---
title: "Earth resistance factors and plate electrode size"
subject: "Industrial Power Distribution and Illumination"
chapter: 2
key: "semester_6/industrial_power_distribution_and_illumination:earth-resistance-plate-50ohm-m"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/earth-resistance-plate-50ohm-m
site: BCT Engineering
---

# Earth resistance factors and plate electrode size

Industrial Power Distribution and Illumination · Chapter 2: Earthing

- Asked in: 2075 Bhadra, 8 marks
- Video: [Types of Earthing used in Electrical system](https://www.youtube.com/watch?v=eMR1rT8p-oA) — ELECTRICAL ENGINEER
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/earth-resistance-plate-50ohm-m
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-2

## Question

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.

## 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.

## Answer

#### Factors influencing 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
For a plate electrode buried vertically (IS 3043):

```text
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:

```text
√(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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which formula is used for the resistance of a vertically buried plate electrode?
   A. R = (rho / 2 pi L) ln(4L/d)
   B. R = rho / A
   C. R = (rho/4) * sqrt(pi/A)
   D. R = rho * sqrt(A) / pi
   Answer: C. IS 3043 gives R = (rho/4) sqrt(pi/A) for a plate with face area A. (Key point: Plate formula)

2. For rho = 50 ohm-m and R = 10 ohm, what plate area is required?
   A. About 4.91 m^2
   B. About 1.25 m^2
   C. About 2.22 m^2
   D. About 15.7 m^2
   Answer: A. A = pi (rho/4R)^2 = pi (50/40)^2 = pi * 1.5625 = 4.91 m^2. (Key point: Plate area result)

3. What is the side of the equivalent square plate?
   A. About 4.91 m
   B. About 1.25 m
   C. About 0.6 m
   D. About 2.22 m
   Answer: D. Side = sqrt(4.909) = 2.216 m. (Key point: Side of square plate)

4. How is the required resistance obtained in practice instead of one huge plate?
   A. Standard 60 cm x 60 cm plates in parallel with charcoal and salt treatment
   B. One thinner plate buried shallower
   C. Leaving the pit dry
   D. Using a plate of insulating material
   Answer: A. Several standard plates in parallel plus soil treatment reach the low resistance without an impractical plate. (Key point: Practical plates)

## Close variants set in other exams

- [Plate earth electrode size numerical](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aplate-electrode-60-ohm) — Industrial Power Distribution and Illumination, chapter 2, asked 4×

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