← JE Departmental
Past Questions — 2025
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.
Temporary connection shall be granted for a maximum period of _______ at a time.
A)
30 days
B)
3 months
C)
6 months
D)
45 days
According to the JERC (Joint Electricity Regulatory Commission) regulations governing the supply code of practice for the North Eastern states (including Mizoram), a temporary connection is provided for specific short-term requirements such as construction work, festivals, or exhibitions. The standard regulatory provision specifies that a temporary connection shall be granted for a maximum period of 6 months at a time. If the requirement extends beyond this period, the consumer must apply for an extension or convert the connection into a permanent one, subject to the approval of the licensee/department.
Voltage drop in a distributor is primarily caused by:
A)
High consumer power factor
B)
Increased frequency
C)
Resistance of the conductor and current flow
D)
Poor generator efficiency
Voltage drop in an electrical distributor is governed by Ohm's Law (V = I * Z), where V is the voltage drop, I is the current flowing through the conductor, and Z is the impedance of the conductor. Impedance consists of resistance (R) and reactance (X). Therefore, the primary causes of voltage drop are the resistance of the conductor material and the amount of current being drawn by the load.
Analysis of other options:
- Option (a): A high power factor actually reduces the current required to deliver the same amount of real power, which typically decreases the voltage drop rather than causing it.
- Option (b): Increased frequency primarily affects the inductive reactance (X_L = 2πfL), but it is not the primary cause of voltage drop in standard distribution systems compared to the fundamental R and I components.
- Option (d): Generator efficiency relates to the energy conversion process at the source and does not directly cause the voltage drop that occurs along the length of the distribution lines.
Ohmmeter is used for measuring:
A)
Voltage
B)
Current
C)
Resistance
D)
None of the above
An Ohmmeter is a specific electrical instrument designed to measure electrical resistance, which is the opposition to the flow of electric current in a circuit. It works by applying a known voltage to the component being tested and measuring the resulting current, then using Ohm's Law (R = V/I) to determine the resistance.
Why other options are incorrect:
- Option (a) Voltage is measured using a Voltmeter.
- Option (b) Current is measured using an Ammeter.
- Option (d) is incorrect because a valid answer exists in option (c).
Which of the following is not an advantage of hydroelectric power plant?
A)
no fuel requirement
B)
low running cost
C)
continuous power source
D)
no standby losses
To determine which option is NOT an advantage of a hydroelectric power plant, let's analyze each point:
1. **No fuel requirement (Option a):** This is a major advantage. Hydroelectric plants use the natural flow of water (gravity), meaning there is no need to purchase, transport, or burn fossil fuels.
2. **Low running cost (Option b):** This is an advantage. Once the initial high capital cost of dam construction is recovered, the operational and maintenance costs are very low compared to thermal plants.
3. **Continuous power source (Option c):** This is NOT a guaranteed advantage. Hydroelectric power is heavily dependent on seasonal rainfall and water availability in the reservoir. During droughts or dry seasons, power generation can drop significantly. Therefore, it is often considered an intermittent or seasonal source rather than a guaranteed 'continuous' source in the same way a base-load nuclear plant might be.
4. **No standby losses (Option d):** This is generally considered an advantage. Unlike thermal plants that must keep boilers hot (standby losses) to respond to load changes, a hydro turbine can be started and synchronized to the grid very quickly with minimal energy waste.
Therefore, 'continuous power source' is the correct answer as it is a limitation/disadvantage rather than an advantage.
As per JERC(M&M) Electricity Supply Code Regulation 2013, LT supply shall be given upto a contracted load of _______ kW.
A)
10
B)
30
C)
50
D)
65
According to the JERC (Joint Electricity Regulatory Commission) Electricity Supply Code Regulations for the North Eastern States (which includes Mizoram), the categorization of supply voltage is based on the contracted load. Low Tension (LT) supply is typically provided to consumers with a contracted load up to 50 kW. For loads exceeding 50 kW, the consumer is generally required to be supplied at High Tension (HT) voltage levels (such as 11 kV) to ensure grid stability and reduce distribution losses. Therefore, the limit for LT supply is 50 kW.
