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Past Questions
The efficiency of a transformer will be maximum when-
A)
copper losses = hysteresis losses
B)
hysteresis losses = eddy current losses
C)
eddy current losses = copper losses
D)
copper losses = iron losses
The efficiency of a transformer is defined as the ratio of output power to input power. The total losses in a transformer consist of two main components:
1. Iron Losses (Core Losses): These include hysteresis and eddy current losses. They are considered constant losses because they depend on the supply voltage and frequency, not the load.
2. Copper Losses (Ohmic Losses): These occur due to the resistance of the windings and are variable losses because they vary with the square of the load current (I²R).
Mathematically, Efficiency (η) = Output / (Output + Constant Losses + Variable Losses). To find the maximum efficiency, we differentiate the efficiency expression with respect to the load current and set it to zero. This calculus derivation proves that maximum efficiency occurs when the variable loss (copper loss) is exactly equal to the constant loss (iron loss).
Analysis of options:
- Option a, b, and c are incorrect because they compare specific components of the core loss or mix core loss with copper loss. Hysteresis and eddy current losses together make up the total iron loss; the condition for maximum efficiency is not based on their individual equality, but on the equality of the total iron loss to the total copper loss.
Four resistors each of 3 ohm are connected in series, the equivalent resistance of the system will be-
A)
4/3 Ohm
B)
12 Ohm
C)
6 Ohm
D)
3/4 Ohm
Step-by-step calculation:
1. Identify the given values: There are four resistors, each with a resistance (R) of 3 ohms.
2. Identify the connection type: The resistors are connected in 'series'.
3. Apply the formula for equivalent resistance (Req) in a series circuit: Req = R1 + R2 + R3 + ... + Rn.
4. Substitute the values: Req = 3 Ω + 3 Ω + 3 Ω + 3 Ω = 12 Ω.
Analysis of options:
- Option (a) 4/3 Ohm: This would be the result if the resistors were connected in parallel (1/Req = 1/3 + 1/3 + 1/3 + 1/3 = 4/3, so Req = 3/4, but 4/3 is the reciprocal).
- Option (b) 12 Ohm: Correct, as per the series addition formula.
- Option (c) 6 Ohm: Incorrect calculation.
- Option (d) 3/4 Ohm: This is the correct equivalent resistance if the resistors were connected in parallel (Req = R/n = 3/4).
The SI unit of Magnetic Flux is-
A)
Hertz
B)
Weber
C)
Ampere
D)
Candela
Magnetic flux (Φ) is a measure of the total magnetic field which passes through a given area. In the International System of Units (SI), the unit of magnetic flux is the Weber (Wb). One Weber is defined as the flux through a surface that is perpendicular to a magnetic field of one tesla (T) over an area of one square meter (m²).
Analysis of other options:
- a. Hertz (Hz): The SI unit of frequency.
- c. Ampere (A): The SI unit of electric current.
- d. Candela (cd): The SI unit of luminous intensity.
A transformer core is laminated to-
A)
reduce hysteresis loss
B)
reduce eddy current losses
C)
reduce copper losses
D)
reduce all above losses
Step-by-step reasoning:
1. **Understanding Eddy Currents**: When an alternating magnetic flux passes through a conductive core, it induces small circulating currents within the bulk of the core material. These are known as eddy currents.
2. **The Problem**: These currents flow in loops and, due to the electrical resistance of the core material, they generate heat (I²R loss), which leads to energy inefficiency and overheating of the transformer.
3. **The Solution (Lamination)**: To prevent these currents from flowing in large loops, the solid core is replaced by a stack of thin sheets (laminations) insulated from each other by a thin layer of varnish or oxide.
4. **Mechanism**: Lamination increases the effective electrical resistance of the path available for eddy currents, thereby restricting their magnitude and significantly reducing the power loss.
Why other options are incorrect:
- **Option a (Hysteresis loss)**: Hysteresis loss depends on the material's magnetic properties (the area of the B-H loop). It is reduced by using high-permeability materials like Silicon Steel, not by laminating the core.
- **Option c (Copper losses)**: Copper losses (I²R losses) occur in the windings due to the resistance of the conductor. Laminating the core has no effect on the resistance of the copper wires.
