3000 Most Asked Transmission & Distribution Objective Questions with Full Explanations Part-8

3000-Electrical Engineering MCQs — Transmission & Distribution Objective

🔹 MCQ 451. The proximity effect is caused by:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Magnetic fields of nearby current-carrying conductors

📚 Detailed Explanation

The proximity effect is the redistribution of alternating current in a conductor caused by the magnetic field of a nearby current-carrying conductor. The neighbouring conductor’s alternating magnetic field induces circulating effects that push current toward some regions of the conductor and away from others. The exact direction depends on whether the currents in the adjacent conductors flow in the same or opposite directions.

🧮 Formula / Key Relationship

\(R_{\text{AC}}\text{ may exceed }R_{\text{DC}}\)

🎯 Exam Point: Nearby alternating magnetic fields alter current distribution.

📌 Remember Points

📌 Nearby alternating magnetic fields alter current distribution.

📌 It is different from skin effect, which is caused by the conductor's own changing field.

📌 Frequency and conductor spacing influence the effect.

📌 It can increase effective AC resistance.

🔹 MCQ 452. Proximity effect generally causes:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Non-uniform current distribution

📚 Detailed Explanation

Because the magnetic field of a nearby conductor changes the current distribution, current density is no longer uniform across the conductor’s cross-section. Some regions carry more current while other regions carry less. This uneven distribution can increase the effective resistance for AC compared with the DC resistance.

🧮 Formula / Key Relationship

\(J=\frac{I}{A}\) only for uniform current density; proximity effect makes \(J\) non-uniform.

🎯 Exam Point: Current density is not uniform across the conductor.

📌 Remember Points

📌 Current density is not uniform across the conductor.

📌 The total current is still carried by the conductor.

📌 The pattern depends on adjacent conductor currents and geometry.

📌 Effective AC resistance can rise.

🔹 MCQ 453. Both skin effect and proximity effect can cause:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Increase in effective AC resistance

📚 Detailed Explanation

Skin effect concentrates AC toward the conductor surface, while proximity effect redistributes current because of neighbouring conductors’ magnetic fields. Both can reduce the effective area available for current flow and therefore increase AC resistance. The magnitude depends on frequency, conductor dimensions, material, and conductor arrangement.

🧮 Formula / Key Relationship

\(R_{\text{AC}}>R_{\text{DC}}\) when skin/proximity effects are significant.

🎯 Exam Point: Skin effect is caused by the conductor's own alternating field.

📌 Remember Points

📌 Skin effect is caused by the conductor's own alternating field.

📌 Proximity effect is caused by nearby conductors.

📌 Both can raise effective AC resistance.

📌 Their severity depends on frequency and conductor geometry.

🔹 MCQ 454. Skin effect is absent ideally in:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. DC steady-state current

📚 Detailed Explanation

In steady-state DC, after switching transients have died away, current is constant rather than alternating. There is no continuously changing magnetic field to produce the usual skin-effect redistribution, so skin effect is ideally absent. During the instant of switching, transient electromagnetic effects may still occur.

🧮 Formula / Key Relationship

\(\frac{dI}{dt}=0\) in steady-state DC; the usual AC skin effect is ideally absent.

🎯 Exam Point: Skin effect is associated with time-varying current.

📌 Remember Points

📌 Skin effect is associated with time-varying current.

📌 Steady-state DC has no continuous alternating field.

📌 Switching transients are separate from steady-state DC.

📌 AC skin effect becomes more significant at higher frequency.

🔹 MCQ 455. The severity of proximity effect depends strongly on:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Distance between nearby conductors

📚 Detailed Explanation

The distance between neighbouring conductors strongly affects how much of one conductor’s magnetic field links with another. Closely spaced conductors generally have stronger mutual magnetic influence, making proximity-effect distortion more significant. Frequency, conductor size, material, and current phase relationship also matter.

🧮 Formula / Key Relationship

\(\text{Proximity effect}\uparrow\text{ as conductor spacing decreases (all else equal)}\)

🎯 Exam Point: Closer conductors usually increase magnetic coupling.

📌 Remember Points

📌 Closer conductors usually increase magnetic coupling.

📌 Frequency also affects proximity effect.

📌 Current direction/phase changes the distribution pattern.

📌 Conductor geometry and material matter too.

🔹 MCQ 456. The primary function of a transmission-line insulator is to:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Electrically isolate the conductor from the supporting structure

📚 Detailed Explanation

An insulator supports the conductor mechanically while electrically separating it from the earthed tower or pole. It must withstand the operating voltage and environmental stresses without allowing an unwanted current path to earth. Insulators therefore provide both electrical insulation and mechanical support.

🧮 Formula / Key Relationship

\(\text{Conductor}\rightarrow\text{Insulator}\rightarrow\text{Earthed support}\)

🎯 Exam Point: Insulators provide electrical isolation and mechanical support.

📌 Remember Points

📌 Insulators provide electrical isolation and mechanical support.

📌 They must withstand operating voltage and weather.

