3000 Most Asked Transmission & Distribution Objection with Full Explanations Part-1

MCQ 1. Electrical power can be transmitted through which systems?

A. Overhead system
B. Underground system
C. Both overhead and underground systems
D. Neither system

📖 View Answer and Detailed Explanation

✅ Correct Answer: C. Both overhead and underground systems

💡 Detailed Explanation

Electrical power can be transmitted using both overhead lines and underground cables.

Overhead transmission is widely used because it is economical and easier to inspect and maintain. Underground transmission is preferred where space, appearance, safety, or environmental conditions make overhead lines unsuitable.

📌 Remember Points

⚡ Power transmission can be carried out through overhead lines as well as underground cables.

MCQ 2. The conductor connecting a consumer's terminals to a distributor is called:

A. Feeder
B. Service main
C. Transmission line
D. Sub-transmission line

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Service main

💡 Detailed Explanation

A service main is the conductor that connects the distributor to the consumer's installation or terminals.

A feeder generally carries power from a substation to a distribution area, while a distributor has tapping points from which consumers are supplied.

📌 Remember Points

⚡ Feeder → Substation to distribution area

⚡ Distributor → Consumer tapping points

⚡ Service main → Distributor to consumer

MCQ 3. A feeder is primarily designed on the basis of:

A. Voltage drop
B. Current-carrying capacity
C. Frequency
D. Power factor

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Current-carrying capacity

💡 Detailed Explanation

A feeder normally carries current from a substation to a distribution area without intermediate consumer tapping. Therefore, its cross-sectional area is mainly selected according to its current-carrying capacity.

📌 Remember Points

⚡ Feeder → Mainly designed based on current-carrying capacity.

⚡ Distributor → Mainly designed based on permissible voltage drop.

MCQ 4. A distributor is primarily designed considering:

A. Current-carrying capacity only
B. Voltage drop
C. Frequency
D. Mechanical strength only

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Voltage drop

💡 Detailed Explanation

A distributor supplies power to consumers at different tapping points. Therefore, maintaining the consumer voltage within permissible limits is very important. Hence, the distributor is primarily designed considering voltage drop.

📌 Remember Points

⚡ Feeder → Current-carrying capacity.

⚡ Distributor → Voltage-drop consideration.

MCQ 5. Which conductor generally has no consumer tapping along its length?

A. Distributor
B. Service main
C. Feeder
D. House wiring

📖 View Answer and Detailed Explanation

✅ Correct Answer: C. Feeder

💡 Detailed Explanation

A feeder normally carries power from a generating station, transmission substation, or distribution substation toward the distribution area. It generally has no consumer tapping along its length.

📌 Remember Points

⚡ Feeder → No intermediate consumer tapping normally.

⚡ Distributor → Consumer tapping is provided.

MCQ 6. A distributor is a conductor from which:

A. No tapping is taken
B. Consumer connections are taken
C. Generators are connected
D. Transmission lines originate

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Consumer connections are taken

💡 Detailed Explanation

A distributor is a conductor from which various consumer connections or tapping points are taken. Its voltage drop is therefore an important design consideration.

📌 Remember Points

⚡ Distributor = conductor having consumer tapping points.

MCQ 7. Which distribution system has the lowest initial cost?

A. Radial system
B. Ring-main system
C. Interconnected system
D. Network system

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Radial system

💡 Detailed Explanation

The radial distribution system has a simple arrangement in which power flows from one source toward the loads. Because of its simple construction and lower conductor requirements, its initial cost is generally the lowest.

Its major disadvantage is comparatively poor reliability because a fault can interrupt supply to downstream consumers.

📌 Remember Points

⚡ Radial → Simple and economical, but lower reliability.

MCQ 8. Which distribution system generally provides the highest reliability?

A. Radial system
B. Ring-main system
C. Interconnected system
D. Single-end system

📖 View Answer and Detailed Explanation

✅ Correct Answer: C. Interconnected system

💡 Detailed Explanation

An interconnected distribution system receives power from more than one source. If one source or section becomes unavailable, supply can often be maintained through another path.

📌 Remember Points

⚡ Interconnected system → Highest reliability among the listed systems.

MCQ 9. In a radial distribution system, power normally flows:

A. From several sources to one load
B. From one source along a single path to loads
C. In a closed loop only
D. From loads to the generating station

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. From one source along a single path to loads

💡 Detailed Explanation

In a radial system, each load is normally supplied through a single electrical path from the source. The arrangement resembles the branches of a tree.

📌 Remember Points

⚡ Radial = One source + one main path toward loads.

MCQ 10. The major disadvantage of a radial distribution system is:

A. High reliability
B. Simple construction
C. Poor reliability
D. High installation cost

📖 View Answer and Detailed Explanation

✅ Correct Answer: C. Poor reliability

💡 Detailed Explanation

Since a radial system generally has only one path from the source to a consumer, a fault on that path can interrupt supply to downstream loads. Therefore, its major disadvantage is poor reliability.

📌 Remember Points

⚡ Main advantage → Simple and economical.

⚡ Main disadvantage → Poor continuity/reliability.

MCQ 11. A ring-main distribution system is normally supplied:

A. At one end only
B. At both ends or through a closed loop
C. Without any transformer
D. Only through DC supply

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. At both ends or through a closed loop

💡 Detailed Explanation

A ring-main system forms a closed electrical path around the distribution area. Supply can reach loads from alternative directions, improving continuity compared with a radial arrangement.

📌 Remember Points

⚡ Ring-main → Closed loop and improved continuity of supply.

MCQ 12. The main advantage of a ring-main system over a radial system is:

A. Lower conductor cost in every case
B. Better continuity of supply
C. No voltage drop
D. Zero power loss

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Better continuity of supply

💡 Detailed Explanation

The ring-main arrangement provides an alternative path for power flow. Consequently, a section can often be isolated during a fault while supply is maintained to other sections.

📌 Remember Points

⚡ Ring-main → Better continuity than radial system.

MCQ 13. An interconnected distribution system receives power from:

A. Only one source
B. More than one source
C. No source
D. Only a DC generator

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. More than one source

💡 Detailed Explanation

An interconnected distribution system is supplied from more than one source. This arrangement provides flexibility and improves reliability and continuity of supply.

📌 Remember Points

⚡ Interconnected = Multiple supply sources.

MCQ 14. Which distribution arrangement is particularly suitable where continuity of supply is important?

A. Radial
B. Ring-main
C. Single-wire
D. Dead-end system

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Ring-main

💡 Detailed Explanation

Ring-main systems provide alternative paths for current flow, making them suitable where continuity of supply is important.

📌 Remember Points

⚡ Important continuity → Ring-main system.

MCQ 15. The voltage drop in a distributor is important because:

A. Consumer voltage must remain within permissible limits
B. Frequency increases with distance
C. Conductor resistance becomes zero
D. Transformer losses disappear

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Consumer voltage must remain within permissible limits

💡 Detailed Explanation

Consumers are required to receive voltage within an acceptable range. Excessive voltage drop along a distributor can result in inadequate voltage at distant consumers.

Voltage drop ≈ I R

📌 Remember Points

⚡ Distributor design is strongly influenced by permissible voltage drop.

MCQ 16. Which of the following is normally a part of a distribution system?

A. Feeder
B. Distributor
C. Service main
D. All of the above

📖 View Answer and Detailed Explanation

✅ Correct Answer: D. All of the above

💡 Detailed Explanation

The basic components of a conventional distribution system include the feeder, distributor, and service main.

