V_ph.
2. Line Current (I) is related to Phase Current (I_ph) by the relation: I = √3 × I_ph, which means I_ph = I / √3.
Substituting these into the power formula:
P = 3 × V × (I / √3) × cosΦ
Since 3 / √3 = √3, the formula simplifies to:
P = √3 × V × I × cosΦ
Comparing this to the options:
- Option (a) 'V I Cos' matches the derived formula (√3VI cosΦ).
- Option (b) and (c) introduce incorrect divisors and subscripts that do not align with standard electrical power formulas for a 3-phase delta system.
Electrical energy is measured in-
A)
Ohm meter
B)
volt meter
C)
ampere meter
D)
energy meter
To determine the correct instrument for measuring electrical energy, we must analyze the function of each device listed in the options:
1. **Ohm meter (Option a):** This instrument is used to measure electrical resistance, which is the opposition to the flow of current in a circuit. Its unit is the Ohm (Ω).
2. **Volt meter (Option b):** This instrument is used to measure electrical potential difference (voltage) between two points in a circuit. Its unit is the Volt (V).
3. **Ampere meter (Option c):** This instrument is used to measure the flow of electric current through a conductor. Its unit is the Ampere (A).
4. **Energy meter (Option d):** Electrical energy is the product of power (voltage × current) and the time for which it is consumed. An energy meter (also known as a kilowatt-hour meter) is specifically designed to integrate power over time to measure the total electrical energy consumed. Its common unit is the kilowatt-hour (kWh).
Therefore, the energy meter is the only instrument among the choices designed to measure electrical energy.
The standard door width for hospital elevators are
A)
1,100mm
B)
1,300mm
C)
1,200mm
D)
1,400mm
In the design of elevators for healthcare facilities (Hospital Elevators), the primary requirement is to accommodate stretchers, hospital beds, and medical personnel simultaneously. Standard architectural and engineering guidelines for hospital bed elevators typically specify a clear door opening width of at least 1,200 mm (approximately 48 inches). This width is essential to ensure that a standard hospital stretcher (which is typically around 700-900mm wide) can pass through easily with accompanying medical staff or equipment without obstruction. Options a (1,100mm) is generally too narrow for efficient stretcher movement, while options b (1,300mm) and d (1,400mm) are non-standard widths for basic hospital requirements, although larger custom elevators may exist, 1,200mm remains the industry standard benchmark for minimum clear opening.
A generator is a machine which converts
A)
electrical energy into mechanical energy
B)
mechanical energy into electrical energy
C)
both (a) and (b)
D)
none of the above
An electric generator is an electromechanical device that operates on the principle of electromagnetic induction (Faraday's Law). It takes mechanical energy (provided by a prime mover such as a turbine, engine, or hand crank) and converts it into electrical energy.
Analysis of options:
- Option (a) describes an electric motor, which converts electrical energy into mechanical energy.
- Option (b) correctly describes a generator, which converts mechanical energy into electrical energy.
- Option (c) is incorrect because a single machine cannot be both a generator and a motor simultaneously in its primary function, although some machines (like synchronous machines) can operate in both modes depending on the input.
- Option (d) is incorrect as option (b) is the standard definition.
The electric motor helps the elevator car to
A)
move downward
B)
move upward
C)
both (a) and (b)
D)
none of above
In an elevator system, the electric motor (typically an AC or DC motor) is the primary prime mover. It drives a hoisting machine (sheave) that moves the cables. To move the elevator car upward, the motor rotates in one direction to pull the car up. To move the elevator car downward, the motor either rotates in the opposite direction or controls the descent (braking/regulating speed) to ensure a safe landing. Regardless of whether the system is traction-based or hydraulic, the motor is essential for initiating and controlling movement in both vertical directions. Therefore, the motor helps the car move both upward and downward.
Three lamps of following ratings are connected in series across 230 V supply. Which of the lamp will glow the brightest?
A)
15 W 230 V lamp
B)
100 W 230 V lamp
C)
60 W 230 V lamp
D)
200 W 230 V lamp
Step-by-step reasoning:
1. **Understanding Resistance**: The power rating of a lamp is given by P = V²/R. Therefore, the resistance of a lamp is R = V²/P. Since all lamps are rated for the same voltage (230V), the lamp with the lowest power rating (15W) will have the highest resistance, and the lamp with the highest power rating (200W) will have the lowest resistance.
2. **Series Connection**: In a series circuit, the current (I) flowing through every component is the same.
3. **Actual Power Dissipation**: The brightness of a lamp depends on the actual power it dissipates when connected to the circuit, which is calculated by P_actual = I² * R.
