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Wenner's method of earth resistivity measurement

Short answer

Wenner’s method uses four electrodes in a line at spacing a; current I is injected via outer electrodes and voltage V measured across inner electrodes to yield resistance R = V/I. For depth b << a, earth resistivity is ρ = 2π a R, eliminating contact resistance.

Explain Wenner's test method for measurement of earth resistivity with diagram and derive the relation for the resistivity.

Wenner's four-electrode method measures the average resistivity of soil by passing current between two outer electrodes and measuring the voltage between two inner electrodes, all equally spaced in a straight line. It removes the effect of electrode contact resistance.

Arrangement

Four electrodes C1, P1, P2, C2 are driven in a straight line at equal spacing aa, to a small depth bb (b≪ab \ll a, usually b<a/20b < a/20). A current II is passed through the outer (current) electrodes C1 and C2 from an earth tester, and the voltage VV between the inner (potential) electrodes P1 and P2 is measured.

        Earth tester (megger)
     +---I----------------------+
     |      +---V---+           |
     |      |       |           |
     C1     P1      P2          C2
  ---|------|-------|-----------|--- ground
     |<--a->|<--a-->|<----a---->|
     depth b (b << a)

Derivation

Treat each electrode as a point source of current at the surface. Current II from a point spreads into the earth over hemispheres. At distance rr the current density and field are

J=I2πr2,E=ρJ=ρI2πr2J = \frac{I}{2\pi r^2}, \qquad E = \rho J = \frac{\rho I}{2\pi r^2}

so the potential at distance rr is

V(r)=∫r∞E dr=ρI2πrV(r) = \int_r^\infty E\,dr = \frac{\rho I}{2\pi r}

Current +I+I enters at C1 and −I-I leaves at C2. Using superposition:

Potential at P1 (distance aa from C1, 2a2a from C2):

VP1=ρI2π(1a−12a)=ρI4πaV_{P1} = \frac{\rho I}{2\pi}\left(\frac{1}{a} - \frac{1}{2a}\right) = \frac{\rho I}{4\pi a}

Potential at P2 (distance 2a2a from C1, aa from C2):

VP2=ρI2π(12a−1a)=−ρI4πaV_{P2} = \frac{\rho I}{2\pi}\left(\frac{1}{2a} - \frac{1}{a}\right) = -\frac{\rho I}{4\pi a}

Voltage between P1 and P2:

V=VP1−VP2=ρI2πaV = V_{P1} - V_{P2} = \frac{\rho I}{2\pi a}

The tester reads R=V/IR = V/I, so

ρ=2πaR\rho = 2\pi a R

with ρ\rho in Ω-m\Omega\text{-m}, aa in m and RR in Ω\Omega.

If the electrode depth bb is not small compared with aa (image of each electrode taken), the general relation is

ρ=4πaR1+2aa2+4b2−aa2+b2\rho = \frac{4\pi a R}{1 + \dfrac{2a}{\sqrt{a^2+4b^2}} - \dfrac{a}{\sqrt{a^2+b^2}}}

which reduces to ρ=2πaR\rho = 2\pi aR when b≪ab \ll a.

Procedure and notes

  • Readings are taken at several spacings (aa = 1, 2, 5, 10 m …) and in different directions. The current spreads roughly to a depth equal to aa, so the change of ρ\rho with aa shows layered soil.
  • Measurements are made in dry season to get the worst (highest) value for earthing design.
  • An AC or reversing-DC tester is used to avoid polarisation and stray currents.

Example: with a=5a = 5 m and R=3.2 ΩR = 3.2\ \Omega, ρ=2π×5×3.2=100.5 Ω-m\rho = 2\pi \times 5 \times 3.2 = 100.5\ \Omega\text{-m}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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