📌 Surface contamination and moisture can affect performance.

📌 The supporting tower is generally earthed.

🔹 MCQ 457. A pin-type insulator is normally mounted on:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. A pin fixed to the supporting structure

📚 Detailed Explanation

A pin-type insulator is mounted on a metal pin secured to a cross-arm or other supporting structure. The conductor is placed in a groove and tied or otherwise secured according to the design. This construction is traditionally used on distribution lines at suitable voltage levels.

🧮 Formula / Key Relationship

\(\text{Conductor}\rightarrow\text{Pin insulator}\rightarrow\text{Cross-arm}\)

🎯 Exam Point: Pin insulators are mounted on pins attached to cross-arms.

📌 Remember Points

📌 Pin insulators are mounted on pins attached to cross-arms.

📌 The conductor is secured in a groove or designed fitting.

📌 Pin type is common in distribution applications.

📌 The pin provides mechanical mounting, not insulation by itself.

🔹 MCQ 458. Suspension insulators are particularly suitable for:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Higher transmission voltages

📚 Detailed Explanation

A suspension-insulator string is especially useful at higher transmission voltages because several disc units can be connected in series to provide the required insulation level. The conductor hangs from the string, and the mechanical load is distributed through the units and fittings.

🧮 Formula / Key Relationship

\(V_{\text{string}}=\sum_{k=1}^{n}V_k\)

🎯 Exam Point: Suspension strings suit higher voltages.

📌 Remember Points

📌 Suspension strings suit higher voltages.

📌 Disc units are connected in series mechanically.

📌 The string carries conductor tensile load.

📌 Unit count depends on voltage, pollution and design.

🔹 MCQ 459. A suspension-insulator string consists of:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Several disc units connected in series

📚 Detailed Explanation

A suspension string is assembled from multiple disc insulator units joined by metal links or fittings. The voltage is shared across the series-connected units, although the actual distribution is not perfectly equal because of stray capacitances to the tower. The number of discs depends on the system voltage and design requirements.

🧮 Formula / Key Relationship

\(V_{\text{string}}=V_1+V_2+\cdots+V_n\)

🎯 Exam Point: A string contains multiple disc units.

📌 Remember Points

📌 A string contains multiple disc units.

📌 The total voltage is the sum of unit voltages.

📌 Practical voltage distribution is unequal.

📌 Stray capacitance to the tower contributes to inequality.

🔹 MCQ 460. One important advantage of suspension insulators is:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Insulator string length can be increased by adding units

📚 Detailed Explanation

A key benefit of a suspension string is that its insulation level can be increased by adding more disc units. This modular construction is practical when voltage ratings change between systems, subject to mechanical, electrical, and clearance requirements. Individual damaged units can also often be replaced rather than replacing an entire rigid insulator.

🧮 Formula / Key Relationship

\(V_{\text{string}}=\sum V_k\); adding units increases the available insulation length.

🎯 Exam Point: Adding disc units can raise the string insulation level.

📌 Remember Points

📌 Adding disc units can raise the string insulation level.

📌 The string is modular and units can often be replaced individually.

📌 Mechanical loading and clearances must be checked.

📌 More units do not guarantee perfectly equal voltage sharing.

🔹 MCQ 461. Strain insulators are mainly used where the conductor is subjected to:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. High tensile stress

📚 Detailed Explanation

Strain insulators are designed for locations where the conductor pulls strongly on the support. They must withstand substantial tensile load while providing electrical insulation. Their arrangement can use suspension discs placed horizontally or in other suitable configurations depending on the line design.

🧮 Formula / Key Relationship

\(\text{Strain insulator: electrical insulation under tensile mechanical load}\)

🎯 Exam Point: Strain means high tensile loading.

📌 Remember Points

📌 Strain means high tensile loading.

📌 Strain insulators also provide electrical isolation.

📌 They are used at line ends and angle points.

📌 Mechanical strength is a key selection criterion.

🔹 MCQ 462. Strain insulators are commonly used at:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Dead ends and sharp angle locations

📚 Detailed Explanation

At a dead end, the conductor tension is not balanced by a continuing span in the same way as at an intermediate support. At sharp line-angle locations, the change in direction creates a significant resultant pull on the structure. Strain-insulator arrangements are used to withstand these tensile forces safely.

🧮 Formula / Key Relationship

\(\text{Resultant pull depends on conductor tension and line angle}\)

🎯 Exam Point: Dead-end locations resist conductor pull.

📌 Remember Points

📌 Dead-end locations resist conductor pull.

📌 Sharp line angles create a resultant force.

📌 Strain strings are arranged for the expected mechanical load.

📌 Tower and fitting strength must also be adequate.

🔹 MCQ 463. Shackle insulators are commonly associated with:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Low-voltage distribution lines

📚 Detailed Explanation

Shackle insulators are compact insulators commonly used on low-voltage distribution lines, often mounted on poles or walls. They can be installed in different orientations to suit the line arrangement and are suitable for relatively modest mechanical and voltage duties.

