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04-Geol-B10 · May 2017

Question 8 of 10: Deriving the Wenner Apparent-Resistivity Formula

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

EGBC National Exam — Geological Engineering, 04-Geol-B10-2 Electrical Methods, 2017-May. Closed book; no calculator permitted. All ten questions require an answer in essay format, with diagrams used wherever appropriate. The exam instructs "choose six (6) of the following ten (10) questions, the first six as they appear in the answer book will be marked, each of equal value, about half an hour each".

Reference texts: Telford, Geldart & Sheriff, Applied Geophysics, 2nd ed. (electrical properties of rocks ch.5; self-potential ch.6; induced polarization ch.9; resistivity ch.8; electromagnetic methods ch.7; magnetotellurics ch.10); Kearey, Brooks & Hill, An Introduction to Geophysical Exploration, 3rd ed. (resistivity arrays, EM systems, MT surveying, ch.8–9); Simpson & Bahr, Practical Magnetotellurics (MT instrumentation and robust/remote-reference processing, ch.2–6).

Question 8: Deriving the Wenner Apparent-Resistivity Formula (Choose 6 of 10 – equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Point-source half-space potential V(r) = Iρ/(2πr); Wenner array C1–P1–P2–C2 collinear with equal spacing a between every adjacent electrode.

Find. The apparent-resistivity formula ρa for the Wenner array in terms of the measured potential difference ΔV, injected current I, and spacing a.

C1 (+I) P1 P2 C2 (−I) a a a r=a (C1→P1)
Wenner geometry used in the derivation: equal spacing a between C1–P1, P1–P2, and P2–C2, so C1–P2 and C2–P1 are both 2a.

Approach

Superpose the potential from the two current electrodes (+I at C1, −I at C2) at each potential electrode, take the difference VP1 − VP2, and solve for ρ.

  1. Potential at P1 from both current electrodes. Distances from C1 and C2 to P1 are a and 2a respectively: $$V_{P1} = \dfrac{I\rho}{2\pi a} - \dfrac{I\rho}{2\pi (2a)}$$
  2. Potential at P2 from both current electrodes. By symmetry the distances from C1 and C2 to P2 are 2a and a respectively: $$V_{P2} = \dfrac{I\rho}{2\pi (2a)} - \dfrac{I\rho}{2\pi a}$$
  3. Measured potential difference. Subtracting and collecting terms: $$\Delta V = V_{P1}-V_{P2} = \dfrac{I\rho}{2\pi}\left(\dfrac{1}{a}-\dfrac{1}{2a}-\dfrac{1}{2a}+\dfrac{1}{a}\right) = \dfrac{I\rho}{2\pi}\left(\dfrac{2}{a}-\dfrac{1}{a}\right) = \dfrac{I\rho}{2\pi a}$$
  4. Solve for the apparent resistivity. Rearranging for ρ (relabelled ρa, the apparent resistivity, since a real half-space is only ever piecewise-homogeneous): $$\boxed{\rho_a = 2\pi a\left(\dfrac{\Delta V}{I}\right)}$$
QuantityResult
Potential difference ΔVIρ/(2πa)
Wenner apparent resistivity ρa2πa(ΔV/I)
Geometric factor k (Wenner)2πa

Why the field crew might prefer Schlumberger for a sounding

To deepen a Wenner sounding, ALL FOUR electrodes must be dug up and replanted farther apart at every reading — slow, and it repeatedly re-exposes the measurement to fresh near-surface lateral inhomogeneity under the moving potential electrodes. In a Schlumberger sounding, the potential pair P1P2 stays fixed (only the outer current electrodes are moved outward), so far fewer electrodes are relocated per sounding, the survey runs faster, and successive readings share the same potential-electrode footprint, improving internal consistency of the sounding curve.

Why the field crew might prefer Wenner along a profile

For a profile at FIXED spacing (not a sounding), the whole four-electrode Wenner set simply advances together along the line at constant a, which is operationally simple and, critically, gives the strongest possible signal (smallest geometric factor 2πa of any array at that separation) with the potential electrodes always close to and symmetric about the current electrodes — giving the best sensitivity to lateral resistivity changes directly beneath the array as it is walked along the line, which is exactly what a profile is trying to resolve.