📌 Remember Points

⚡ Distribution system → Feeder + Distributor + Service main.

MCQ 17. The conductor connecting a substation to the distribution area is generally called:

A. Service main
B. Feeder
C. House wire
D. Neutral wire

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Feeder

💡 Detailed Explanation

A feeder carries electrical power from a substation toward the distribution area. Consumer tapping is normally not provided along its length.

📌 Remember Points

⚡ Substation → Feeder → Distribution area.

MCQ 18. The current in an ideal feeder without intermediate tapping is approximately:

A. Constant along its length
B. Zero at the sending end
C. Zero at the receiving end
D. Continuously increasing toward the source

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Constant along its length

💡 Detailed Explanation

In an ideal feeder with no intermediate consumer tapping, the same load current flows through the entire feeder. Hence its current is approximately constant along its length.

📌 Remember Points

⚡ Feeder current → Approximately constant when there are no intermediate taps.

MCQ 19. The current in a distributor generally:

A. Remains constant everywhere
B. Varies along its length
C. Is always zero
D. Is independent of consumer loading

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Varies along its length

💡 Detailed Explanation

Since consumer loads are connected at different points of a distributor, current changes as we move along its length. Current generally decreases toward the end as different loads are tapped off.

📌 Remember Points

⚡ Distributor current → Varies along the length because of consumer tapping.

MCQ 20. The service main supplies power from:

A. Generator to transformer
B. Distributor to consumer
C. Transformer to transmission tower
D. Substation to substation

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Distributor to consumer

💡 Detailed Explanation

The service main is the final connection between the distributor and the consumer installation.

📌 Remember Points

⚡ Service main → Distributor to consumer.

MCQ 21. Which system is generally preferred in densely populated urban areas where appearance and safety are important?

A. Overhead system
B. Underground system
C. Bare-wire system
D. Open conductor system

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Underground system

💡 Detailed Explanation

Underground distribution is generally preferred in densely populated areas because it improves appearance, reduces visual obstruction and can provide greater protection against accidental contact with live conductors.

📌 Remember Points

⚡ Urban areas → Underground distribution is often preferred.

MCQ 22. One major disadvantage of an underground distribution system is:

A. Poor appearance
B. High installation cost
C. Low safety
D. Large number of visible conductors

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. High installation cost

💡 Detailed Explanation

Underground cables require excavation, ducts or cable trenches and specialized installation. Therefore, their initial installation cost is generally higher than that of overhead lines.

📌 Remember Points

⚡ Underground → Better appearance but higher initial cost.

MCQ 23. An important advantage of an underground distribution system is:

A. Easy fault location
B. Low initial cost
C. Better appearance
D. Easy installation everywhere

📖 View Answer and Detailed Explanation

✅ Correct Answer: C. Better appearance

💡 Detailed Explanation

Underground cables are not exposed on poles and towers. Consequently, there are fewer visible conductors and supporting structures, giving the installation a cleaner appearance.

📌 Remember Points

⚡ Major visible advantage of underground system → Better appearance.

MCQ 24. Which distribution system is generally easier to extend at relatively low initial cost?

A. Overhead system
B. Underground system
C. Cable tunnel system
D. Submarine cable system

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Overhead system

💡 Detailed Explanation

Overhead systems can generally be extended more easily by adding poles, towers, conductors and associated equipment. They also normally have lower initial installation costs than underground systems.

📌 Remember Points

⚡ Overhead → Economical and comparatively easy to extend.

MCQ 25. The principal disadvantage of overhead lines in cities is:

A. No possibility of faults
B. Requirement of large right-of-way and visual obstruction
C. Zero maintenance
D. Complete immunity from weather

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Requirement of large right-of-way and visual obstruction

💡 Detailed Explanation

Overhead lines require poles or towers and adequate clearances. In densely populated cities, obtaining the required right-of-way can be difficult, and the visible structures can also create visual obstruction.

📌 Remember Points

⚡ Urban overhead-line problem → Right-of-way and visual obstruction.

MCQ 26. Which material is most commonly used for overhead transmission conductors?

A. Aluminium and its alloys
B. Tungsten
C. Lead
D. Cast iron

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Aluminium and its alloys

💡 Detailed Explanation

Aluminium and its alloys are widely used because aluminium provides good electrical conductivity while having relatively low weight. This gives favourable electrical and mechanical characteristics for overhead lines.

📌 Remember Points

⚡ Aluminium → Light weight + good conductivity.

MCQ 27. ACSR stands for:

A. Aluminium Conductor Steel Reinforced
B. Aluminium Cable Steel Resistance
C. Aluminium Core Steel Rod
D. Alloy Conductor Steel Resistance

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Aluminium Conductor Steel Reinforced

💡 Detailed Explanation

ACSR means Aluminium Conductor Steel Reinforced. It consists of aluminium strands around a steel core.

📌 Remember Points

⚡ ACSR = Aluminium Conductor Steel Reinforced.

MCQ 28. The steel core in an ACSR conductor mainly provides:

A. Electrical insulation
B. Mechanical strength
C. Higher capacitance
D. Thermal insulation

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Mechanical strength

💡 Detailed Explanation

The steel core of an ACSR conductor provides the required mechanical strength to withstand tension and mechanical loads such as wind and the weight of the conductor.

📌 Remember Points

⚡ Aluminium → Main current-carrying material.

⚡ Steel core → Mechanical strength.

MCQ 29. The outer strands of a conventional ACSR conductor are generally made of:

A. Copper
B. Aluminium
C. Steel
D. Lead

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Aluminium

💡 Detailed Explanation

Conventional ACSR consists of a central steel core surrounded by aluminium strands. Aluminium provides most of the electrical conductivity while steel provides mechanical strength.

📌 Remember Points

⚡ ACSR outer strands → Aluminium.

MCQ 30. The steel core of an ACSR conductor is mainly responsible for:

A. Carrying all the current
B. Increasing mechanical strength
C. Providing insulation
D. Reducing conductor diameter to zero

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Increasing mechanical strength

💡 Detailed Explanation

The steel core increases the tensile strength of the ACSR conductor, allowing it to withstand mechanical tension and sag-related requirements.

📌 Remember Points

⚡ Steel core = Mechanical strength.

MCQ 31. Which of the following is an important property of a good overhead-line conductor?

A. High resistivity
B. Low conductivity
C. High tensile strength
D. Very high weight

📖 View Answer and Detailed Explanation

✅ Correct Answer: C. High tensile strength

💡 Detailed Explanation

An overhead conductor must withstand mechanical forces caused by its own weight, wind, temperature changes and other loading conditions. Therefore, high tensile strength is desirable.

📌 Remember Points

⚡ Good conductor → High conductivity + adequate tensile strength.

MCQ 32. A desirable conductor for transmission lines should have:

A. High electrical conductivity and high tensile strength
B. High resistivity and low strength
C. Very high density only
D. Low conductivity only

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. High electrical conductivity and high tensile strength

💡 Detailed Explanation

A transmission conductor should carry current efficiently, requiring high electrical conductivity. It must also withstand mechanical tension, requiring high tensile strength.

📌 Remember Points

⚡ Electrical requirement → High conductivity.

⚡ Mechanical requirement → High tensile strength.