🧮 Formula / Key Relationship

\(\text{Shackle insulator}\rightarrow\text{low-voltage distribution}\)

🎯 Exam Point: Shackle insulators are common on low-voltage distribution lines.

📌 Remember Points

📌 Shackle insulators are common on low-voltage distribution lines.

📌 They can be mounted in suitable orientations.

📌 They are compact compared with high-voltage strings.

📌 Voltage and mechanical ratings must match the application.

🔹 MCQ 464. The dielectric strength of an insulating material represents its ability to:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Withstand electric stress without breakdown

📚 Detailed Explanation

Dielectric strength is the maximum electric field an insulating material can withstand before electrical breakdown occurs under specified test conditions. If the electric stress exceeds the material’s capability, the insulation may puncture or otherwise fail. Actual performance also depends on thickness, defects, moisture, temperature, and test conditions.

🧮 Formula / Key Relationship

\(E_{\text{avg}}=\frac{V}{d}\)

🎯 Exam Point: Dielectric strength is an electric-field limit.

📌 Remember Points

📌 Dielectric strength is an electric-field limit.

📌 Breakdown occurs when the material's withstand capability is exceeded.

📌 Thickness and defects influence the actual withstand voltage.

📌 Common units include kV/mm and MV/m.

🔹 MCQ 465. The unit of dielectric strength is commonly:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. kV/mm

📚 Detailed Explanation

Dielectric strength expresses voltage withstand per unit thickness. For a simple uniform field, an approximate value can be calculated as E = V/d, where V is the voltage across the material and d is its thickness. For example, 30 kV across 5 mm corresponds to an average field of 6 kV/mm, assuming a uniform field.

🧮 Formula / Key Relationship

\(E_{\text{avg}}=\frac{V}{d}\quad\text{(uniform-field approximation)}\)

🎯 Exam Point: Average field is estimated by voltage divided by thickness.

📌 Remember Points

📌 Average field is estimated by voltage divided by thickness.

📌 Use consistent units for V and d.

📌 The actual field may be non-uniform near edges.

📌 Dielectric strength depends on material and test conditions.

🔹 MCQ 466. Flashover of an insulator occurs when:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Discharge travels over the external surface of the insulator

📚 Detailed Explanation

Flashover occurs when an electrical discharge travels along the outside surface of an insulator, often through the surrounding air and across a contaminated or wet surface. It can momentarily connect the energized conductor to the earthed structure. Pollution, moisture, surface creepage distance, and overvoltage affect flashover performance.

🧮 Formula / Key Relationship

\(\text{Flashover path: external surface/air}\)

🎯 Exam Point: Flashover occurs over an external surface/air path.

📌 Remember Points

📌 Flashover occurs over an external surface/air path.

📌 Wetness and pollution can lower surface withstand.

📌 It can cause a line-to-earth fault.

📌 It differs from internal puncture.

🔹 MCQ 467. Puncture of an insulator means:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Breakdown occurs through the body of the insulating material

📚 Detailed Explanation

Puncture is an internal dielectric failure through the body of the insulating material. It can permanently damage an insulator and compromise its insulating properties. A puncture path is different from a flashover path, which occurs over the surface or through surrounding air.

🧮 Formula / Key Relationship

\(\text{Puncture path: through insulating material}\)

🎯 Exam Point: Puncture occurs through the insulating body.

📌 Remember Points

📌 Puncture occurs through the insulating body.

📌 It can permanently damage an insulator.

📌 Flashover is external; puncture is internal.

📌 Insulation coordination aims to avoid both.

🔹 MCQ 468. Which is generally more desirable for an insulator?
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Flashover before puncture

📚 Detailed Explanation

It is generally preferable for an insulator to flash over externally before it punctures internally. External flashover is often less permanently damaging and may clear after the fault is interrupted, whereas puncture can leave the unit permanently defective. Protection and insulation design aim to reduce both risks.

🧮 Formula / Key Relationship

\(V_{\text{flashover}}

🎯 Exam Point: Flashover is generally less permanently damaging than puncture.

📌 Remember Points

📌 Flashover is generally less permanently damaging than puncture.

📌 Puncture can render the unit unusable.

📌 Both withstand levels must meet design requirements.

📌 Surface condition influences flashover performance.

🔹 MCQ 469. String efficiency is associated with:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Voltage distribution across individual suspension-insulator units

📚 Detailed Explanation

String efficiency measures how evenly the total voltage is shared among the units of a suspension-insulator string. Stray capacitance between metal fittings and the tower causes unequal sharing, so the unit nearest the conductor commonly has the highest voltage stress. Better uniformity means higher string efficiency.

🧮 Formula / Key Relationship

\(\eta_{\text{string}}=\frac{V_{\text{string}}}{n\,V_{\max}}\times100\%\)

🎯 Exam Point: String efficiency measures voltage-sharing uniformity.

📌 Remember Points

📌 String efficiency measures voltage-sharing uniformity.

📌 A lower maximum unit voltage improves efficiency.

📌 Shunt capacitance causes unequal distribution.

