ElectroHub

Earth Pit Resistance Calculator (IS 3043)

For a driven rod the resistance is ρ/(2πL) × [ln(8L/d) − 1]. Read where the geometry sits: length is outside the logarithm and diameter is inside it. Doubling the length nearly halves the resistance; doubling the diameter buys about ten per cent. Driving deeper always beats driving fatter, and that is the whole design rule in one line.

Earth pit / electrode resistance

For a driven rod, R = ρ / (2πL) × [ln(8L/d) − 1]. Length sits outside the logarithm and diameter inside it, which is the entire design rule in one line: driving deeper works, driving fatter barely does.

One electrode
Ideal R / n
Group resistance
Short of the 5 Ω target. About 12 rods would reach it, spaced at least one rod length apart.
  • Soil resistivity is the input that dominates the answer and it is the one nobody knows accurately. It varies by a factor of ten between wet and dry season on the same site, so a pit that measures 4 Ω in August can measure 12 Ω in April.
  • This is a design estimate. IS 3043 expects the installed electrode to be measured — by fall-of-potential or a clamp meter — and re-measured periodically, because that is the only number that counts.
Common trap: Four rods do not give a quarter of the resistance. Each rod sits inside its neighbours' voltage funnel, so the honest figure is roughly R × 1.36 / 4 — about a third, not a quarter. Designing on the ideal R/n is how an installation misses its target by 40 %.
Rod formula after Dwight, as reduced in IS 3043 and IEEE 142; plate formula R = ρ/4 × √(π/A). Soil resistivity is a seasonal figure that can move by a factor of ten on one site, so this is a design estimate. IS 3043 expects the installed pit to be measured by fall-of-potential and re-measured periodically.

The formula

Rod: R = ρ/(2πL) × [ln(8L/d) − 1] ; Plate: R = ρ/4 × √(π/A) ; n rods: R × F / n

ρ
soil resistivity, Ω·m — 10 in marsh, 100 in loam, 1000 in dry sand
L
buried length of the rod, metres
d
rod diameter, metres
A
plate face area, m²
F
interference factor for a group — 1.36 for four rods, and always above 1

Worked example

A 3 m rod, 25 mm diameter, in loam of 100 Ω·m.

  1. ρ/(2πL) = 100 / (2π × 3) = 5.305
  2. ln(8L/d) = ln(8 × 3 / 0.025) = ln(960) = 6.867
  3. R = 5.305 × (6.867 − 1)
  4. Four such rods: 31.1 × 1.36 / 4 = 10.6 Ω, not the 7.8 Ω that R/4 suggests

31.1 Ω for one rod. Reaching 5 Ω in this soil takes about twelve rods, not six.

Where you'll use it

Designing an earthing system before it is dug, and explaining why an installed pit missed its target. This sizes the electrode; the protective conductor that connects to it is a separate calculation. Soil resistivity moves by a factor of ten between wet and dry season, so IS 3043 expects the finished pit to be measured by fall-of-potential, never assumed.

The laws behind it

Parts this applies to

Resistor (fixed)Thermistor (NTC / PTC)Photoresistor (LDR)Resettable Fuse (PTC / Polyfuse)Temperature Sensor (thermocouple/RTD/IC)Humidity Sensor (DHT etc)Gas / Air-Quality Sensor (MQ)Load Cell / Strain Gauge

Related calculators

Earthing Conductor Size CalculatorS = √(I² t) / kShort Circuit Fault Current CalculatorI_sc = I_fl × 100 / %ZResistivity Calculator (R = ρL/A)R = ρ × L / AMCB & RCCB Selection CalculatorI_b ≤ I_n ≤ I_z