MCQ 33. Copper has traditionally been used as an overhead conductor because of its:

A. High resistivity
B. High conductivity
C. Very low tensile strength
D. Extremely low melting point

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. High conductivity

💡 Detailed Explanation

Copper has high electrical conductivity and therefore offers low resistance for a given cross-sectional area. This is one of the principal reasons for its traditional use as a conductor.

📌 Remember Points

⚡ Copper → Excellent electrical conductivity.

MCQ 34. Aluminium is widely used in transmission lines mainly because it:

A. Has high conductivity-to-weight ratio
B. Has zero resistance
C. Is heavier than copper
D. Is an insulator

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Has high conductivity-to-weight ratio

💡 Detailed Explanation

Aluminium is considerably lighter than copper while still providing good electrical conductivity. Its favourable conductivity-to-weight ratio makes it very useful for overhead transmission lines.

📌 Remember Points

⚡ Aluminium → Light weight + good conductivity.

MCQ 35. Which material is commonly used for the earth wire of an overhead transmission line?

A. Galvanized steel
B. Glass
C. Porcelain
D. Rubber

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Galvanized steel

💡 Detailed Explanation

Galvanized steel is commonly used for overhead earth or shield wires because it offers good mechanical strength and corrosion resistance.

📌 Remember Points

⚡ Earth wire → Protects the line and equipment from lightning-related effects.

MCQ 36. The primary purpose of an earth wire on an overhead transmission line is to:

A. Carry normal load current
B. Protect the line against lightning
C. Increase frequency
D. Reduce conductor resistance

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Protect the line against lightning

💡 Detailed Explanation

An overhead earth wire, also called a shield wire, is installed above the phase conductors to intercept lightning strokes and provide a path toward earth, thereby reducing the likelihood of direct lightning strikes on the phase conductors.

📌 Remember Points

⚡ Earth/shield wire → Lightning protection.

MCQ 37. Which of the following is used to support overhead conductors?

A. Pole
B. Tower
C. Both A and B
D. Transformer oil

📖 View Answer and Detailed Explanation

✅ Correct Answer: C. Both A and B

💡 Detailed Explanation

Overhead conductors may be supported by poles in distribution systems and by towers in transmission systems.

📌 Remember Points

⚡ Distribution → Poles are commonly used.

⚡ Transmission → Towers are commonly used.

MCQ 38. Steel towers are mainly used for:

A. Domestic wiring
B. High-voltage transmission lines
C. Low-voltage house connections
D. Telephone handsets

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. High-voltage transmission lines

💡 Detailed Explanation

Steel lattice towers provide high mechanical strength and are widely used for supporting high-voltage and extra-high-voltage transmission conductors.

📌 Remember Points

⚡ Steel lattice towers → High-voltage transmission.

MCQ 39. RCC poles are commonly used in:

A. Distribution systems
B. Extra-high-voltage transmission only
C. Underground cables
D. Transformer windings

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Distribution systems

💡 Detailed Explanation

RCC (Reinforced Cement Concrete) poles are commonly used in electrical distribution systems because they provide adequate mechanical strength and durability.

📌 Remember Points

⚡ RCC poles → Commonly used in distribution networks.

MCQ 40. The main function of a pole or tower is to:

A. Generate electricity
B. Support conductors and maintain required clearances
C. Increase system frequency
D. Convert AC into DC

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Support conductors and maintain required clearances

💡 Detailed Explanation

Poles and towers mechanically support the conductors and maintain the required electrical clearances from ground, structures and other conductors.

📌 Remember Points

⚡ Pole/Tower → Mechanical support + required electrical clearance.

MCQ 41. The device used to electrically insulate a conductor from its support is called:

A. Relay
B. Insulator
C. Fuse
D. Isolator

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Insulator

💡 Detailed Explanation

An insulator electrically separates the conductor from its supporting structure while mechanically supporting the conductor.

📌 Remember Points

⚡ Insulator → Electrical insulation + mechanical support.

MCQ 42. Which property is essential for an overhead-line insulator?

A. High mechanical strength
B. Low insulation resistance
C. High leakage current
D. High conductivity

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. High mechanical strength

💡 Detailed Explanation

An overhead-line insulator must support the conductor mechanically while withstanding mechanical forces caused by conductor weight, wind and other loads. Hence, high mechanical strength is essential.

📌 Remember Points

⚡ Insulator → High mechanical strength is essential.

MCQ 43. An insulator should have:

A. High dielectric strength
B. Low mechanical strength
C. High conductivity
D. Low insulation resistance

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. High dielectric strength

💡 Detailed Explanation

Dielectric strength is the ability of an insulating material to withstand electric stress without electrical breakdown. An overhead-line insulator therefore needs high dielectric strength.

📌 Remember Points

⚡ Good insulator → High dielectric strength + high insulation resistance.

MCQ 44. Which insulator is commonly used for lower-voltage distribution lines?

A. Shackle insulator
B. Suspension insulator only
C. Disc insulator only
D. Bushing insulator

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Shackle insulator

💡 Detailed Explanation

Shackle insulators are commonly used in low-voltage and lower-voltage distribution applications, especially at bends and ends of distribution lines.

📌 Remember Points

⚡ Shackle insulator → Commonly associated with lower-voltage distribution.

MCQ 45. Pin insulators are generally mounted:

A. Below the conductor support
B. On cross-arms or supports
C. Inside transformers
D. Inside circuit breakers

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. On cross-arms or supports

💡 Detailed Explanation

Pin insulators are mounted on poles or cross-arms. The conductor is secured to the upper portion of the insulator while the insulator electrically separates the conductor from the supporting structure.

📌 Remember Points

⚡ Pin insulator → Mounted on pole/cross-arm support.

MCQ 46. Suspension insulators are particularly suitable for:

A. High-voltage transmission lines
B. House wiring
C. Low-voltage lamp circuits
D. Battery terminals

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. High-voltage transmission lines

💡 Detailed Explanation

Suspension insulator strings are widely used on high-voltage transmission lines. Additional disc units can be added to increase the insulation level for higher voltages.

📌 Remember Points

⚡ High voltage → Suspension insulator strings.

MCQ 47. Strain insulators are mainly used at:

A. Straight intermediate points only
B. Dead ends and sharp deviations
C. Transformer oil tanks
D. Underground cable joints only

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Dead ends and sharp deviations

💡 Detailed Explanation

At dead ends and locations where the line changes direction significantly, the conductor experiences considerable mechanical tension. Strain insulators are designed to withstand these mechanical forces.

📌 Remember Points

⚡ Strain insulator → Dead ends, corners and locations with high conductor tension.

MCQ 48. A string of suspension insulators consists of:

A. Several insulating units connected mechanically
B. One copper conductor only
C. Several transformers
D. Several circuit breakers

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Several insulating units connected mechanically

💡 Detailed Explanation

A suspension insulator string is formed by connecting multiple individual insulating units mechanically in series. More units can be added when a higher system voltage requires a greater insulation level.

📌 Remember Points

⚡ Suspension string = Multiple insulating units connected mechanically.

MCQ 49. String efficiency is associated with:

A. Suspension insulator strings
B. Transformers only
C. Underground cables only
D. Motors only

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Suspension insulator strings

💡 Detailed Explanation

String efficiency describes how uniformly the voltage is distributed across the individual units of a suspension insulator string. Due to stray capacitance, the voltage distribution is generally not uniform.