- **Option d (All above losses)**: Since lamination specifically targets eddy currents and does not affect hysteresis or copper losses, this option is incorrect.
As per JERC(M&M) Electricity Supply Code Regulation 2013, any of the following documents or its attested copies, shall not be considered as acceptable proof of identity for individual in respect to application of new service connection-
A)
Electoral identity card
B)
Driving Licence
C)
Land Settlement Certificate (LSC)
D)
Ration Card
Under the JERC (M&M) Electricity Supply Code Regulations, when applying for a new service connection, the applicant must provide 'Proof of Identity' (POI) and 'Proof of Ownership/Occupancy' (Proof of Address).
1. Electoral Identity Card, Driving Licence, and Ration Card are standard government-issued documents that verify the identity of an individual (who the person is), making them acceptable as Proof of Identity (POI).
2. A Land Settlement Certificate (LSC) is a document used to prove ownership or legal possession of a piece of land. Therefore, it serves as 'Proof of Ownership' or 'Proof of Address', not as a 'Proof of Identity'.
Since the question asks which document shall NOT be considered as acceptable proof of identity, the Land Settlement Certificate is the correct answer because its purpose is to prove property rights, not personal identity.
In aluminium conductors, steel core is provided to-
A)
compensate for skin effect
B)
neutralise proximity effect
C)
reduce line inductance
D)
increase the tensile strength
The question refers to ACSR (Aluminium Conductor Steel Reinforced) conductors.
1. **Reasoning for the correct option**: Aluminium is an excellent conductor of electricity due to its low weight and cost, but it has relatively low mechanical strength (tensile strength). In long-span transmission lines, the conductor must support its own weight and withstand external loads like wind and ice without excessive sagging or breaking. By placing a core of high-strength galvanized steel at the center, the conductor gains the necessary mechanical strength (tensile strength) to withstand these stresses, while the outer aluminium strands carry the bulk of the electrical current.
2. **Why other options are incorrect**:
- **a. Compensate for skin effect**: Skin effect is a phenomenon where current tends to flow on the outer surface of a conductor. A steel core does not eliminate this; in fact, since the current flows primarily in the outer aluminium layers, the steel core is strategically placed in the center where current density is lowest.
- **b. Neutralise proximity effect**: Proximity effect is caused by the magnetic fields of adjacent conductors. A steel core does not neutralize this effect.
- **c. Reduce line inductance**: Line inductance depends on the spacing between conductors and the radius of the conductor. Adding a steel core does not significantly reduce the overall line inductance of the transmission system.
A SWG is used to measure-
A)
length of the wire
B)
Cross sectional area of the wire
C)
Diameter of the wire
D)
Insulation of the wire
SWG stands for Standard Wire Gauge. It is a standardized system used primarily in the UK and other Commonwealth countries to specify the diameter of electrical wires.
1. How it works: The SWG number is inversely proportional to the diameter; as the gauge number increases, the diameter of the wire decreases.
2. Why option 'c' is correct: The primary purpose of a wire gauge (like SWG or AWG) is to provide a standard reference for the physical thickness (diameter) of the conductor.
3. Why other options are incorrect:
- Option 'a' (length): Length is measured using a measuring tape or ruler, not a gauge.
- Option 'b' (Cross sectional area): While the area can be calculated using the diameter (Area = πr²), the SWG specifically defines the diameter. Area is usually expressed in mm².
- Option 'd' (Insulation): Insulation properties are measured using an insulation resistance tester (Megger), not a wire gauge.
Distribution transformers are usually designed to have maximum efficiency-
A)
at 50% of full load
B)
at 75% of full load
C)
near full load
D)
near no load
Option A is correct: at 50% of full load.
Five capacitors each of 5 µF are connected in series, the equivalent capacitance of the system will be-
A)
5 µF
B)
25 µF
C)
10 µF
D)
1 µF
To find the equivalent capacitance (C_eq) of capacitors connected in series, we use the reciprocal formula: 1/C_eq = 1/C1 + 1/C2 + ... + 1/Cn.