Meggar is an instrument uses for:
A)
measuring current
B)
measuring voltage
C)
measuring power
D)
testing insulation
A Megger (short for Mega-ohmmeter) is a specialized instrument used to measure very high electrical resistance, typically in the range of mega-ohms (MΩ). Its primary application is to test the insulation resistance of electrical equipment such as cables, motor windings, and transformers to ensure that the insulation has not degraded and that there is no leakage current. It works by applying a high DC voltage to the insulation and measuring the resulting small leakage current to calculate resistance using Ohm's Law (R = V/I).
Why other options are incorrect:
- Option (a): Current is measured using an Ammeter.
- Option (b): Voltage is measured using a Voltmeter.
- Option (c): Power is measured using a Wattmeter.
Copper loss in the transformer can be determined by:
A)
Short Circuit Test
B)
Open Circuit Test
C)
Both (a) and (b)
D)
None of the above
In a transformer, losses are primarily categorized into Core (Iron) losses and Copper losses.
1. **Copper Loss**: This is the $I^2R$ loss occurring in the primary and secondary windings due to the resistance of the conductors. To measure this, a **Short Circuit (SC) Test** is performed. In this test, the low-voltage side is short-circuited, and a reduced voltage is applied to the high-voltage side until the rated current flows. Since the applied voltage is very low (typically 2-12% of rated voltage), the magnetic flux in the core is minimal, making the core losses negligible. Therefore, the power measured by the wattmeter during this test represents almost exclusively the copper losses.
2. **Core Loss (Iron Loss)**: This consists of hysteresis and eddy current losses. These are determined by the **Open Circuit (OC) Test**. In this test, the transformer is operated at rated voltage and frequency with one winding left open. Since the current drawn is very small (no-load current), the copper losses are negligible, and the wattmeter reading represents the core losses.
Therefore, option (a) is correct, and option (b) is incorrect because it measures iron losses.
A small, continuous current to keep a fully charged battery healthy is known as:
A)
Fast charging
B)
Equalizing charge
C)
Trickle charging
D)
Boost charging
Step-by-step reasoning: 1. The question asks for the term describing a small, continuous current used to maintain a fully charged battery. 2. Trickle charging is a method of charging a battery with a very low current (trickle) that is just enough to compensate for the battery's self-discharge rate, thereby keeping it fully charged and 'healthy' over long periods. 3. Analyzing other options: (a) Fast charging involves high currents to charge a battery quickly, which can cause overheating if applied to a full battery. (b) Equalizing charge is a deliberate overcharge used to balance the voltage of individual cells in a battery bank. (d) Boost charging is a high-current charge used to quickly bring a battery up to a certain state of charge or to overcome a low-voltage state. Therefore, option (c) is the only correct definition.
As per JERC(M&M) Electricity Supply Code Regulation 2013, timeline for releasing new electricity connection in urban areas is:
A)
7 days
B)
2 weeks
C)
30 days
D)
None of the above
According to the Joint Electricity Regulatory Commission (JERC) for the States of Goa and Mizoram (M&M) Electricity Supply Code Regulations, there are specific timelines mandated for the release of new electricity connections to ensure consumer protection and service efficiency. For urban areas, the standard timeline prescribed for the release of a new connection (from the date of application and payment of necessary fees) is 30 days.
- Option (a) 7 days and (b) 2 weeks are too short and do not align with the regulatory framework for standard urban installations which involve site inspection, feasibility checks, and infrastructure deployment.
- Option (c) 30 days is the regulatory benchmark set by JERC for urban areas to ensure the Distribution Licensee provides the service within a reasonable timeframe.
Which type of turbine is commonly used for high head hydroelectric power plants?
A)
Kaplan turbine
B)
Francis turbine
C)
Pelton wheel
D)
Propeller turbine
Hydroelectric turbines are classified based on the 'head' (the vertical distance the water falls) and the flow rate.
1. Pelton Wheel (Impulse Turbine): These are specifically designed for high-head applications (typically above 300 meters). They use high-velocity jets of water that strike buckets on the wheel, converting the kinetic energy of the water into mechanical energy.