String efficiency = $\frac{\text{Total voltage across string}} {\text{Number of units} \times \text{maximum voltage across one unit}} \times 100\%$

📌 Remember Points

⚡ String efficiency → Suspension insulator strings.

⚡ Higher string efficiency → More uniform voltage distribution.

MCQ 50. Ideally, the voltage distribution across the units of a suspension insulator string should be:

A. Highly unequal
B. Uniform
C. Zero
D. Random

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Uniform

💡 Detailed Explanation

Ideally, each unit of a suspension insulator string should withstand an equal share of the total voltage. Therefore, the voltage distribution should be uniform.

In an actual string, stray capacitance causes unequal voltage distribution, with the unit nearest the conductor generally experiencing the highest voltage stress.

📌 Remember Points

⚡ Ideal voltage distribution → Uniform.

⚡ Actual string → Voltage distribution is unequal because of capacitance effects.

MCQ 51. In a practical suspension-insulator string, voltage distribution is:

A. Uniform
B. Unequal
C. Always zero
D. Independent of capacitance

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Unequal

💡 Detailed Explanation

In a practical suspension-insulator string, the voltage does not divide equally among all the insulator discs. Therefore, the voltage distribution is unequal.

The main reason is the presence of stray capacitance between the metal fittings of the insulator string and the tower or earth. These capacitances cause different units of the string to experience different voltage stresses.

The insulator unit nearest the conductor generally experiences the highest voltage stress.

$$V_{\text{string}}=\sum V_i$$

Since the individual voltages $V_1,V_2,\ldots,V_n$ are not equal, the voltage distribution is non-uniform.

📌 Remember Points

Practical insulator-string voltage distribution is unequal.
Stray capacitance is the major reason.
The conductor-side unit has the highest electrical stress.

MCQ 52. The unequal voltage distribution across a suspension-insulator string is mainly caused by:

A. Stray capacitance
B. Conductor resistance only
C. Transformer oil
D. Load frequency only

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Stray capacitance

💡 Detailed Explanation

A suspension-insulator string consists of several disc units connected in series. Ideally, the total voltage would divide equally among the discs. However, in practice, each disc has capacitance not only to the adjacent disc but also to the tower and earth.

These additional capacitances are called stray or shunt capacitances. Because of them, the voltage distribution becomes non-uniform.

$$V_{\text{total}}=V_1+V_2+\cdots+V_n$$

The unit nearest the conductor generally receives the largest portion of the total voltage.

📌 Remember Points

Unequal distribution → mainly due to stray capacitance.
Highest stress → conductor-side disc.
String efficiency measures how uniformly voltage is distributed.

MCQ 53. The unit nearest the conductor in a suspension-insulator string generally experiences:

A. The lowest voltage stress
B. The highest voltage stress
C. Zero voltage
D. No electric field

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. The highest voltage stress

💡 Detailed Explanation

The voltage distribution across a practical suspension-insulator string is non-uniform because of stray capacitances.

The insulator disc nearest the high-voltage conductor is subjected to the greatest voltage. Hence, it experiences the highest electrical stress.

This is important because excessive electrical stress on one disc can cause premature insulation failure.

$$\text{String efficiency} = \frac{V_{\text{string}}} {nV_{\max}}\times100\%$$

A perfectly uniform distribution would give a string efficiency of $100\%$.

📌 Remember Points

Conductor-side disc → highest voltage stress.
Tower-side disc → comparatively lower voltage stress.
Non-uniform distribution reduces string efficiency.

MCQ 54. String efficiency can be improved by:

A. Increasing unequal voltage distribution
B. Using grading arrangements
C. Removing all insulators
D. Reducing insulation strength

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Using grading arrangements

💡 Detailed Explanation

String efficiency is improved when the voltage is distributed more uniformly among the insulator units.

Grading arrangements such as capacitance grading and guard rings are used to control the electric field and improve voltage distribution.

$$\eta_s= \frac{V_{\text{string}}} {nV_{\max}}\times100\%$$

As the maximum voltage across any individual disc is reduced, string efficiency increases.

📌 Remember Points

Better voltage distribution → higher string efficiency.
Grading arrangements reduce voltage concentration.
Guard rings are commonly used for voltage grading.

MCQ 55. Corona is associated primarily with:

A. Ionization of air around a high-voltage conductor
B. Mechanical failure of a pole
C. Transformer oil circulation
D. Battery charging

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Ionization of air around a high-voltage conductor

💡 Detailed Explanation

Corona is a phenomenon associated with ionization of the air surrounding a high-voltage conductor when the electric field near the conductor becomes sufficiently high.

The strong electric field accelerates free electrons in the surrounding air. These electrons collide with air molecules and may produce further ionization.

$$E=\frac{V}{d}$$

When the electric field reaches a sufficiently high value, ionization begins and corona may appear.

📌 Remember Points

Corona → electrical discharge around a high-voltage conductor.
It involves ionization of surrounding air.
Corona is important in EHV transmission systems.

MCQ 56. Corona is more likely to occur when:

A. Electric field around the conductor becomes sufficiently high
B. Voltage is exactly zero
C. The conductor is completely buried
D. Frequency is zero

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Electric field around the conductor becomes sufficiently high

💡 Detailed Explanation

Corona starts when the electric field at the conductor surface becomes sufficiently high to ionize the surrounding air.

The electric field is influenced by conductor radius, conductor spacing, voltage and atmospheric conditions.

A smaller conductor radius produces a higher surface electric field for the same operating voltage.

$$E_{\text{surface}}\propto\frac{1}{r}$$

Therefore, increasing the effective conductor radius is an important method of reducing corona.

📌 Remember Points

High surface electric field promotes corona.
Larger effective conductor radius reduces corona tendency.
Bundled conductors increase effective radius.

MCQ 57. Corona can cause:

A. Power loss
B. Radio interference
C. Audible noise
D. All of the above

📖 View Answer and Detailed Explanation

✅ Correct Answer: D. All of the above

💡 Detailed Explanation

Corona is undesirable in high-voltage transmission systems because it produces several effects.

First, energy is consumed in maintaining the ionization process, causing corona power loss. Corona can also generate electromagnetic disturbances that produce radio interference.

In addition, corona can produce a characteristic hissing or crackling sound known as audible noise.

📌 Remember Points

Corona causes power loss.
Corona may cause radio interference.
Corona may produce audible noise.

MCQ 58. Corona is strongly affected by:

A. Conductor size
B. Conductor surface condition
C. Atmospheric conditions
D. All of the above

📖 View Answer and Detailed Explanation

✅ Correct Answer: D. All of the above

💡 Detailed Explanation

Corona depends on several physical and environmental factors.

Conductor size: A larger conductor has a larger effective radius and therefore a lower surface electric field.

Surface condition: Rough, dirty or irregular conductor surfaces can locally increase electric-field intensity.

Atmospheric conditions: Air density, pressure, temperature, rain, fog and humidity can influence corona.

📌 Remember Points

Conductor radius affects corona.
Rough surfaces increase local electric-field concentration.
Weather and atmospheric conditions influence corona.

MCQ 59. Increasing conductor diameter generally:

A. Increases corona tendency
B. Reduces corona tendency
C. Has no effect on corona
D. Eliminates resistance completely

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Reduces corona tendency

💡 Detailed Explanation

Increasing conductor diameter increases the conductor radius. Consequently, the electric-field intensity at the conductor surface decreases.