📌 The conductor-end unit usually has the highest stress.

🔹 MCQ 470. Ideally, if voltage is distributed equally over all units of an insulator string, string efficiency is:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. 100%

📚 Detailed Explanation

String efficiency is defined as the total string voltage divided by the product of the number of units and the maximum voltage across any unit, multiplied by 100%. If every unit shares voltage equally, the maximum unit voltage equals the average voltage, so the ratio is 1 and efficiency is 100%.

🧮 Formula / Key Relationship

\(\eta_{\text{string}}=\frac{nV_{\text{unit}}}{nV_{\text{unit}}}\times100\%=100\%\)

🎯 Exam Point: Ideal equal voltage sharing gives 100% string efficiency.

📌 Remember Points

📌 Ideal equal voltage sharing gives 100% string efficiency.

📌 Practical strings are usually below 100%.

📌 The highest unit voltage appears in the denominator.

📌 Grading arrangements can improve efficiency.

🔹 MCQ 471. In a practical suspension-insulator string, the voltage distribution is usually:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Unequal

📚 Detailed Explanation

In a practical suspension string, each metal fitting has capacitance not only through the insulator unit but also to the earthed tower. These shunt capacitances cause the voltages across the discs to differ. Consequently, the voltage distribution is unequal, with the conductor-end unit typically under the greatest stress.

🧮 Formula / Key Relationship

\(\text{Shunt capacitance}\rightarrow\text{unequal unit voltages}\)

🎯 Exam Point: Practical voltage sharing is unequal.

📌 Remember Points

📌 Practical voltage sharing is unequal.

📌 Fittings have stray capacitance to the tower.

📌 Shunt capacitance diverts capacitive current.

📌 The conductor-end disc is usually most stressed.

🔹 MCQ 472. In a suspension-insulator string, the disc nearest the conductor generally experiences:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Higher voltage stress

📚 Detailed Explanation

The disc nearest the line conductor usually has the greatest voltage across it because the capacitive currents to the tower accumulate along the string and distort the voltage division. This makes the conductor-end unit the most electrically stressed and an important consideration in string design.

🧮 Formula / Key Relationship

\(V_{\text{conductor-end unit}}\text{ is usually the highest}\)

🎯 Exam Point: The line-end unit usually has the highest voltage stress.

📌 Remember Points

📌 The line-end unit usually has the highest voltage stress.

📌 Stray capacitance to earth causes the inequality.

📌 The tower-end unit commonly has lower stress.

📌 Grading rings can reduce the inequality.

🔹 MCQ 473. String efficiency can be improved by:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Using grading arrangements

📚 Detailed Explanation

Grading arrangements are used to make the voltage across each unit more nearly equal. Methods include capacitance grading with appropriately selected units and the use of grading or guard rings to reshape the electric field and capacitance distribution. More uniform sharing reduces the maximum stress on any one unit.

🧮 Formula / Key Relationship

\(\eta_{\text{string}}=\frac{V_{\text{string}}}{nV_{\max}}\times100\%\)

🎯 Exam Point: Grading aims for more uniform voltage distribution.

📌 Remember Points

📌 Grading aims for more uniform voltage distribution.

📌 Guard rings modify electric-field distribution.

📌 String efficiency rises as sharing becomes more uniform.

📌 Insulation and mechanical requirements still apply.

🔹 MCQ 474. A guard ring fitted to an insulator string is used mainly to:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Improve voltage distribution

📚 Detailed Explanation

A guard ring, also called a grading ring in many arrangements, is fitted near the conductor end of an insulator string. It changes the electric-field and capacitance distribution so that voltage is shared more evenly across the discs. Correct design also helps control local electric stress and corona.

🧮 Formula / Key Relationship

\(\text{Guard ring}\rightarrow\text{more uniform electric-field distribution}\)

🎯 Exam Point: Guard rings are fitted near the line end of the string.

📌 Remember Points

📌 Guard rings are fitted near the line end of the string.

📌 They reshape the electric field.

📌 They can reduce excessive local stress and corona.

📌 They do not carry normal load current.

🔹 MCQ 475. Capacitance between insulator metal fittings and the tower is commonly called:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Shunt capacitance

📚 Detailed Explanation

Capacitance between the metal fittings of an insulator string and the earthed tower is called shunt capacitance. It is in parallel with the relevant part of the string’s electrical path and diverts capacitive current to earth. This is a principal cause of unequal voltage distribution across the discs.

🧮 Formula / Key Relationship

\(I_C=\omega CV\quad\text{for a capacitive path}\)

🎯 Exam Point: Shunt capacitance is between fittings and tower/earth.

📌 Remember Points

📌 Shunt capacitance is between fittings and tower/earth.

📌 It is distinct from capacitance through the insulator units.

📌 It contributes to unequal voltage distribution.

📌 Capacitive current depends on frequency and voltage.

🔹 MCQ 476. The presence of shunt capacitance in an insulator string causes:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Unequal voltage distribution

📚 Detailed Explanation

Shunt capacitance allows some capacitive current to flow from the metal fittings toward the earthed tower. Because this current varies from point to point along the string, each unit carries a different capacitive current and develops a different voltage. The result is unequal voltage distribution.

