---
title: "Plate earth electrode size numerical"
subject: "Industrial Power Distribution and Illumination"
chapter: 2
key: "semester_6/industrial_power_distribution_and_illumination:plate-electrode-60-ohm"
source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/plate-electrode-60-ohm
site: BCT Engineering
---

# Plate earth electrode size numerical

Industrial Power Distribution and Illumination · Chapter 2: Earthing

- Asked in: 2069 Bhadra, 4 marks
- 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/plate-electrode-60-ohm
- Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-2

## Question

Calculate the size of the earth electrode for plate earthing system, if the soil resistivity is 60 Ω and required earth resistance is 8 Ω.

## Short answer

Using the plate electrode formula R = (ρ/4)√(π/A) with ρ = 60 Ω·m and required R = 8 Ω, the required plate area is A = π × (ρ / 4R)² = 11.04 m², corresponding to a square plate side of 3.32 m. A single plate of this size is impractical; practical installations use multiple standard plates with treatment.

## Answer

For a plate electrode buried vertically, IS 3043 gives:

```text
R = (ρ/4) × √(π/A)
```
R = earth resistance (Ω), ρ = soil resistivity (Ω-m), A = area of one face of the plate (m²).

The soil resistivity is taken as **60 Ω-m** (the unit is printed as Ω). Required R = 8 Ω.

```text
√(π/A) = 4R/ρ
A = π × (ρ / 4R)²
  = π × (60 / 32)²
  = π × (1.875)²
  = π × 3.5156
  = 11.04 m²
Side of square plate = √11.04 = 3.32 m
```

Check: R = (60/4) × √(π/11.04) = 15 × 0.5333 = 8.0 Ω.

**Answer: Area of plate ≈ 11.04 m², i.e. a square plate of about 3.32 m × 3.32 m.**

A single plate of this size is not practical. Standard plates (60 cm × 60 cm × 6.3 mm GI) are used in parallel, with charcoal and salt treatment of the pit, or deeper rod/pipe electrodes, to obtain 8 Ω.

## Check yourself

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

1. How is plate area found from R = (rho/4) sqrt(pi/A)?
   A. A = pi (4R/rho)^2
   B. A = pi (rho/4R)^2
   C. A = rho/(4R)
   D. A = 4R/(pi rho)
   Answer: B. sqrt(pi/A) = 4R/rho, so A = pi (rho/4R)^2. (Key point: Formula rearrangement)

2. For rho = 60 ohm-m and R = 8 ohm, what is the area?
   A. About 3.32 m^2
   B. About 11.04 m^2
   C. About 1.875 m^2
   D. About 24.1 m^2
   Answer: B. A = pi (60/32)^2 = pi * 3.5156 = 11.04 m^2. (Key point: Area result)

## Close variants set in other exams

- [Equipment earthing objective and plate electrode size](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aearthing-objective-plate-electrode) — Industrial Power Distribution and Illumination, chapter 2, asked 3×
- [Earth resistance factors and plate electrode size](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aearth-resistance-plate-50ohm-m) — Industrial Power Distribution and Illumination, chapter 2, asked 2×

## Related in Industrial Power Distribution and Illumination

- [Earthing objectives, factors and plate earthing size](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aearthing-objectives-plate-40ohm-m) — Industrial Power Distribution and Illumination, chapter 2
- [Earth resistance factors and rod earthing diameter](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aearth-resistance-rod-diameter) — Industrial Power Distribution and Illumination, chapter 2
- [Equipment earthing objective and earth resistance factors](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aearthing-objective-factors) — Industrial Power Distribution and Illumination, chapter 2, asked 2×
- [Reducing earth resistance and pipe earthing](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Areduce-earth-resistance-pipe-earthing) — Industrial Power Distribution and Illumination, chapter 2
- [Earth resistance factors and lightning arrester earthing](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aearth-resistance-lightning-arrester) — Industrial Power Distribution and Illumination, chapter 2
- [Low earth resistance and equipment earthing methods](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Alow-earth-resistance-methods) — Industrial Power Distribution and Illumination, chapter 2

## Related in other subjects

- [Soil resistivity and reducing earth resistance](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Asoil-resistivity-earth-resistance) — Switchgear and Protection, chapter 5
- [Chemical earthing and earth resistance](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Achemical-earthing-resistance) — Switchgear and Protection, chapter 5
- [Tower footing resistance of 16 Ω electrode design](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Atower-footing-16-ohm-electrodes) — High Voltage Engineering, chapter 2, asked 3×
- [Tower footing resistance of 20 Ω electrode design](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Atower-footing-20-ohm-electrodes) — High Voltage Engineering, chapter 2, asked 3×