Given:
- Number of capacitors (n) = 5
- Capacitance of each capacitor (C) = 5 µF
Since all capacitors are identical, the formula simplifies to:
1/C_eq = n * (1/C)
1/C_eq = 5 * (1/5 µF)
1/C_eq = 1 µF⁻¹
C_eq = 1 / 1 µF⁻¹ = 1 µF.
Analysis of options:
- Option (a) 5 µF: This would be the result if only one capacitor was present or if they were in parallel and we divided by 5.
- Option (b) 25 µF: This is the result if the capacitors were connected in parallel (C_eq = n * C = 5 * 5 = 25 µF).
- Option (c) 10 µF: This is mathematically incorrect for this configuration.
- Option (d) 1 µF: This is the correct result for 5 identical 5 µF capacitors in series.
The capacity of battery is expressed in terms of-
A)
Ampere
B)
Volt-hour
C)
Ampere-hour
D)
Volt-Ampere hour
The capacity of a battery refers to the total amount of electric charge it can store and deliver over a period of time. Mathematically, Charge (Q) = Current (I) × Time (t). In the context of batteries, the standard unit for this is the Ampere-hour (Ah). For example, a battery with a capacity of 100 Ah can theoretically provide a current of 1 Ampere for 100 hours, or 10 Amperes for 10 hours.
Analysis of other options:
- Option (a) Ampere: This is the unit of electric current, not capacity.
- Option (b) Volt-hour: This is not a standard unit for battery capacity.
- Option (d) Volt-Ampere hour: This is equivalent to Watt-hour (Wh), which measures the total energy stored in the battery, not the charge capacity. While energy is important, 'battery capacity' by standard technical definition refers to the charge capacity in Ah.
Transformer ratings are expressed in terms of-
A)
kVA
B)
kW
C)
kVAr
D)
kWh
Transformer ratings are expressed in kVA (kilo-Volt-Ampere) because the losses in a transformer depend on the current (copper losses) and the voltage (core losses), but not on the phase angle between them (power factor).
1. **Copper Losses (I²R):** These depend on the load current flowing through the windings, regardless of whether the load is resistive, inductive, or capacitive.
2. **Iron/Core Losses:** These depend on the supply voltage and frequency, which remain constant for a given rating.
3. **Power Factor Independence:** Since the manufacturer does not know the power factor of the load that the end-user will connect to the transformer, the rating cannot be expressed in kW (which is real power and depends on the power factor: kW = kVA × cosφ).
Therefore, the apparent power (kVA) is the only universal measure for rating a transformer.
Why other options are incorrect:
- **b. kW (kilo-Watt):** This is real power. It depends on the load power factor, which is unknown at the time of transformer manufacture.
- **c. kVAr (kilo-Volt-Ampere Reactive):** This represents reactive power, which is only a component of the total power and does not represent the full capacity of the transformer.
- **d. kWh (kilo-Watt Hour):** This is a unit of energy consumption over time, not a rating of power capacity.
Highest system voltage in Mizoram is-
A)
220kV
B)
132kV
C)
66kV
D)
None of the above
In the state of Mizoram, the power transmission network is managed by the Power and Electricity Department. The state is integrated with the North Eastern Regional Grid. The highest transmission voltage level currently utilized for the main backbone of the state's transmission system is 132kV. While higher voltages like 220kV or 400kV are used in the National Grid and in neighboring states (like Assam), the internal highest system voltage for distribution and transmission within Mizoram's primary infrastructure is 132kV, which is then stepped down to 66kV, 33kV, and 11kV for local distribution. Therefore, option (b) is the correct answer.
Which of the following device is used to measure power in A.C. circuits?
A)
Ammeter
B)
Wattmeter
C)
Voltmeter
D)
Ohmmeter
Step-by-step reasoning:
1. The question asks for the device used to measure 'power' in an A.C. (Alternating Current) circuit.
2. Power in an A.C. circuit is the product of Voltage (V), Current (I), and the Power Factor (cos φ).
3. A Wattmeter is specifically designed to measure active power (real power) in watts. It typically consists of two coils: a current coil (connected in series) and a pressure/voltage coil (connected in parallel), allowing it to account for the phase difference between voltage and current.
