--- title: "Earth resistance factors and rod earthing diameter" subject: "Industrial Power Distribution and Illumination" chapter: 2 key: "semester_6/industrial_power_distribution_and_illumination:earth-resistance-rod-diameter" source: https://www.bctengineer.com/q/semester_6/industrial_power_distribution_and_illumination/earth-resistance-rod-diameter site: BCT Engineering --- # Earth resistance factors and rod earthing diameter Industrial Power Distribution and Illumination · Chapter 2: Earthing - Asked in: 2076 Bhadra, 4 marks - Video: [Earthing System Design & Calculation | Grid Earthing vs Rod Earthing Explained with Example](https://www.youtube.com/watch?v=3gRRihaIiW4) — Rashid iqubal - 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-rod-diameter - Whole chapter: https://www.bctengineer.com/md/semester_6/industrial_power_distribution_and_illumination/chapter-2 ## Question What are the factors affecting the earth resistance? Calculate the diameter of Rod Earthing system, so that the resistance is 10 Ω. Use the length of rod in 5 m and take the soil resistivity to be 65 Ω-m. ## Short answer Soil type, moisture, salts, temperature, electrode size/depth, number/spacing, contact condition, and corrosion affect earth resistance. Using R = (ρ/(2πL)) ln(4L/d), solving for d gives d = 4L / exp(R·2πL/ρ) = 0.159 m (about 15.9 cm). ## Answer #### 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. #### Diameter of rod electrode For a rod (pipe) electrode driven vertically (IS 3043, written in SI units): ```text R = (ρ / 2πL) × ln(4L / d) ``` R = earth resistance (Ω), ρ = soil resistivity (Ω-m), L = length of rod (m), d = diameter of rod (m). Given R = 10 Ω, ρ = 65 Ω-m, L = 5 m: ```text ln(4L/d) = R × 2πL / ρ = 10 × 2π × 5 / 65 = 4.833 4L/d = e^4.833 = 125.61 d = 4L / 125.61 = 20 / 125.61 = 0.1592 m ``` Check: R = (65 / (2π×5)) × ln(20/0.1592) = 2.069 × 4.833 = 10.0 Ω. **Answer: Diameter of rod ≈ 0.159 m (about 15.9 cm).** Such a thick rod is not practical. Since resistance depends only weakly (logarithmically) on diameter, in practice a standard rod (16–25 mm) is made longer, or several rods are used in parallel, or the soil is treated, to get 10 Ω. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which formula gives the resistance of a vertical rod electrode used here? A. R = (rho/4) * sqrt(pi/A) B. R = (rho / 2 pi L) * ln(4L/d) C. R = rho * L / d D. R = rho * 2 pi L * ln(d/4L) Answer: B. The rod formula is R = (rho / 2 pi L) ln(4L/d), with L the length and d the diameter. (Key point: Rod resistance formula) 2. For R = 10 ohm, rho = 65 ohm-m and L = 5 m, what is ln(4L/d)? A. About 4.83 B. About 2.07 C. About 125.6 D. About 0.16 Answer: A. ln(4L/d) = R * 2 pi L / rho = 10 * 2 pi * 5 / 65 = 4.833. (Key point: Solving for ln(4L/d)) 3. What diameter results from the calculation? A. About 1.59 m B. About 15.9 mm C. About 0.159 m D. About 2.5 cm Answer: C. 4L/d = e^4.833 = 125.6, so d = 20/125.6 = 0.159 m. (Key point: Final diameter) 4. Why is such a thick rod not used in practice to reach 10 ohm? A. Resistance depends strongly on diameter, so a thicker rod is hazardous B. Thick rods always corrode faster C. Resistance depends only weakly (logarithmically) on diameter, so longer or parallel rods or soil treatment are used instead D. Thin rods give lower resistance than thick rods Answer: C. Because of the log dependence, diameter changes little; length, multiple rods or treatment are more effective. 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