🧮 Formula / Key Relationship

\(V_1\ne V_2\ne\cdots\ne V_n\quad\text{in a practical string}\)

🎯 Exam Point: Shunt capacitance is a cause of unequal voltage sharing.

📌 Remember Points

📌 Shunt capacitance is a cause of unequal voltage sharing.

📌 The units are not equally stressed in a conventional string.

📌 Grading arrangements can improve uniformity.

📌 Line-end unit stress is an important design consideration.

🔹 MCQ 477. An earth wire on an overhead transmission line is primarily intended to:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Protect phase conductors from direct lightning strokes

📚 Detailed Explanation

The overhead earth wire, or shield wire, is installed to intercept direct lightning strokes that might otherwise strike the phase conductors. It is connected to towers and earth so that lightning current can be conducted toward ground. It does not normally carry load current.

🧮 Formula / Key Relationship

\(\text{Lightning current}\rightarrow\text{shield wire/tower}\rightarrow\text{earth}\)

🎯 Exam Point: The earth wire is also called a shield wire.

📌 Remember Points

📌 The earth wire is also called a shield wire.

📌 It intercepts many direct lightning strokes.

📌 Lightning current is conducted toward earth through tower grounding.

📌 It does not normally carry load current.

🔹 MCQ 478. The earth wire is normally installed:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Above the phase conductors

📚 Detailed Explanation

The earth wire is placed above the phase conductors so it can shield them from lightning strokes arriving from above. Its position and the shield angle are selected to provide an appropriate protective zone. Tower earthing and insulation coordination are also important for overall performance.

🧮 Formula / Key Relationship

\(\text{Earth wire is positioned above the phase conductors}\)

🎯 Exam Point: The shield wire is above the phase conductors.

📌 Remember Points

📌 The shield wire is above the phase conductors.

📌 Its position defines a protective shielding zone.

📌 Shielding is not perfect under all conditions.

📌 Tower earthing is also important.

🔹 MCQ 479. A shield angle is associated with:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Lightning protection of transmission lines

📚 Detailed Explanation

Shield angle is the angle used to describe the geometric protection provided by the overhead earth wire relative to the phase conductors and tower. A smaller, appropriately designed shield angle generally improves shielding effectiveness, although the final design depends on tower geometry and lightning performance criteria.

🧮 Formula / Key Relationship

\(\text{Shield angle}=\text{geometric angle used to assess shielding}\)

🎯 Exam Point: Shield angle is a geometric lightning-protection parameter.

📌 Remember Points

📌 Shield angle is a geometric lightning-protection parameter.

📌 Smaller angles generally improve shielding, all else equal.

📌 Tower geometry affects the design.

📌 It is unrelated to power factor.

🔹 MCQ 480. A tower footing resistance should generally be:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Low

📚 Detailed Explanation

Tower footing resistance is the resistance between the tower earthing system and remote earth. A low value helps lightning current discharge into the ground and limits the rise of tower potential. The required value depends on soil resistivity, line voltage, tower design, and utility standards.

🧮 Formula / Key Relationship

\(R_{\text{footing}}\text{ should be kept low}\)

🎯 Exam Point: Tower footing resistance is measured relative to remote earth.

📌 Remember Points

📌 Tower footing resistance is measured relative to remote earth.

📌 Lower resistance generally improves lightning performance.

📌 Soil resistivity affects the achievable value.

📌 Earthing design must follow utility standards.

🔹 MCQ 481. A low tower footing resistance helps:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Provide an effective path for lightning current to earth

📚 Detailed Explanation

When lightning strikes the shield wire or tower, a low-resistance tower earthing path helps carry the impulse current into the soil and limits the tower’s voltage rise. If tower potential rises excessively, the voltage across the insulator string may cause a back-flashover to a phase conductor.

🧮 Formula / Key Relationship

\(V_{\text{tower rise}}\approx I_{\text{lightning}}R_{\text{footing}}\quad\text{(simplified)}\)

🎯 Exam Point: Low footing resistance helps dissipate lightning current.

📌 Remember Points

📌 Low footing resistance helps dissipate lightning current.

📌 It limits tower potential rise.

📌 Excessive tower rise can cause back-flashover.

📌 It cannot eliminate every lightning outage.

🔹 MCQ 482. A lightning arrester is connected between:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Line and earth

📚 Detailed Explanation

A lightning arrester is connected from the line terminal to earth, usually as close as practicable to the equipment being protected. During a surge it provides a conductive path to ground, limiting the voltage appearing across transformer or line insulation. Correct leads and earthing connections are important because lead inductance can add voltage.

🧮 Formula / Key Relationship

\(\text{Line}\;-\;\text{Lightning arrester}\;-\;\text{Earth}\)

🎯 Exam Point: Arresters are connected line-to-earth.

📌 Remember Points

📌 Arresters are connected line-to-earth.