Evaluating other options:
- a. Ammeter: Used to measure electric current (I) in Amperes.
- c. Voltmeter: Used to measure electric potential difference (V) in Volts.
- d. Ohmmeter: Used to measure electrical resistance (R) in Ohms.
Therefore, the correct device for measuring power is the Wattmeter.
The corona is considerably affected by which of the following?
A)
Size of the conductor
B)
Shape of the conductor
C)
Surface condition of the conductor
D)
All of the above
Corona discharge occurs when the electric field intensity at the surface of a conductor exceeds the dielectric strength of the surrounding air. The critical disruptive voltage (the voltage at which corona begins) is influenced by several factors:
1. **Size of the conductor (Option a):** The electric field intensity is inversely proportional to the radius of the conductor. A larger conductor radius reduces the potential gradient at the surface, thereby increasing the critical disruptive voltage and reducing the likelihood of corona.
2. **Shape of the conductor (Option b):** The distribution of the electric field depends on the geometry. Smooth, circular conductors have a uniform field, whereas sharp edges, protrusions, or non-circular shapes create points of high electric field concentration, which trigger corona at lower voltages.
3. **Surface condition of the conductor (Option c):** The presence of scratches, dust, moisture, or oxidation on the conductor surface creates local points of high stress (irregularities), which lower the critical disruptive voltage and enhance corona discharge.
Since all three factors significantly influence the onset and intensity of corona, the correct answer is 'All of the above'.
Which element of hydroelectric power plant prevents the penstock from water hammer phenomenon?
A)
Valves and Gates
B)
Draft tubes
C)
Spillway
D)
Surge Tank
Step-by-step reasoning: 1. Water Hammer Phenomenon: This occurs in a hydroelectric plant when there is a sudden change in the velocity of water flowing through the penstock (e.g., due to a sudden closure of the turbine gates). This creates a high-pressure shock wave that can rupture the penstock pipes. 2. Function of a Surge Tank: A surge tank is a storage reservoir or tank placed between the main reservoir and the turbine. When the turbine gates close suddenly, the excess water rises into the surge tank instead of slamming against the penstock walls, thereby absorbing the pressure surge and protecting the pipeline. 3. Analysis of other options: (a) Valves and Gates are used to control flow but their sudden operation is actually the cause of water hammer. (b) Draft tubes are used to recover kinetic energy from the water exiting the turbine to increase efficiency. (c) Spillways are used to release excess water from the dam reservoir to prevent overtopping, not to protect the penstock from pressure surges.
Silicon steel used in laminations mainly reduces-
A)
hysteresis loss
B)
eddy current losses
C)
copper losses
D)
all of the above
In a transformer core, core losses consist of two main components: Hysteresis loss and Eddy current loss.
1. **Hysteresis Loss:** This occurs due to the friction between magnetic domains during the reversal of magnetization in each AC cycle. Silicon steel (a silicon-alloyed steel) is used because it has a narrow hysteresis loop (low coercivity and low retentivity), which significantly reduces the energy dissipated as heat during these cycles. Therefore, the addition of silicon specifically targets the reduction of hysteresis loss.
2. **Eddy Current Loss:** These are circulating currents induced in the core material. While silicon increases the electrical resistivity of the steel (which helps), the primary method used to reduce eddy current losses is the use of **laminations** (thin sheets insulated from each other) to break the current paths.
3. **Copper Loss:** These are $I^2R$ losses occurring in the windings, not the core. The material of the core (silicon steel) has no effect on copper losses.
Since the question asks specifically what the use of **Silicon steel** (the material) reduces, the primary answer is hysteresis loss. (Note: While silicon also increases resistivity to help with eddy currents, the 'laminations' part of the core design handles the bulk of eddy current reduction, whereas the 'silicon' part of the material handles the hysteresis reduction).
Ammeter is used for measuring-
A)
Voltage
B)
Current
C)
Resistance
D)
None of the above
An ammeter is a specialized instrument designed specifically to measure the electric current flowing through a circuit. To measure current, an ammeter must be connected in series with the load so that the full current flows through the device. It is designed with very low internal resistance to ensure that it does not significantly affect the current it is intended to measure.