2. Francis Turbine (Reaction Turbine): These are versatile and commonly used for medium-head applications (typically 30 to 300 meters).
3. Kaplan and Propeller Turbines (Reaction Turbines): These are designed for low-head applications (typically below 30 meters) where the flow rate is high.
Since the question asks for the turbine used for 'high head' plants, the Pelton wheel is the correct choice.
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.
Five capacitors each of 2µF are connected in series, the equivalent capacitance of the system will be:
A)
2/5µF
B)
10µF
C)
10µF
D)
1µF
Step-by-step calculation:
1. **Identify the given values**: We have five capacitors, each with a capacitance (C) of 2µF.
2. **Identify the connection type**: The capacitors are connected in series.
3. **Formula for Series Capacitance**: For capacitors connected in series, the reciprocal of the equivalent capacitance (C_eq) is the sum of the reciprocals of the individual capacitances:
1/C_eq = 1/C1 + 1/C2 + 1/C3 + 1/C4 + 1/C5
4. **Substitute the values**: Since all capacitors are identical (C = 2µF):
1/C_eq = 1/2 + 1/2 + 1/2 + 1/2 + 1/2
1/C_eq = 5/2
5. **Solve for C_eq**: Invert the fraction to find the equivalent capacitance:
C_eq = 2/5 µF (which is equal to 0.4 µF).
Analysis of options:
- Option (a) 2/5µF is the mathematically correct result.
- Option (b) and (c) 10µF would be the result if the capacitors were connected in parallel (C_eq = 5 * 2µF).
- Option (d) 1µF is incorrect.
A Maxwell Bridge is used for measuring:
A)
Inductance
B)
Capacitance
C)
Frequency
D)
Dissipation Factor
A Maxwell Bridge (specifically the Maxwell-Wien Bridge) is an AC bridge circuit used to measure an unknown inductance in terms of a known capacitance. It works by balancing the inductive reactance of the unknown coil against the capacitive reactance of a standard capacitor.
1. Why 'Inductance' is correct: The bridge is designed to measure the inductance (L) and the quality factor (Q) of a coil. It is particularly useful for measuring medium-Q coils (Q between 1 and 10).
2. Why 'Capacitance' is incorrect: While a capacitor is used in the bridge as a reference, bridges like the Schering Bridge are specifically used for measuring capacitance.
3. Why 'Frequency' is incorrect: Frequency is typically measured using frequency meters or counters, not via a Maxwell Bridge balance method.
4. Why 'Dissipation Factor' is incorrect: While the bridge can determine the loss (resistance) of the coil, the Dissipation Factor is a characteristic primarily associated with capacitors and is measured using a Schering Bridge.
Four resistors each of 5 ohm are connected in Parallel, the equivalent resistance of the system will be:
A)
4/5 Ohm
B)
10 Ohm
C)
25 Ohm
D)
5/4 Ohm
Step-by-step calculation:
1. Given: Four resistors (n = 4), each with a resistance (R) of 5 ohms.
2. The resistors are connected in parallel. The formula for the equivalent resistance (Req) of resistors in parallel is: 1/Req = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn.
3. Since all resistors are identical (R = 5 Ω), the formula simplifies to: 1/Req = n/R.
4. Substituting the values: 1/Req = 4 / 5.
5. To find Req, take the reciprocal of both sides: Req = 5 / 4 Ohm (or 1.25 Ohm).
Analysis of options:
- Option (a) 4/5 Ohm: This is the value of 1/Req, not Req.
- Option (b) 10 Ohm: Incorrect calculation.
- Option (c) 25 Ohm: Incorrect calculation.
- Option (d) 5/4 Ohm: Correct result of the parallel resistance formula.
Which of the following does not change in a transformer?
A)
Current
B)
Voltage
C)
Frequency
D)
All of the above
A transformer is a static electrical device that transfers electrical energy between two or more circuits through electromagnetic induction.