📌 They are placed close to protected equipment where practical.

📌 Lead length and inductance affect protection.

📌 They are normally high impedance at system voltage.

🔹 MCQ 483. The main purpose of a lightning arrester is to:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Divert surge current to earth and limit overvoltage

📚 Detailed Explanation

A lightning arrester limits transient overvoltage by conducting surge current to earth and clamping the voltage to a safer level. Under normal system voltage it draws only a small leakage current; during an overvoltage, its impedance drops sharply. This protects insulation from damaging surge stress.

🧮 Formula / Key Relationship

\(\text{Surge arrester: limits }V_{\text{surge}}\text{ and diverts }I_{\text{surge}}\)

🎯 Exam Point: The arrester clamps transient voltage.

📌 Remember Points

📌 The arrester clamps transient voltage.

📌 It diverts surge current to earth.

📌 It should not conduct heavily in normal service.

📌 Correct rating and earthing are essential.

🔹 MCQ 484. A surge diverter should normally have:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. High impedance during normal voltage and low impedance during surge conditions

📚 Detailed Explanation

A surge diverter must have high resistance/impedance at normal operating voltage so that it does not carry significant current continuously. When a surge raises the voltage, it becomes highly conductive, allowing surge current to flow to earth and limiting the overvoltage. After the surge, it returns to its high-impedance state.

🧮 Formula / Key Relationship

\(Z_{\text{normal}}\text{ high};\quad Z_{\text{surge}}\text{ low}\)

🎯 Exam Point: Normal operation: high impedance.

📌 Remember Points

📌 Normal operation: high impedance.

📌 Surge condition: low impedance.

📌 The arrester returns to its normal state after the surge.

📌 It is not intended to carry full load current continuously.

🔹 MCQ 485. A metal-oxide surge arrester commonly uses:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Zinc oxide

📚 Detailed Explanation

Modern metal-oxide surge arresters commonly use zinc-oxide (ZnO) varistor blocks. These blocks have a strongly nonlinear voltage-current characteristic, so a relatively small current flows at normal voltage while a much larger current can flow during a surge.

🧮 Formula / Key Relationship

\(\text{Metal-oxide varistor material: }ZnO\)

🎯 Exam Point: Zinc oxide is the common metal-oxide varistor material.

📌 Remember Points

📌 Zinc oxide is the common metal-oxide varistor material.

📌 The V-I characteristic is nonlinear.

📌 Normal leakage current is relatively small.

📌 Surge current is conducted strongly.

🔹 MCQ 486. The nonlinear characteristic of a metal-oxide arrester allows it to:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Conduct heavily during overvoltage while remaining relatively nonconductive at normal voltage

📚 Detailed Explanation

The nonlinear voltage-current characteristic of zinc-oxide blocks lets the arrester remain relatively nonconductive at normal voltage but conduct strongly when surge voltage rises. The arrester therefore clamps the overvoltage while carrying the surge current to earth. It must be selected for the system’s continuous operating voltage and required protective level.

🧮 Formula / Key Relationship

\(\text{Nonlinear }V\!-\!I\text{ characteristic}\Rightarrow\text{surge clamping}\)

🎯 Exam Point: Nonlinearity enables voltage clamping.

📌 Remember Points

📌 Nonlinearity enables voltage clamping.

📌 Normal leakage should remain low.

📌 Arrester rating must match system voltage.

📌 Surge current is diverted to earth.

🔹 MCQ 487. A surge on an overhead transmission line may be caused by:
📖 View Answer and Detailed Explanation
✅ Correct Answer: C. Both A and B

📚 Detailed Explanation

Lightning can inject a steep voltage/current impulse into an overhead line. Switching operations—such as energizing or de-energizing lines, transformers, or reactors—can also produce transient overvoltages. Both are important sources of surges, although their wave shapes, durations, and magnitudes can differ.

🧮 Formula / Key Relationship

\(\text{Surges: lightning and switching transients}\)

🎯 Exam Point: Lightning and switching can both create surges.

📌 Remember Points

📌 Lightning and switching can both create surges.

📌 Transient waveforms differ by source.

📌 Surge arresters and insulation coordination address overvoltage risk.

📌 Surges are not limited to one voltage level.

🔹 MCQ 488. Switching surges become particularly important in:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. EHV systems

📚 Detailed Explanation

At extra-high voltage (EHV), switching operations can produce significant overvoltages because of the large stored electromagnetic energy and the electrical characteristics of long lines and connected equipment. Switching-surge control may therefore influence insulation coordination, circuit-breaker design, and the use of surge arresters or controlled switching.

🧮 Formula / Key Relationship

\(\text{Switching overvoltage is a key EHV insulation-coordination concern}\)

🎯 Exam Point: Switching surges matter particularly in EHV systems.

📌 Remember Points

📌 Switching surges matter particularly in EHV systems.

📌 They influence insulation coordination.

📌 Controlled switching and arresters may help limit overvoltage.

📌 Lightning surges remain relevant too.