Analysis of other options:
- Option (a) Voltage: Measured using a Voltmeter, which is connected in parallel.
- Option (c) Resistance: Measured using an Ohmmeter.
- Option (d) None of the above: Incorrect, as option (b) is the standard definition.
How many Consumer Category are there in the prevailing Tariff?
A)
6
B)
8
C)
10
D)
12
In the prevailing electricity tariff structure implemented by the Power and Electricity Department, Government of Mizoram, consumers are categorized based on the nature of their electricity usage to ensure equitable billing. The 8 standard consumer categories typically include: 1. Domestic (Residential), 2. Commercial, 3. Industrial, 4. Agricultural, 5. Public Lighting/Street Lighting, 6. Government/Institutional, 7. Temporary Connections, and 8. Special Categories (such as Low Tension/High Tension distinctions or specific subsidized groups). This classification allows the department to apply different tariff rates (slabs) based on the socio-economic impact and energy intensity of the category.
Earth wire or ground wire is made of-
A)
Copper
B)
Aluminium
C)
Iron
D)
Galvanised Steel
In the context of electrical power distribution and transmission (which is the focus of a Junior Engineer exam for a Power Electricity Department), 'Earth wire' or 'Ground wire' typically refers to the overhead shield wire installed at the topmost position of transmission towers to protect phase conductors from lightning strikes. These wires are primarily made of Galvanised Steel (GS) because they require high mechanical tensile strength to span long distances between towers and must be resistant to atmospheric corrosion. While copper is an excellent conductor, it is too expensive and lacks the necessary mechanical strength for overhead spans. Aluminium is used for phase conductors but is not the standard for overhead ground wires due to strength requirements. Plain iron would corrode rapidly. Therefore, Galvanised Steel is the industry standard for overhead earth wires.
Which of the following material is not used for overhead line insulators?
A)
Porcelain
B)
Glass
C)
PVC
D)
Polymer
Overhead line insulators must possess high dielectric strength, high mechanical strength, and the ability to withstand harsh environmental conditions (UV radiation, temperature fluctuations, and pollution) without degrading.
1. Porcelain (Option a): This is the most traditional material used for insulators due to its excellent electrical insulation properties and mechanical strength.
2. Glass (Option b): Toughened glass is widely used for suspension and strain insulators because it is durable and allows for easy visual inspection of defects.
3. Polymer (Option d): Also known as composite insulators, these use a silicone rubber housing over a fiberglass core. They are lightweight, hydrophobic, and resistant to pollution.
4. PVC (Polyvinyl Chloride) (Option c): PVC is a thermoplastic used primarily for cable insulation (low voltage) and conduits. It is not suitable for overhead line insulators because it lacks the necessary mechanical strength to support heavy conductors and degrades rapidly when exposed to sunlight (UV radiation) and outdoor weather, leading to cracking and electrical failure.
Why is copper commonly used for electrical wires?
A)
It’s a poor conductor
B)
It’s an excellent conductor
C)
It’s an insulator
D)
It’s a superconductor
Copper is widely used for electrical wiring because it possesses a very high electrical conductivity, meaning it allows electric current to flow through it with very low resistance. This minimizes energy loss (power dissipation) in the form of heat.
Analysis of options:
- Option (a) is incorrect because a poor conductor would resist current flow, leading to inefficiency and overheating.
- Option (b) is correct as copper is one of the best known electrical conductors available for commercial use.
- Option (c) is incorrect because insulators (like rubber or plastic) block the flow of electricity and are used for coating wires, not for the conductive core.
- Option (d) is incorrect because a superconductor is a material that has zero electrical resistance, typically only at extremely low temperatures (near absolute zero); copper is a conductor, not a superconductor.
Which of the following is a non-technical loss in the power system?
A)
Conductor resistance loss
B)
Insulation breakdown loss
C)
Meter tampering
D)
Core losses
In power distribution, losses are categorized into two main types: Technical Losses and Non-Technical Losses (Commercial Losses).
1. Technical Losses: These are losses inherent to the physical properties of the electrical system. They occur due to the laws of physics (e.g., Ohm's Law).
- Option (a) Conductor resistance loss (I²R loss) is a technical loss.