Therefore, the conductor requires a higher voltage before corona begins.

$$E_{\text{surface}}\propto\frac{1}{r}$$

Hence, a larger radius means a lower surface electric field and a reduced tendency for corona formation.

📌 Remember Points

Diameter ↑ → radius ↑.
Radius ↑ → surface electric field ↓.
Therefore, corona tendency ↓.

MCQ 60. Bundled conductors are used in EHV transmission mainly to:

A. Increase corona effects
B. Reduce corona and improve transmission performance
C. Eliminate mechanical strength
D. Reduce the number of phases to one

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Reduce corona and improve transmission performance

💡 Detailed Explanation

In EHV transmission systems, each phase may consist of two or more subconductors separated by spacers. This arrangement is called a bundled conductor.

The bundle behaves electrically like a conductor having a larger effective radius. This reduces the electric-field intensity at the surface and therefore reduces corona.

$$E_{\text{surface}}\downarrow \quad\Rightarrow\quad \text{Corona tendency}\downarrow$$

Bundling also improves the transmission characteristics of EHV lines.

📌 Remember Points

Bundling is particularly important in EHV transmission.
It increases the effective conductor radius.
Corona and associated losses are reduced.

MCQ 61. A bundle conductor consists of:

A. Several subconductors per phase
B. Several transformers per phase
C. Several poles per phase
D. Several insulators connected electrically in parallel

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Several subconductors per phase

💡 Detailed Explanation

A bundled conductor is formed by using two or more separate subconductors for one electrical phase.

The subconductors are maintained at a fixed spacing using mechanical spacers.

For example, a phase may use two, three or four subconductors depending upon the voltage level and transmission requirements.

📌 Remember Points

Bundle conductor = multiple subconductors belonging to one phase.
Spacers maintain the physical separation.
Used extensively in EHV transmission.

MCQ 62. One major advantage of bundled conductors is:

A. Increased corona loss
B. Reduced corona loss
C. Increased conductor resistance
D. Reduced mechanical stability

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Reduced corona loss

💡 Detailed Explanation

The main electrical advantage of bundled conductors is reduction of corona effects.

Multiple subconductors increase the effective radius of the phase conductor. The surface electric-field intensity therefore decreases.

$$E_{\text{surface}}\propto\frac{1}{r_{\text{effective}}}$$

As $r_{\text{effective}}$ increases, corona loss decreases.

📌 Remember Points

Bundle → effective radius ↑.
Surface electric field ↓.
Corona loss ↓.

MCQ 63. The sag of an overhead conductor is provided mainly to:

A. Maintain mechanical safety and accommodate thermal expansion
B. Eliminate all conductor resistance
C. Increase voltage frequency
D. Stop current flow

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Maintain mechanical safety and accommodate thermal expansion

💡 Detailed Explanation

An overhead conductor cannot be installed perfectly straight between two supports. A certain amount of downward curvature called sag is intentionally provided.

Sag allows the conductor to accommodate changes in length caused by temperature variation. It also prevents excessive mechanical tension from being transferred to towers and supports.

$$\text{Sag}\propto\frac{wL^2}{T}$$

where $w$ is conductor weight per unit length, $L$ is span and $T$ is horizontal tension.

📌 Remember Points

Sag is necessary for mechanical safety.
Temperature rise increases conductor length.
Proper sag prevents excessive tension.

MCQ 64. If the temperature of an overhead conductor increases, its sag generally:

A. Decreases
B. Increases
C. Becomes zero
D. Remains exactly unchanged

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Increases

💡 Detailed Explanation

When the temperature of an overhead conductor rises, the conductor undergoes thermal expansion.

$$\Delta L=\alpha L\Delta T$$

Here, $\alpha$ is the coefficient of thermal expansion. The conductor becomes longer, resulting in greater downward sag.

Therefore:

$$T\uparrow\text{ (temperature)} \Rightarrow L\uparrow \Rightarrow \text{sag}\uparrow$$

📌 Remember Points

Temperature ↑ → conductor length ↑.
Conductor length ↑ → sag generally ↑.
High-temperature conditions must be considered during line design.

MCQ 65. The shape assumed by a uniformly loaded flexible conductor suspended between two supports is approximately:

A. Circle
B. Catenary
C. Rectangle
D. Ellipse

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Catenary

💡 Detailed Explanation

A flexible conductor suspended between two supports under its own uniformly distributed weight assumes a curve called a catenary.

$$y=a\cosh\left(\frac{x}{a}\right)-a$$

For practical transmission-line calculations where sag is small compared with the span, the catenary can be approximated by a parabola.

📌 Remember Points

Exact conductor curve → catenary.
Small-sag approximation → parabolic curve.

MCQ 66. For a conductor with small sag compared with span, the sag curve is often approximated as:

A. Parabolic
B. Circular only
C. Rectangular
D. Hyperbolic only

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Parabolic

💡 Detailed Explanation

The exact curve of a conductor under uniformly distributed weight is a catenary. However, for normal overhead-line calculations, sag is usually small compared with the span.

Under this condition, the catenary can be approximated by a parabola.

$$d=\frac{wL^2}{8T}$$

where $d$ is sag, $w$ is conductor weight per unit length, $L$ is span and $T$ is horizontal tension.

📌 Remember Points

Exact curve → catenary.
Small sag approximation → parabola.
Parabolic sag formula → $d=\frac{wL^2}{8T}$.

MCQ 67. Sag is measured as the:

A. Vertical distance between the support level and the lowest point of conductor
B. Horizontal distance between poles
C. Conductor diameter
D. Tower height

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Vertical distance between the support level and the lowest point of conductor

💡 Detailed Explanation

Sag is the vertical distance between the level of the supports and the lowest point of the conductor.

For a level span, the lowest point occurs approximately at the centre of the span.

$$d=\frac{wL^2}{8T}$$

📌 Remember Points

Sag is a vertical distance.
It is not the span length.
It is usually measured from support level to the lowest conductor point.

MCQ 68. For a level span with uniform loading, the lowest point of the conductor is normally:

A. At the midpoint of the span
B. At one support
C. Outside the span
D. At the top of the tower

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. At the midpoint of the span

💡 Detailed Explanation

When the two supports are at the same level and the conductor loading is uniform, the conductor curve is symmetrical.

Therefore, the lowest point of the conductor occurs at the midpoint of the span.

$$d=\frac{wL^2}{8T}$$

📌 Remember Points

Level supports + uniform loading → lowest point at midpoint.
Unequal support levels require a different sag analysis.

MCQ 69. Increasing conductor tension, while other conditions remain constant, generally:

A. Increases sag
B. Reduces sag
C. Has no effect on sag
D. Eliminates the span

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Reduces sag

💡 Detailed Explanation

For a level span with uniformly distributed conductor weight, parabolic sag is given by:

$$d=\frac{wL^2}{8T}$$

From this equation:

$$d\propto\frac{1}{T}$$

Therefore, when conductor tension $T$ increases, sag decreases.

📌 Remember Points

Tension ↑ → sag ↓.
Tension ↓ → sag ↑.
Very high tension is undesirable because it increases mechanical stress.

MCQ 70. Excessively high conductor tension may:

A. Improve mechanical safety without limit
B. Increase mechanical stress on conductors and supports
C. Reduce all line losses to zero
D. Eliminate wind loading

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Increase mechanical stress on conductors and supports

💡 Detailed Explanation

Although increasing conductor tension reduces sag, excessive tension is mechanically dangerous.

