BS 7430 rod formula, NEC 250.53(A)(2)4 min read

Worked Example: Ground Rod Resistance — Does a 3 m Rod Reach 25 Ω in 100 Ω·m Soil?

A single 3 m rod misses the common 25 Ω target; doubling its length gets there, while a fatter rod barely helps.

Scenario

Soil resistivity100 Ω·m (uniform)
Rod16 mm diameter, driven vertically
Target25 Ω (NEC 250.53(A)(2) single-electrode exception)

Step-by-step calculation

Step 1: Resistance of a 3 m rod

R = ρ / (2πL) × [ln(8L/d) − 1]
100 / (2π × 3) × [ln(8 × 3 / 0.016) − 1]
R = 33.5 Ω — above 25 Ω

Step 2: Resistance of a 6 m rod (or two coupled 3 m sections)

Result summary

CheckRequirementActualStatus
3 m × 16 mm rod≤ 25 Ω33.5 Ω✗ FAIL
6 m × 16 mm rod≤ 25 Ω18.6 Ω✓ PASS
A single 3 m rod gives 33.5 Ω, so under NEC 250.53(A)(2) a supplemental electrode would be needed. Driving the rod to 6 m gives 18.6 Ω and meets the 25 Ω exception on its own.

Key insight: Resistance falls almost in proportion to rod length but only weakly with diameter, because diameter appears inside a logarithm. Deeper rods also tend to reach moister, lower-resistivity soil, so in practice the gain is often larger than this uniform-soil estimate.

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Frequently asked questions

Is two rods side by side the same as one rod twice as long?

Only if they are spaced at least about one rod length apart. Closer rods share the same soil and their combined resistance is noticeably higher than half of one rod.

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