- Option (b) Insulation breakdown loss (leakage current) is a technical loss.
- Option (d) Core losses (hysteresis and eddy current losses in transformers) are technical losses.
2. Non-Technical Losses: These are losses that do not involve the physical dissipation of energy as heat but rather the failure to account for or bill the energy consumed. These are often referred to as 'commercial losses'.
- Option (c) Meter tampering involves illegally manipulating the energy meter to record less electricity than is actually consumed. This is a human-induced, non-physical loss of revenue for the utility company, making it a classic example of a non-technical loss. Other examples include theft of electricity (hooking) and billing errors.
Transformer cores are built up from laminations rather than from solid metal so that:
A)
oil penetrates thecore more easily
B)
eddy current loss is reduced
C)
less insulation is required for the windings
D)
turn ratio is higher than voltage ratio
Step-by-step reasoning:
1. **Understanding Eddy Currents**: When a transformer core is subjected to an alternating magnetic field, circulating currents are induced within the conductive bulk of the core material. These are known as eddy currents.
2. **The Problem with Solid Cores**: In a solid metal core, there is a large, uninterrupted path for these currents to flow. Since the core has some electrical resistance, these currents generate heat (I²R loss), leading to significant energy waste and potential overheating.
3. **The Solution (Laminations)**: By constructing the core from thin sheets (laminations) coated with an insulating layer (like varnish), the path for the eddy currents is physically interrupted.
4. **Effect of Laminations**: The resistance to the flow of eddy currents is greatly increased because the current is forced to flow in very narrow paths within each thin lamination. Since the magnitude of the eddy current is proportional to the square of the thickness of the lamination, reducing the thickness drastically reduces the total eddy current loss.
Analysis of other options:
- Option (a) is incorrect because oil penetration is related to cooling and insulation, not the structural choice of laminations over solid metal.
- Option (c) is incorrect because the insulation of windings is independent of whether the core is laminated or solid.
- Option (d) is incorrect because the turn ratio and voltage ratio are determined by the number of turns in the primary and secondary windings, not the physical construction of the core material.
The All day (energy) efficiency is the term related to:
A)
Power Transformer
B)
Distribution transformer
C)
Current transformer
D)
Voltage transformer
The 'All-day efficiency' is a specific metric used for distribution transformers because they are not loaded at full capacity throughout the 24-hour period. Unlike power transformers, which typically operate at a nearly constant load, distribution transformers experience fluctuating loads (peaking during the evening and dropping significantly at night).
1. **Definition**: All-day efficiency is the ratio of the total energy delivered to the load over 24 hours to the total energy supplied to the transformer (including core losses and copper losses) over the same period.
2. **Why Distribution Transformers?**: Since core losses (iron losses) occur as long as the transformer is energized, regardless of the load, calculating efficiency based on a single point in time is misleading. All-day efficiency accounts for these constant losses over a full day of varying load.
3. **Why not others?**:
- **Power Transformers (a)**: These are designed for maximum efficiency at full load or near full load because they operate consistently.
- **Current (c) and Voltage (d) Transformers**: These are instrument transformers used for metering and protection; their performance is measured by accuracy classes and ratio errors, not by energy efficiency over a 24-hour cycle.
An analog ammeter is:
A)
a recording instrument
B)
a controlling instrument
C)
an absolute instrument
D)
an indicating instrument
Electrical measuring instruments are classified based on their function into three main categories: 1. Indicating Instruments: These provide a visual representation of the magnitude of the quantity being measured at that specific instant using a pointer and a scale (e.g., an analog ammeter, voltmeter). 2. Recording Instruments: These keep a continuous record of the variations of the quantity over a period of time (e.g., a chart recorder). 3. Integrating Instruments: These measure the total amount of energy or quantity over a period (e.g., an energy meter). An analog ammeter uses a pointer to indicate the current flowing through a circuit, making it an indicating instrument. Option (b) 'controlling instrument' refers to a component within the meter (like a spring) rather than the type of meter itself, and option (c) 'absolute instrument' refers to instruments that give the magnitude of the quantity without calibration against another instrument (e.g., a tangent galvanometer), which is not the general classification for a standard analog ammeter.