High tension increases the mechanical stress in the conductor and transfers greater forces to towers, poles, clamps and foundations.

$$\sigma=\frac{T}{A}$$

where $\sigma$ is tensile stress, $T$ is tension and $A$ is the conductor cross-sectional area.

Hence, conductor tension must be selected as a compromise between acceptable sag and acceptable mechanical stress.

📌 Remember Points

High tension → low sag.
Excessive tension → high mechanical stress.
Transmission-line design requires an optimum tension.

MCQ 71. The electrical resistance of a conductor increases with:

A. Increase in cross-sectional area
B. Increase in length
C. Decrease in resistivity
D. Decrease in length

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Increase in length

💡 Detailed Explanation

The resistance of a uniform conductor is given by:

$$R=\rho\frac{L}{A}$$

Therefore, resistance is directly proportional to length:

$$R\propto L$$

If the length doubles while all other parameters remain unchanged, the resistance also doubles.

📌 Remember Points

Length ↑ → resistance ↑.
Area ↑ → resistance ↓.
Resistivity depends on material and temperature.

MCQ 72. The resistance of a conductor is given by:

A. $R=\rho A/l$
B. $R=\rho l/A$
C. $R=l/\rho A$
D. $R=A/\rho l$

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. $R=\rho l/A$

💡 Detailed Explanation

The resistance of a uniform conductor is:

$$R=\rho\frac{L}{A}$$

where:

$R$ = resistance in ohms
$\rho$ = resistivity of material
$L$ = conductor length
$A$ = cross-sectional area

Thus, resistance increases with length and decreases with cross-sectional area.

📌 Remember Points

Basic formula: $R=\rho L/A$.
Unit of resistance → ohm $(\Omega)$.
Unit of resistivity → $\Omega\cdot m$.

MCQ 73. In the expression $R=\rho l/A$, $\rho$ represents:

A. Conductance
B. Resistivity
C. Reactance
D. Capacitance

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Resistivity

💡 Detailed Explanation

In the resistance equation:

$$R=\rho\frac{L}{A}$$

$\rho$ represents the resistivity of the conductor material.

Resistivity indicates how strongly a material opposes the flow of electric current.

$$\rho=R\frac{A}{L}$$

📌 Remember Points

$\rho$ → resistivity.
Unit → $\Omega\cdot m$.
Low resistivity materials are good electrical conductors.

MCQ 74. Increasing the cross-sectional area of a conductor:

A. Increases its resistance
B. Decreases its resistance
C. Has no effect on resistance
D. Makes resistance infinite

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Decreases its resistance

💡 Detailed Explanation

From the fundamental resistance equation:

$$R=\rho\frac{L}{A}$$

Resistance is inversely proportional to cross-sectional area:

$$R\propto\frac{1}{A}$$

Therefore, increasing conductor area decreases resistance.

📌 Remember Points

Conductor area ↑ → resistance ↓.
Larger conductors can carry greater current.
Large conductor size also helps reduce $I^2R$ losses.

MCQ 75. The resistance of an overhead conductor generally increases with:

A. Decrease in temperature
B. Increase in temperature
C. Decrease in length
D. Increase in cross-sectional area

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Increase in temperature

💡 Detailed Explanation

For metallic conductors, resistance generally increases with temperature.

$$R_T=R_0[1+\alpha(T-T_0)]$$

Here, $R_0$ is resistance at reference temperature, $\alpha$ is the temperature coefficient of resistance and $T-T_0$ is the temperature rise.

Therefore, when conductor temperature increases, its resistance increases.

📌 Remember Points

Metal conductor: temperature ↑ → resistance ↑.
Higher resistance results in higher $I^2R$ loss for the same current.

MCQ 76. The inductance of a transmission line is mainly associated with:

A. Magnetic flux linkages produced by line current
B. Transformer oil
C. Mechanical sag only
D. Insulator color

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Magnetic flux linkages produced by line current

💡 Detailed Explanation

Current flowing through a transmission-line conductor produces a magnetic field around the conductor. The magnetic flux links with the conductor and produces inductance.

$$L=\frac{\lambda}{I}$$

where $L$ is inductance, $\lambda$ is flux linkage and $I$ is current.

Therefore, transmission-line inductance is fundamentally associated with magnetic flux produced by current.

📌 Remember Points

Current → magnetic field.
Magnetic flux linkages → inductance.
Inductance produces inductive reactance.

MCQ 77. The capacitance of an overhead transmission line is mainly due to:

A. Electric field between conductors and between conductors and ground
B. Mechanical tension
C. Conductor weight
D. Pole resistance only

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Electric field between conductors and between conductors and ground

💡 Detailed Explanation

Whenever there is a potential difference between conductors, an electric field exists between them. This causes the transmission line to possess capacitance.

Capacitance exists between phase conductors and also between conductors and earth.

$$C=\frac{Q}{V}$$

where $C$ is capacitance, $Q$ is stored charge and $V$ is potential difference.

📌 Remember Points

Inductance → magnetic field.
Capacitance → electric field.
Transmission-line capacitance depends strongly on conductor geometry and spacing.

MCQ 78. The principal electrical parameters of a transmission line are:

A. Resistance only
B. Resistance and inductance only
C. Resistance, inductance and capacitance
D. Capacitance only

📖 View Answer and Detailed Explanation

✅ Correct Answer: C. Resistance, inductance and capacitance

💡 Detailed Explanation

A practical transmission line has several electrical characteristics. The three principal parameters emphasized in basic transmission-line analysis are resistance, inductance and capacitance.

$$R=\rho\frac{L}{A}$$

$$X_L=\omega L$$

$$X_C=\frac{1}{\omega C}$$

For a complete distributed-parameter model, shunt conductance $G$ is also considered.

📌 Remember Points

Basic line parameters → $R,L,C$.
Complete distributed model → $R,L,C,G$.
Resistance causes real-power loss.
Inductance and capacitance affect reactive behavior.

MCQ 79. The four commonly considered distributed parameters of a transmission line include:

A. R, L, C and G
B. R, X, P and Q only
C. V, I, P and Q only
D. E, B, H and D only

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. R, L, C and G

💡 Detailed Explanation

A practical transmission line is modeled using four distributed parameters:

R → series resistance per unit length
L → series inductance per unit length
C → shunt capacitance per unit length
G → shunt conductance per unit length

These parameters are distributed continuously along the line rather than being concentrated at a single point.

$$Z'=R+j\omega L$$

$$Y'=G+j\omega C$$

📌 Remember Points

Four distributed parameters → R, L, C, G.
Series parameters → $R,L$.
Shunt parameters → $C,G$.

MCQ 80. In the transmission-line equivalent circuit, $R$ represents:

A. Resistance per unit length
B. Reactance per unit length
C. Capacitance per unit length
D. Conductance only

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Resistance per unit length

💡 Detailed Explanation

In the distributed transmission-line model, $R$ represents the conductor's series resistance per unit length.

$$R=\rho\frac{L}{A}$$

The total resistance over a length $L$ is obtained by multiplying the resistance per unit length by the line length.

📌 Remember Points

$R$ → series resistance.
$L$ → series inductance.
$C$ → shunt capacitance.
$G$ → shunt conductance.

