- Asked 2/20
- 2075 Bhadra · 8 marks
Earth resistance factors and plate electrode size
Short answer
Soil type, moisture, salts, temperature, electrode size/depth, number/spacing, contact condition, and corrosion influence earth resistance. For a plate, R = (ρ/4)√(π/A) gives area A = π(ρ/(4R))² = π(50/40)² ≈ 4.91 m², i.e., a square plate of about 2.22 m × 2.22 m.
Explain the different factors influencing the Earth Resistance. Calculate the size of the earth electrode for the Plate Earthing system, if the soil resistivity is 50 Ω-m and required earth resistance is 10 Ω.
Factors influencing earth resistance
- Type of soil: clay, loam and garden soil have low resistivity; sand, gravel and rock have very high resistivity.
- Moisture content: resistivity falls sharply as moisture rises. Dry soil has very high resistivity, so earth resistance is highest in the dry season.
- Dissolved salts (chemical content): salts make soil water conductive. Adding salt, charcoal or bentonite lowers the resistance.
- Temperature: resistivity rises as the soil cools, very sharply when soil water freezes.
- Size and shape of electrode: a larger plate or longer rod has more contact with soil and lower resistance. Rod length matters much more than diameter.
- Depth of electrode: deeper electrodes reach moist, stable soil, so resistance is lower and changes less with season.
- Number and spacing of electrodes: electrodes in parallel lower the resistance, provided they are spaced at least about twice their length apart.
- Contact with soil: loose, dry backfill increases resistance; compacted, treated backfill reduces it.
- Condition of electrode and joints: corrosion and loose connections increase resistance with time.
Size of plate electrode
For a plate electrode buried vertically (IS 3043):
R = (ρ/4) × √(π/A)
R = earth resistance (Ω), ρ = soil resistivity (Ω-m), A = area of one face of the plate (m²).
Given ρ = 50 Ω-m, R = 10 Ω:
√(π/A) = 4R/ρ
A = π × (ρ / 4R)²
= π × (50 / 40)²
= π × 1.5625
= 4.909 m²
Side of square plate = √4.909 = 2.216 m
Check: R = (50/4) × √(π/4.909) = 12.5 × 0.8 = 10.0 Ω.
Answer: Plate area ≈ 4.91 m², i.e. a square plate of about 2.22 m × 2.22 m.
In practice a single plate this big is not used. Standard plates (60 cm × 60 cm) are used in parallel, with charcoal and salt treatment of the pit, to reach the required resistance.
✓ Answer checked by Claude Opus 5.5 (high effort)
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