🔹 MCQ 489. Travelling waves on transmission lines are associated with:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Sudden changes in voltage and current

📚 Detailed Explanation

A travelling wave is a voltage and current disturbance that propagates along a line after a sudden event, such as a lightning stroke, switching operation, or fault. The wave moves along the line and is partly reflected or transmitted when it encounters a change in surge impedance, such as a line termination or junction.

🧮 Formula / Key Relationship

\(v=\frac{1}{\sqrt{LC}}\)

🎯 Exam Point: Travelling waves follow sudden voltage/current changes.

📌 Remember Points

📌 Travelling waves follow sudden voltage/current changes.

📌 Lightning, switching and faults can launch waves.

📌 Impedance changes cause reflection/transmission.

📌 They are transient electromagnetic phenomena.

🔹 MCQ 490. The velocity of propagation of a wave on an ideal overhead transmission line is approximately:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. The speed of light

📚 Detailed Explanation

For an ideal overhead line, the propagation velocity is approximately v = 1/√(LC), where L and C are the line inductance and capacitance per unit length. For an overhead line in air, this is typically close to the speed of light, about 3 × 10^8 m/s, with the exact value depending on line parameters.

🧮 Formula / Key Relationship

\(v=\frac{1}{\sqrt{LC}}\approx3\times10^8\ \text{m/s}\text{ for an ideal overhead line}\)

🎯 Exam Point: Overhead-line wave velocity is near light speed.

📌 Remember Points

📌 Overhead-line wave velocity is near light speed.

📌 An ideal estimate uses v = 1/sqrt(LC).

📌 Exact velocity depends on line parameters.

📌 It is much faster than sound.

🔹 MCQ 491. The surge impedance of an overhead line is generally of the order of:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Hundreds of ohms

📚 Detailed Explanation

Surge impedance is the ratio of travelling-wave voltage to current and, for an ideal lossless line, is Z₀ = √(L/C), where L and C are inductance and capacitance per unit length. Typical overhead-line values are on the order of a few hundred ohms, much larger than milliohms and much smaller than megaohms.

🧮 Formula / Key Relationship

\(Z_0=\sqrt{\frac{L}{C}}\)

🎯 Exam Point: Surge impedance is voltage divided by travelling-wave current.

📌 Remember Points

📌 Surge impedance is voltage divided by travelling-wave current.

📌 For a lossless line Z0 = sqrt(L/C).

📌 Overhead-line values are typically hundreds of ohms.

📌 Do not confuse surge impedance with DC resistance.

🔹 MCQ 492. The natural loading of a transmission line is also called:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Surge impedance loading

📚 Detailed Explanation

Surge impedance loading (SIL) is the loading at which the line’s reactive power generated by its capacitance approximately balances the reactive power absorbed by its inductance. For a three-phase line, a common expression is SIL = V_LL²/Z₀, where V_LL is line-to-line voltage and Z₀ is surge impedance, under the usual idealized assumptions.

🧮 Formula / Key Relationship

\(\mathrm{SIL}=\frac{V_{LL}^{\,2}}{Z_0}\quad\text{(three-phase, idealized)}\)

🎯 Exam Point: Natural loading and SIL are synonymous.

📌 Remember Points

📌 Natural loading and SIL are synonymous.

📌 SIL is not the thermal current limit.

📌 The usual three-phase formula uses line-to-line voltage.

📌 Actual system conditions may change operating limits.

🔹 MCQ 493. The power factor of a purely capacitive load is:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Zero leading

📚 Detailed Explanation

For a purely capacitive AC load, current leads voltage by 90°. Real power is P = VI cos φ, and with φ = 90° leading, cos 90° = 0, so the ideal power factor is zero leading. The load exchanges reactive energy with the source but consumes no average real power in the ideal model.

🧮 Formula / Key Relationship

\(\mathrm{pf}=\cos\phi=\cos90^\circ=0\ \text{(leading)}\)

🎯 Exam Point: Capacitor current leads voltage by 90 degrees.

📌 Remember Points

📌 Capacitor current leads voltage by 90 degrees.

📌 Ideal real power is zero for a purely reactive load.

📌 Power factor is zero leading.

📌 Real capacitors may have losses.

🔹 MCQ 494. The power factor of a purely inductive load is:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Zero lagging

📚 Detailed Explanation

For a purely inductive AC load, current lags voltage by 90°. Since P = VI cos φ and cos 90° = 0, the ideal power factor is zero lagging. An ideal inductor stores energy in its magnetic field during part of the cycle and returns it to the source during another part.

🧮 Formula / Key Relationship

\(\mathrm{pf}=\cos\phi=\cos90^\circ=0\ \text{(lagging)}\)

🎯 Exam Point: Inductor current lags voltage by 90 degrees.

📌 Remember Points

📌 Inductor current lags voltage by 90 degrees.

📌 Ideal real power is zero for a purely reactive load.

📌 Power factor is zero lagging.

📌 Real inductors have winding/core losses.