MCQ 81. The primary purpose of transmitting electrical power at high voltage is to:

A. Increase $I^2R$ losses
B. Reduce current for a given power transfer
C. Increase conductor resistance
D. Reduce system frequency

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Reduce current for a given power transfer

💡 Detailed Explanation

For a three-phase transmission system:

$$P=\sqrt{3}V_LI_L\cos\phi$$

Therefore:

$$I_L=\frac{P}{\sqrt{3}V_L\cos\phi}$$

For constant power and power factor, increasing voltage reduces line current.

Since copper loss is:

$$P_{\text{loss}}=3I_L^2R$$

reducing current produces a large reduction in transmission losses.

📌 Remember Points

Transmission voltage ↑ → current ↓ for the same power.
Current ↓ → $I^2R$ loss ↓.
This is the fundamental reason for high-voltage transmission.

MCQ 82. For a fixed transmitted power, increasing transmission voltage generally:

A. Increases current
B. Reduces current
C. Makes current infinite
D. Does not affect current

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Reduces current

💡 Detailed Explanation

For a three-phase system:

$$P=\sqrt{3}V_LI_L\cos\phi$$

Rearranging:

$$I_L=\frac{P}{\sqrt{3}V_L\cos\phi}$$

Thus, for constant $P$ and $\cos\phi$, current is inversely proportional to voltage.

$$I_L\propto\frac{1}{V_L}$$

📌 Remember Points

Voltage ↑ → current ↓.
Voltage doubled → current becomes approximately half.
Lower current means lower copper loss.

MCQ 83. Copper loss in a transmission line is proportional to:

A. $I$
B. $I^2$
C. $1/I$
D. $V^2$ only

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. $I^2$

💡 Detailed Explanation

The power dissipated in the resistance of a conductor is given by:

$$P_{\text{loss}}=I^2R$$

Therefore:

$$P_{\text{loss}}\propto I^2$$

This quadratic relationship is why reducing transmission current is so effective in reducing power loss.

📌 Remember Points

Copper loss → $I^2R$.
Current doubles → loss becomes four times.
Current halves → loss becomes one-fourth.

MCQ 84. The power loss due to resistance in a line is:

A. $IR$
B. $I^2R$
C. $IR^2$
D. $V/R^2$

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. $I^2R$

💡 Detailed Explanation

Electrical power dissipated in a resistor is:

$$P=VI$$

Since $V=IR$:

$$P=I(IR)=I^2R$$

Hence the transmission-line resistive loss is:

$$P_{\text{loss}}=I^2R$$

📌 Remember Points

Basic power formula → $P=VI$.
Resistive loss → $I^2R$.
Transmission losses are strongly affected by current.

MCQ 85. If the transmission voltage is doubled while transmitted power remains constant and power factor is unchanged, the line current ideally becomes:

A. Twice
B. Half
C. Four times
D. Unchanged

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Half

💡 Detailed Explanation

For constant transmitted power and constant power factor:

$$P=\sqrt{3}VI\cos\phi$$

Therefore:

$$I\propto\frac{1}{V}$$

If voltage becomes $2V$:

$$I_2=\frac{I_1}{2}$$

Hence, current becomes half.

📌 Remember Points

For constant power: $I\propto1/V$.
Voltage ×2 → current ÷2.
This is why power is transmitted at high voltage.

MCQ 86. If line current is reduced to half, the $I^2R$ loss becomes:

A. Half
B. One-fourth
C. Twice
D. Four times

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. One-fourth

💡 Step-by-Step Solution

Initial loss:

$$P_1=I^2R$$

New current:

$$I_2=\frac{I}{2}$$

Therefore:

$$P_2=\left(\frac{I}{2}\right)^2R$$

$$P_2=\frac{I^2R}{4}$$

Hence:

$$P_2=\frac{P_1}{4}$$

The loss becomes one-fourth of its original value.

📌 Remember Points

Loss varies as $I^2$.
Current half → loss one-fourth.
Current double → loss four times.

MCQ 87. High-voltage transmission helps to reduce:

A. Current and $I^2R$ losses
B. Frequency
C. System voltage
D. Conductor length

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Current and $I^2R$ losses

💡 Detailed Explanation

For a fixed amount of transmitted power:

$$I=\frac{P}{\sqrt{3}V\cos\phi}$$

Increasing voltage decreases current. Since transmission loss varies as $I^2$, the reduction in current produces a much larger reduction in resistive loss.

$$P_{\text{loss}}=3I^2R$$

📌 Remember Points

High voltage → low current.
Low current → low $I^2R$ loss.
High-voltage transmission improves efficiency.

MCQ 88. Transmission efficiency is the ratio of:

A. Receiving-end power to sending-end power
B. Sending-end power to receiving-end power
C. Losses to receiving-end power
D. Reactive power to active power

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Receiving-end power to sending-end power

💡 Detailed Explanation

Transmission efficiency indicates how much of the power supplied at the sending end is successfully delivered to the receiving end.

$$\eta=\frac{P_R}{P_S}\times100\%$$

Since transmission losses are always present in a practical line:

$$P_S=P_R+P_{\text{loss}}$$

Therefore, receiving-end power is normally less than sending-end power.

📌 Remember Points

Efficiency = useful output / input × 100.
For transmission: $\eta=P_R/P_S\times100\%$.
Higher losses → lower efficiency.

MCQ 89. Transmission efficiency is normally expressed as:

A. $P_s/P_r\times100$
B. $P_r/P_s\times100$
C. $P_{\text{loss}}/P_s\times100$
D. $Q/P\times100$

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. $P_r/P_s\times100$

💡 Step-by-Step Explanation

Transmission efficiency is defined as receiving-end active power divided by sending-end active power.

$$\eta=\frac{P_R}{P_S}\times100\%$$

Also:

$$P_S=P_R+P_{\text{loss}}$$

Therefore, the efficiency must normally be less than $100\%$.

📌 Remember Points

$P_R$ → receiving-end power.
$P_S$ → sending-end power.
$\eta=P_R/P_S\times100\%$.

MCQ 90. Voltage regulation of a transmission line is related to the change in:

A. Receiving-end voltage with load condition
B. Frequency only
C. Conductor diameter only
D. Tower height only

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Receiving-end voltage with load condition

💡 Detailed Explanation

Voltage regulation describes the change in receiving-end voltage as the load changes, usually from no-load to full-load conditions while keeping the sending-end voltage constant.

$$\%VR= \frac{V_{R,\text{no-load}}-V_{R,\text{full-load}}} {V_{R,\text{full-load}}} \times100$$

The value depends upon the line impedance and load power factor.

📌 Remember Points

Voltage regulation concerns receiving-end voltage.
It depends on line impedance and load conditions.
Power factor strongly affects regulation.

MCQ 91. Ferranti effect is mainly observed in:

A. Long transmission lines under light-load or no-load conditions
B. Short lines under heavy load only
C. DC circuits only
D. Low-voltage house wiring

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Long transmission lines under light-load or no-load conditions

💡 Detailed Explanation

The Ferranti effect occurs mainly in long AC transmission lines when they are lightly loaded or unloaded.

The distributed capacitance of the line draws a charging current even when the load current is very small.

Due to the interaction between line capacitance and inductance, the receiving-end voltage can rise above the sending-end voltage.

$$V_R>V_S$$

📌 Remember Points

Ferranti effect → mainly long AC lines.
Most noticeable at light/no load.
Receiving voltage can exceed sending voltage.