4. **Comparison**: Since the current (I) is constant for all lamps in series, the power dissipated is directly proportional to the resistance (P ∝ R).
5. **Conclusion**: The 15W lamp has the highest resistance among the given options. Consequently, it will dissipate the most power (I²R) when connected in series, making it glow the brightest. Conversely, the 200W lamp has the lowest resistance and will glow the dimmest.
The unit of e.m.f is
A)
Watt
B)
Volt
C)
Ampere
D)
Kilowatt-hour
Step-by-step reasoning:
1. Understanding e.m.f: e.m.f stands for Electromotive Force. Despite the word 'force', it is not a mechanical force but rather the energy provided by a cell or generator per unit charge.
2. Definition: e.m.f is defined as the potential difference between the terminals of a source when no current is flowing.
3. Unit Identification: Since e.m.f is a measure of electrical potential difference, its SI unit is the Volt (V).
Analysis of other options:
- a. Watt: This is the unit of Power (P = VI).
- c. Ampere: This is the unit of Electric Current (I).
- d. Kilowatt-hour: This is a unit of Electrical Energy (E = P × t).
According to I.E. rules, the earth resistance for low and medium voltage systems should be
A)
less than 1 ohm
B)
less than 3 ohm
C)
less than 4 ohm
D)
less than 5 ohm
According to the Indian Electricity (I.E.) Rules and standard electrical engineering practices for safety and protection, the earth resistance for low and medium voltage systems—particularly for power stations, large substations, and critical industrial installations—should be kept as low as possible to ensure that fault currents are quickly diverted to the ground, triggering protective devices (like circuit breakers or fuses) to operate. While different types of earthing (like pipe or plate) have varying practical limits, the authoritative standard for high-reliability systems and the general benchmark for 'ideal' low/medium voltage system earthing in competitive exams based on I.E. rules is typically cited as less than 1 ohm. This ensures that the touch and step potentials remain within safe limits during a fault.
The S.I. Unit of luminous flux is-
A)
lux
B)
lumen
C)
candela
D)
none of above
The question asks for the S.I. unit of luminous flux.
1. **Luminous Flux** is defined as the total amount of visible light emitted by a source per unit of time. The S.I. unit for this quantity is the **lumen (lm)**. Mathematically, 1 lumen = 1 candela × 1 steradian.
2. **Analysis of Options:**
- **a. lux:** This is the S.I. unit of *illuminance* (luminous flux per unit area). 1 lux = 1 lumen/m².
- **b. lumen:** This is the correct S.I. unit for *luminous flux*.
- **c. candela:** This is the S.I. base unit of *luminous intensity* (the amount of light emitted in a specific direction).
- **d. none of above:** Incorrect, as option 'b' is the correct answer.
Resistance of 250 V,100W lamp will be-
A)
0.4ohm
B)
25ohm
C)
625ohm
D)
2.4 ohm
To find the resistance of the lamp, we use the electrical power formula that relates Power (P), Voltage (V), and Resistance (R):
Formula: P = V² / R
Given values:
Voltage (V) = 250 V
Power (P) = 100 W
Rearranging the formula to solve for Resistance (R):
R = V² / P
Step-by-step calculation:
1. Square the voltage: 250² = 250 * 250 = 62,500
2. Divide by the power: 62,500 / 100 = 625
Therefore, the resistance is 625 ohms.
Analysis of options:
- Option (a) 0.4 ohm: Incorrect. This would be the result if one incorrectly divided power by voltage squared (100/62500).
- Option (b) 25 ohm: Incorrect.
- Option (c) 625 ohm: Correct based on the calculation.
- Option (d) 2.4 ohm: Incorrect.
The formula Power= VI Cos is applicable to 3-phase circuits having-
A)
star connections
B)
delta connections
C)
port above connections
D)
none of the above connections
The formula for total active power in a balanced 3-phase system is P = √3 × V_L × I_L × cos φ, where V_L is the line-to-line voltage and I_L is the line current.
1. In a Star (Y) connection: The total power is 3 × V_phase × I_phase × cos φ. Since V_L = √3 × V_phase and I_L = I_phase, substituting these into the formula gives P = 3 × (V_L/√3) × I_L × cos φ = √3 × V_L × I_L × cos φ.
2. In a Delta (Δ) connection: The total power is 3 × V_phase × I_phase × cos φ. Since V_L = V_phase and I_L = √3 × I_phase, substituting these into the formula gives P = 3 × V_L × (I_L/√3) × cos φ = √3 × V_L × I_L × cos φ.
Because the formula P = √3 VI cos φ depends on line values (V_L and I_L), it is universally applicable to both star and delta connections. Note: 'port above connections' is a typographical error in the exam paper for 'both above connections'.