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Earthing objectives, factors and plate earthing size

Short answer

Soil type, moisture, salts, temperature, electrode size/depth, number/spacing, and contact condition affect earth resistance. Assuming a required 5 Ω, R = (ρ/4)√(π/A) gives plate area A = π(40/20)² ≈ 12.57 m² (square ~3.54 m); for 10 Ω, area ≈ 3.14 m².

What are the main objectives of an equipment earthing? Explain in brief the various factors affecting the earth resistance. Calculate the size of plate earthing of the building if soil resistivity is 40 Ω m.

Main objectives of equipment earthing

  1. Protect people from electric shock: keeps exposed metal parts near earth potential when insulation fails.
  2. Ensure quick operation of protection: the low-resistance path allows a large fault current, so fuses/MCBs/relays clear the fault fast.
  3. Protect equipment and buildings from fire caused by leakage and arcing.
  4. Discharge lightning surges and static charges safely.

Factors affecting earth resistance

  1. Type of soil: clay, loam and garden soil have low resistivity; sand, gravel and rock have very high resistivity.
  2. Moisture content: resistivity falls sharply as moisture rises. Dry soil has very high resistivity, so earth resistance is highest in the dry season.
  3. Dissolved salts (chemical content): salts make soil water conductive. Adding salt, charcoal or bentonite lowers the resistance.
  4. Temperature: resistivity rises as the soil cools, very sharply when soil water freezes.
  5. 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.
  6. Depth of electrode: deeper electrodes reach moist, stable soil, so resistance is lower and changes less with season.
  7. Number and spacing of electrodes: electrodes in parallel lower the resistance, provided they are spaced at least about twice their length apart.
  8. Contact with soil: loose, dry backfill increases resistance; compacted, treated backfill reduces it.
  9. Condition of electrode and joints: corrosion and loose connections increase resistance with time.

Size of plate electrode

The required earth resistance is not given. It is assumed to be 5 Ω, a common limit for building installations.

Using IS 3043:

R = (ρ/4) × √(π/A)
A = π × (ρ / 4R)²
  = π × (40 / (4×5))²
  = π × 2²
  = 12.57 m²
Side = √12.57 = 3.54 m

Answer: Plate area ≈ 12.57 m² (a square plate of about 3.54 m × 3.54 m) for R = 5 Ω. For R = 10 Ω, the same formula gives A = π × (40/40)² = 3.14 m² (about 1.77 m × 1.77 m).

Because such large plates are impractical, standard 60 cm × 60 cm plates are used in parallel, with charcoal and salt treatment, to reach the required value.

✓ Answer checked by Claude Opus 5.5 (high effort)

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