MCQ 92. In the Ferranti effect, the receiving-end voltage can become:

A. Lower than zero
B. Greater than the sending-end voltage
C. Always equal to the sending-end voltage
D. Independent of line capacitance

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Greater than the sending-end voltage

💡 Detailed Explanation

Ferranti effect is characterized by an increase in receiving-end voltage under light-load or no-load conditions on long AC lines.

$$V_R>V_S$$

The principal reason is the charging current associated with the distributed line capacitance.

The effect becomes more pronounced as the line becomes longer.

📌 Remember Points

Ferranti effect → $V_R>V_S$.
Associated with line capacitance.
Strongest under light/no-load conditions.

MCQ 93. Ferranti effect is primarily associated with the:

A. Capacitance of the transmission line
B. Mechanical strength of the tower
C. Resistance of the pole
D. Weight of the conductor

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Capacitance of the transmission line

💡 Detailed Explanation

Transmission-line capacitance causes a charging current to flow even when the external load current is very small.

$$I_C=j\omega CV$$

This charging current interacts with the line inductance and causes the receiving-end voltage to rise under light-load conditions.

📌 Remember Points

Ferranti effect → capacitance phenomenon.
Charging current exists due to line capacitance.
Longer line → greater capacitance effect.

MCQ 94. Ferranti effect becomes more significant as line:

A. Length increases
B. Length decreases toward zero
C. Resistance becomes zero only
D. Frequency becomes zero

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Length increases

💡 Detailed Explanation

The capacitance of a transmission line increases with its length. Consequently, the charging current also increases.

$$C_{\text{total}}=C' L$$

where $C'$ is capacitance per unit length and $L$ is line length.

Therefore, long transmission lines experience a stronger Ferranti effect than short lines.

📌 Remember Points

Line length ↑ → total capacitance ↑.
Capacitance ↑ → charging current ↑.
Ferranti effect becomes more significant.

MCQ 95. A short transmission line is commonly analyzed by:

A. Neglecting shunt capacitance
B. Neglecting series resistance
C. Neglecting all parameters
D. Considering only capacitance

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Neglecting shunt capacitance

💡 Detailed Explanation

For a short transmission line, the effect of shunt capacitance is relatively small and is commonly neglected in the basic model.

The line is then represented primarily by its series impedance:

$$Z=R+jX_L$$

Hence:

$$V_S=V_R+IZ$$

📌 Remember Points

Short line → shunt capacitance usually neglected.
Series impedance is retained.
Basic model → series $R+jX$.

MCQ 96. In a medium transmission line, which parameter is normally considered in the equivalent circuit?

A. Shunt capacitance
B. Mechanical sag only
C. Tower weight only
D. Insulator color

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Shunt capacitance

💡 Detailed Explanation

In medium transmission lines, the effect of shunt capacitance becomes significant enough that it cannot normally be ignored.

Therefore, the equivalent circuit includes both the series impedance and shunt admittance.

$$Z=R+jX$$

$$Y=G+jB$$

Common models include the nominal-$\pi$ and nominal-T models.

📌 Remember Points

Short line → capacitance usually neglected.
Medium line → shunt capacitance considered using lumped models.
Long line → distributed parameters are used.

MCQ 97. A long transmission line is normally represented using:

A. Lumped parameters only at one point
B. Distributed parameters
C. Resistance only
D. Capacitance only

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Distributed parameters

💡 Detailed Explanation

In a long transmission line, resistance, inductance, capacitance and conductance are distributed continuously along the entire length.

Therefore, a distributed-parameter model is used instead of treating all parameters as concentrated at one location.

$$Z'=R+j\omega L$$

$$Y'=G+j\omega C$$

The propagation constant and characteristic impedance are important quantities in long-line analysis.

$$\gamma=\sqrt{Z'Y'}$$

$$Z_c=\sqrt{\frac{Z'}{Y'}}$$

📌 Remember Points

Long line → distributed parameters.
Parameters → $R,L,C,G$.
Propagation constant → $\gamma=\sqrt{Z'Y'}$.

MCQ 98. The ABCD parameters of a transmission line are used to relate:

A. Sending-end and receiving-end voltage and current
B. Mechanical tension and sag only
C. Pole height and conductor weight
D. Frequency and temperature only

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Sending-end and receiving-end voltage and current

💡 Detailed Explanation

ABCD parameters, also called transmission parameters, provide a convenient relationship between sending-end and receiving-end quantities.

$$V_S=AV_R+BI_R$$

$$I_S=CV_R+DI_R$$

In matrix form:

$$\begin{bmatrix}V_S\\I_S\end{bmatrix} = \begin{bmatrix}A&B\\C&D\end{bmatrix} \begin{bmatrix}V_R\\I_R\end{bmatrix}$$

📌 Remember Points

ABCD parameters → transmission parameters.
They relate sending-end and receiving-end quantities.
Basic equations: $V_S=AV_R+BI_R$, $I_S=CV_R+DI_R$.

MCQ 99. The purpose of a transmission line is primarily to:

A. Transfer electrical power from generating stations to load centers
B. Generate electrical power
C. Store electrical energy permanently
D. Convert electrical energy directly into mechanical energy

📖 View Answer and Detailed Explanation

✅ Correct Answer: A. Transfer electrical power from generating stations to load centers

💡 Detailed Explanation

A transmission system transfers bulk electrical power from generating stations to major load centers and substations.

The overall electrical-power system can broadly be represented as:

$$\text{Generation} \rightarrow \text{Transmission} \rightarrow \text{Distribution} \rightarrow \text{Consumers}$$

Transmission lines are therefore an essential link between generation and distribution networks.

📌 Remember Points

Generation produces electrical power.
Transmission transfers bulk power over longer distances.
Distribution delivers power to consumers.

MCQ 100. Which sequence correctly represents the conventional path from a generating station to a consumer?

A. Consumer → distributor → feeder → generating station
B. Generating station → transmission → distribution → consumer
C. Consumer → transmission → generating station
D. Distributor → generating station → consumer → feeder

📖 View Answer and Detailed Explanation

✅ Correct Answer: B. Generating station → transmission → distribution → consumer

💡 Detailed Explanation

Electrical power follows a structured path from the generating station to the final consumer.

$$\boxed{ \text{Generating Station} \rightarrow \text{Transmission} \rightarrow \text{Distribution} \rightarrow \text{Consumer} }$$

At the generating station, electrical energy is produced. The generated voltage is then stepped up using a transformer for economical transmission.

The transmission network carries bulk power over long distances. Substations then reduce the voltage to appropriate distribution levels.

Finally, the distribution network delivers electrical power to individual consumers.

The reason for transmitting power at high voltage can be understood from the three-phase power equation:

$$P=\sqrt{3}V_LI_L\cos\phi$$

For a fixed power:

$$I_L=\frac{P}{\sqrt{3}V_L\cos\phi}$$

Therefore, increasing the transmission voltage reduces the current. Since transmission loss is proportional to $I^2R$, the reduction in current significantly reduces power losses.

$$P_{\text{loss}}=3I_L^2R$$

📌 Remember Points

Generation → Transmission → Distribution → Consumer.
Transmission carries bulk electrical power over long distances.
High voltage reduces transmission current.
Lower current reduces $I^2R$ losses.
Step-up transformers are used before long-distance transmission.
Step-down transformers are used before distribution and utilization.

3000 Most Asked Transmission & Distribution Objection with Full Explanations Part-1

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