1. **Voltage (Option b):** The primary purpose of a transformer is to change the voltage level. Depending on the turns ratio (Np/Ns), it can either increase the voltage (Step-up transformer) or decrease it (Step-down transformer).
2. **Current (Option a):** According to the principle of conservation of energy (assuming an ideal transformer where Power In = Power Out), P = V × I. Since the voltage changes, the current must change inversely to maintain the power balance. Therefore, current changes in a transformer.
3. **Frequency (Option c):** A transformer operates on the principle of Faraday's Law of Electromagnetic Induction. The alternating magnetic flux is created by the alternating current of the source. The rate of change of this flux (which determines the frequency) remains constant from the primary winding to the secondary winding. There is no mechanism in a transformer to alter the number of cycles per second of the AC waveform.
Therefore, the frequency remains constant regardless of whether the transformer is step-up or step-down.
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. **Why Steel Core?**: 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 stresses like wind and ice loading without excessive sagging or breaking. Steel has a much higher tensile strength than aluminium; therefore, a steel core is placed at the center to provide the necessary mechanical support.
2. **Analysis of other options**:
- **Option a (Skin Effect)**: Skin effect is a phenomenon where current flows primarily on the outer surface of a conductor. A steel core does not compensate for this; in fact, the current flows primarily in the outer aluminium strands anyway.
- **Option b (Proximity Effect)**: Proximity effect is caused by the magnetic fields of adjacent conductors. A steel core does not neutralize this effect.
- **Option c (Line Inductance)**: Adding a steel core does not significantly reduce line inductance. Inductance is primarily determined by the spacing between conductors and the geometric mean radius (GMR) of the conductor.
Earthing or grounding is used:
A)
to provide a high resistance path to the ground
B)
to provide a low resistance path to the ground
C)
to provide flow of positive, negative and zero sequence currents
D)
none of the above
Earthing (or grounding) is a safety measure used in electrical installations to protect personnel and equipment. The primary objective is to ensure that in the event of an insulation failure or a fault (such as a short circuit to the chassis), the fault current has a path of least resistance to the earth.
1. Why Option (b) is correct: By providing a 'low resistance path', the fault current is encouraged to flow through the earthing conductor rather than through a human body (which has high resistance). This high current flow quickly triggers protective devices like circuit breakers or fuses, isolating the faulty circuit and preventing electric shocks or fires.
2. Why Option (a) is incorrect: A high resistance path would impede the flow of fault current, preventing protective devices from tripping and potentially leaving the equipment chassis energized (live), which is extremely dangerous.
3. Why Option (c) is incorrect: While zero sequence currents do flow during ground faults, the primary purpose of earthing is not to 'provide flow' for all sequence components (positive, negative, and zero) as a general operational goal, but specifically to provide a safe return path for fault currents to trigger protection.
Which component of a hydroelectric power plant converts the kinetic energy of water into mechanical energy?
A)
Generator
B)
Turbine
C)
Transformer
D)
Dam
In a hydroelectric power plant, the energy conversion process happens in stages:
1. The Dam stores water, creating potential energy.
2. As water flows through the penstock, potential energy is converted into kinetic energy.
3. The Turbine is the specific component where the flowing water (kinetic energy) strikes the blades, causing the shaft to rotate, thereby converting kinetic energy into mechanical energy.
4. The Generator then converts this mechanical energy into electrical energy.
Why other options are incorrect:
- a. Generator: Converts mechanical energy into electrical energy.
- c. Transformer: Changes the voltage level of the electricity produced; it does not convert kinetic energy.
- d. Dam: Stores water to create a head (potential energy); it does not perform the conversion to mechanical energy.
Connecting cells in parallel increases the:
A)
Voltage
B)
Ampere-hour capacity (Ah)
C)
Internal resistance
D)
Charging time
In battery configurations, cells can be connected in series or parallel.
1. Parallel Connection: When cells are connected in parallel (positive terminal to positive, negative to negative), the total voltage remains the same as a single cell, but the total current capacity (Ampere-hours) is the sum of the capacities of all individual cells. For example, two 2Ah cells in parallel result in a 4Ah capacity. Therefore, option (b) is correct.