🔹 MCQ 495. The dielectric used in a conventional overhead line is primarily:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Air

📚 Detailed Explanation

In a conventional overhead line, the phase conductors are separated from one another and from the earthed support mainly by air. Air acts as the insulating dielectric, while insulator units provide mechanical support and maintain the required clearances. Contamination and moisture can reduce surface insulation performance.

🧮 Formula / Key Relationship

\(\text{Main overhead-line dielectric}=\text{air}\)

🎯 Exam Point: Air is the main dielectric around overhead conductors.

📌 Remember Points

📌 Air is the main dielectric around overhead conductors.

📌 Insulators support conductors and isolate them from the tower.

📌 Clearance is important for voltage withstand.

📌 Pollution and moisture can affect insulator surfaces.

🔹 MCQ 496. Underground cables have higher capacitance than overhead lines mainly because:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Conductors are closer together and surrounded by dielectric material

📚 Detailed Explanation

Cable conductors are typically closer together than overhead-line conductors, and the space between them is filled with insulation whose permittivity is much higher than that of air. Since capacitance increases with permittivity and generally increases as conductor spacing decreases, underground cables have greater capacitance.

🧮 Formula / Key Relationship

\(C\propto\frac{\varepsilon}{\text{conductor spacing}}\quad\text{(qualitative relationship)}\)

🎯 Exam Point: Cable conductors are closer together than overhead conductors.

📌 Remember Points

📌 Cable conductors are closer together than overhead conductors.

📌 Cable insulation has higher permittivity than air.

📌 Higher capacitance increases AC charging current.

📌 Exact capacitance depends on cable construction.

🔹 MCQ 497. Compared with overhead lines of similar voltage rating, underground cables generally have:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Higher capacitance

📚 Detailed Explanation

For a comparable voltage rating, cable conductors are close together and separated by a solid or fluid dielectric, while overhead conductors are farther apart with air as the main dielectric. These differences increase cable capacitance. The resulting charging current can be important for long AC cables.

🧮 Formula / Key Relationship

\(C\propto\varepsilon\quad\text{and generally increases as conductor spacing decreases}\)

🎯 Exam Point: Underground cables generally have higher capacitance.

📌 Remember Points

📌 Underground cables generally have higher capacitance.

📌 Solid dielectric has higher permittivity than air.

📌 Close spacing increases capacitance.

📌 Charging current matters on long AC cables.

🔹 MCQ 498. One important limitation of underground cables for very high-voltage long-distance AC transmission is:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Large charging current

📚 Detailed Explanation

The capacitive charging current of an AC cable is I_c = ωCV = 2πfCV per phase for a simplified lumped-capacitance model. Because underground cables have high capacitance, a long high-voltage AC cable can draw substantial charging current even when delivering little useful load. This consumes current capacity and creates reactive-power and voltage-control challenges.

🧮 Formula / Key Relationship

\(I_c=\omega CV=2\pi fCV\quad\text{(per phase, simplified model)}\)

🎯 Exam Point: Charging current increases with frequency, capacitance and voltage.

📌 Remember Points

📌 Charging current increases with frequency, capacitance and voltage.

📌 Long high-voltage AC cables can draw substantial charging current.

📌 Charging current uses part of cable current capacity.

📌 HVDC can be advantageous for very long cable links in suitable cases.

🔹 MCQ 499. The main advantage of underground transmission over overhead transmission is:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Better protection from environmental exposure and improved appearance

📚 Detailed Explanation

Underground cables are less exposed to wind, storms, falling branches, and visual clutter, so they can improve appearance and protect the circuit from many environmental disturbances. However, they still require insulation, jointing, thermal management, and fault protection, and their installation is usually more expensive than overhead lines.

🧮 Formula / Key Relationship

\(\text{Underground cable: improved appearance/environmental protection, higher installation cost}\)

🎯 Exam Point: Underground cables improve appearance and reduce exposure to many weather events.

📌 Remember Points

📌 Underground cables improve appearance and reduce exposure to many weather events.

📌 They still need insulation, joints and thermal design.

📌 Installation costs are often higher.

📌 Fault repair can take longer.

🔹 MCQ 500. The main disadvantage of underground cables compared with overhead lines is generally:
📖 View Answer and Detailed Explanation
✅ Correct Answer: A. Higher installation cost and more difficult fault location/repair

📚 Detailed Explanation

Underground cable installation usually requires expensive cable, insulation, trenching or ducting, joint bays, and specialized termination equipment. When a fault occurs, locating it and repairing or replacing the affected section can take longer than on an overhead line, where conductors and supports are more accessible. Cable reliability can be good, but repair logistics are often more difficult.

🧮 Formula / Key Relationship

\(\text{Total cost}=\text{installation}+\text{maintenance}+\text{fault repair}\)

🎯 Exam Point: Cable installation often needs trenching/ducts and specialized joints.

📌 Remember Points

📌 Cable installation often needs trenching/ducts and specialized joints.

📌 Fault location and repair can be difficult.

📌 Overhead lines are usually easier to inspect and repair.

📌 Whole-life cost depends on local conditions.

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