Why this topic is important for ECET?
In power transmission, sag and insulators are core concepts. Sag directly affects conductor safety, clearance, and economy, while insulators ensure reliable operation of overhead lines. In ECET 2026, 1–2 questions are almost guaranteed from these topics, either numerical (sag calculation) or conceptual (types of insulators).
📘 Concept Notes
1. Sag in Transmission Lines
- Definition: Sag is the vertical distance between the conductor (at the midpoint of the span) and the line joining two supports.
- Proper sag ensures safety, economy, and prevents conductor breakage.
Factors affecting sag:
- Conductor weight – Heavier conductors → more sag.
- Span length – Longer span → more sag.
- Tension in conductor – Higher tension → less sag.
- Wind and ice loading – Increase effective conductor weight.
Types of Supports:
- Equal level supports.
- Unequal level supports (sag different at each side).
2. Insulators in Transmission Lines
- Function: Provide insulation between live conductor and supporting tower/pole.
- Requirements: High mechanical strength, high electrical resistance, weather resistance.
Types of insulators:
- Pin-type insulator – Up to 33 kV.
- Suspension-type insulator – For higher voltages (>33 kV).
- Strain insulator – At dead-ends of lines.
- Shackle insulator – Low-voltage distribution lines.
Insulator Failures:
- Flashover: Surface breakdown of air around insulator.
- Puncture: Current passes through insulator body (permanent damage).
⚙️ Formulas
- Sag (for level supports, no wind/ice):
Sag with wind loading:
Tension in conductor:
Where:
→ weight per unit length of conductor (N/m)
→ span length (m)
→ conductor tension (N)
→ wind force per unit length (N/m)
🔟 10 MCQs
Q1. Sag in transmission line increases when:
a) Tension increases
b) Span decreases
c) Conductor weight increases
d) Temperature decreases
Q2. Sag is minimum when:
a) Span is long
b) Tension is high
c) Weight is more
d) Wind pressure increases
Q3. A conductor of weight 1 N/m, span length 200 m, and tension 2000 N. Calculate sag.
a) 1.25 m
b) 2.5 m
c) 5 m
d) 10 m
Q4. Pin insulators are generally used up to:
a) 11 kV
b) 33 kV
c) 66 kV
d) 132 kV
Q5. Suspension insulators are preferred when voltage is:
a) Below 11 kV
b) Up to 33 kV
c) Above 33 kV
d) Up to 132 kV
Q6. Flashover means:
a) Current through conductor
b) Breakdown of air around insulator
c) Puncture through insulator body
d) Mechanical failure
Q7. For equal supports, sag is proportional to:
a)
b) 
c) 
d) 
Q8. A span is 300 m, weight of conductor is 2 N/m, tension = 3000 N. Find sag.
a) 5 m
b) 7.5 m
c) 10 m
d) 15 m
Q9. Which insulator is used at the dead end of transmission line?
a) Pin
b) Suspension
c) Strain
d) Shackle
Q10. The main reason for sag in conductor is:
a) Magnetic effect
b) Weight of conductor
c) Corona discharge
d) Tower vibration
✅ Answer Key
| Q.No | Answer |
|---|---|
| 1 | c |
| 2 | b |
| 3 | a |
| 4 | b |
| 5 | c |
| 6 | b |
| 7 | b |
| 8 | c |
| 9 | c |
| 10 | b |
🧠 Explanations
- Q1: Sag increases with conductor weight → (c).
- Q2: Higher tension → smaller sag → (b).
- Q3:
→ (a). - Q4: Pin insulator rating up to 33 kV → (b).
- Q5: Suspension used above 33 kV → (c).
- Q6: Flashover = air breakdown → (b).
- Q7: Formula shows
→ (b). - Q8:
→ (c). - Q9: Strain insulators used at dead ends → (c).
- Q10: Sag mainly due to conductor’s weight → (b).
🎯 Motivation / Why Practice Matters
Sag and insulators are high-weightage transmission line topics in ECET 2026. Questions are often direct formula-based or conceptual. Practicing daily helps you answer within 10–20 seconds, saving time for tougher numerical questions. Fast recall of formulas like 
gives a big edge over other students.
📲 CTA
👉 Join our WhatsApp Group for ECET 2026 updates, free notes & mock tests:
🔗 Join Now