2. Why other options are incorrect:
- (a) Voltage: Voltage increases when cells are connected in series, not parallel.
- (c) Internal resistance: Connecting cells in parallel actually decreases the total internal resistance (similar to resistors in parallel), which allows for higher current discharge.
- (d) Charging time: While the total energy increases, the charging time depends on the charger's current. However, the primary electrical characteristic increased by the parallel configuration is the capacity (Ah).
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. Several factors influence the 'Critical Disruptive Voltage' (the voltage at which corona begins):
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 voltage gradient at the surface, thereby reducing the likelihood of corona. This is why bundled conductors are used in high-voltage lines.
2. **Shape of the conductor (Option b):** The curvature of the conductor affects the concentration of the electric field. Sharp edges or non-circular shapes create points of high electric field intensity, which lowers the critical disruptive voltage and promotes corona.
3. **Surface condition of the conductor (Option c):** A smooth conductor has a higher critical disruptive voltage. Conversely, a rough, scratched, or dirty surface creates local points of high stress (irregularities), which facilitate the ionization of air and increase corona loss.
Since all three factors—size, shape, and surface condition—directly impact the onset and intensity of corona, the correct answer is 'All of the above'.
Which conductor is commonly used in 11kV line?
A)
Weasel
B)
Gnat
C)
Panther
D)
Grasshopper
In electrical transmission and distribution, ACSR (Aluminum Conductor Steel Reinforced) conductors are named using a standardized nomenclature (often based on animal names) to denote their size and capacity. For 11kV distribution lines, which are typically used for secondary transmission or primary distribution, a conductor with a relatively small cross-sectional area is required. The 'Weasel' conductor is a standard small-sized ACSR conductor commonly utilized for 11kV lines due to its appropriate current-carrying capacity and mechanical strength for short-span distribution. 'Panther' and 'Grasshopper' are typically larger conductors used for higher voltage transmission lines (like 33kV, 66kV, or 132kV) where higher current capacity and greater mechanical strength are necessary to support longer spans.
Which of the following distribution systems is more reliable?
A)
Radial system
B)
Tree system
C)
Ring main system
D)
All are equally reliable
In electrical distribution, reliability refers to the ability of the system to maintain power supply to consumers even during a fault.
1. **Radial System (Option a):** This is the simplest system where power flows from a single source to consumers along a single path. If a fault occurs at any point in the line, all consumers downstream of that fault lose power. Therefore, it has the lowest reliability.
2. **Tree System (Option b):** This is essentially an extension of the radial system with multiple branches. While it covers more area, it still lacks redundancy; a fault on a main branch disconnects all subsequent sub-branches.
3. **Ring Main System (Option c):** In this system, the distribution transformer is fed by a loop (ring) that returns to the source. This provides two paths for the current to reach any given point. If a fault occurs in one section of the ring, that section can be isolated, and power can still be supplied to all consumers from the alternative path. This redundancy significantly increases the reliability of the system.
Therefore, the Ring Main System is the most reliable among the given options.
The frictional loss in a transformer is:
A)
10 %
B)
50 %
C)
0 %
D)
More than 50 %
In a transformer, losses are primarily categorized into Core losses (Hysteresis and Eddy current losses) and Copper losses (I²R losses). Unlike rotating electrical machines (such as motors or generators), a transformer is a static device with no moving parts. Frictional losses occur due to the mechanical friction of bearings and rotating components, and windage losses occur due to air resistance against a rotating rotor. Since a transformer has no rotating parts, the frictional and windage losses are zero. Therefore, the frictional loss in a transformer is 0%.
The SI unit of Power is:
A)
Hertz
B)
Watt
C)
Ampere
D)
Candela
Power is defined as the rate at which work is done or energy is transferred over time. In the International System of Units (SI), the unit of power is the Watt (W), named after James Watt. One Watt is defined as one Joule per second (1 W = 1 J/s).
Analysis of other options:
- a. Hertz (Hz): The SI unit of frequency (cycles per second).
- c. Ampere (A): The SI unit of electric current.
- d. Candela (cd): The SI unit of luminous intensity.