Electrical Methods: Conduction in Rocks, DC Resistivity, Apparent Chargeability, Layered-Earth Sounding and Interpretation

This chapter answers the Part B2 heading “Electrical Methods” of the Geology and Geophysics (GG) paper: “Conduction of electricity through rocks, electrical conductivities of metals, non-metals, rock-forming minerals and different rocks, concepts of D.C. resistivity measurement and depth of investigation; Apparent Resistivity and Apparent Chargeability, Theory of Reciprocity, Sounding and Profiling, Various electrode arrangements, application of linear filter theory, Sounding curves over multi-layered earth, interpretation of resistivity field data, Principles of equivalence and suppression.” It teaches conduction mechanisms and Archie’s law, the point-source potential and the geometric factors K, depth of investigation, sounding against profiling, reciprocity, induced polarisation, Dar Zarrouk parameters and curve types, the filter method, and equivalence and suppression. The numericals are ρa = K ΔV/I for Wenner, Schlumberger and dipole-dipole arrays, Archie resistivity, longitudinal resistivity of a stack, and chargeability.

1. Conduction through rocks, resistivities of earth materials and Archie’s law

Resistivity ρ (Ω·m) is the reciprocal of conductivity σ (S/m). Three mechanisms carry current. Electronic conduction runs through metals, graphite and many sulphides and oxides by free electrons. In metals conductivity falls as temperature rises. Metals and graphite are the best conductors, while ore minerals such as pyrite, chalcopyrite, galena and magnetite are semiconductors with a wide spread of values. Electrolytic conduction is by ions in pore water, and conductivity rises with temperature, about 2 percent per °C. Dielectric displacement in insulators matters only at high frequency. Rock-forming minerals (quartz, feldspar, mica, calcite, pyroxene, olivine) and sulphur are insulators when dry, so a rock conducts mainly through the water in its pores and cracks: resistivity falls with porosity, saturation and salinity and rises with compaction and cementation.

Approximate resistivities (ranges overlap and depend on water, clay and texture)
MaterialResistivity (Ω·m)What controls it
Native copperabout 2 × 10⁻⁸Free electrons
Massive sulphide, graphiteabout 10⁻⁵ to 1Connected conducting grains
Sea waterabout 0.2Dissolved salts
Wet clayabout 1 to 30Water and clay surface conduction
Fresh groundwater, saturated sand and graveltens to a few hundredPorosity, water quality
Dry crystalline rock (granite, gneiss)10³ to 10⁶Fractures and their water
Dry quartzabove 10¹⁰Insulator

Archie’s law for a clean, water-saturated rock: ρ₀ = a ρw φ^(−m), so the formation factor is F = ρ₀/ρw = a φ^(−m), with a ≈ 1 and cementation exponent m ≈ 2 (about 1.3 for loose sand to 2.5 for tight rock). At partial saturation Sw, ρt = ρ₀ Sw^(−n), with n ≈ 2. Example: ρw = 0.5 Ω·m, φ = 0.2, a = 1, m = 2 give ρ₀ = 0.5/0.04 = 12.5 Ω·m; at Sw = 0.5 the rock reads ρt = 12.5/0.25 = 50 Ω·m, so draining water raises resistivity.

⚠️ Archie fails in shaly rock
Clay minerals add surface conduction in parallel with the pore water, so shaly rock is more conductive than Archie predicts and a low resistivity may mean clay, not water. Also remember the temperature rule is opposite for metals and for electrolytes.

2. DC resistivity: point source, geometric factor, apparent resistivity and depth of investigation

A point electrode at the surface of a uniform half-space, injecting current I, spreads it over hemispheres of area 2πr², so the potential is V = ρ I/(2π r) (a full space gives 4π r²). With current electrodes A and B and potential electrodes M and N, superposition gives ΔV = (ρ I/2π)(1/AM − 1/BM − 1/AN + 1/BN), hence ρ = K ΔV/I with the geometric factor K = 2π/(1/AM − 1/BM − 1/AN + 1/BN). Over a real earth the same formula defines the apparent resistivity ρa: the resistivity of the uniform half-space that would give the same ΔV/I with that array. It is a weighted average over the volume sensed, not a mean of the layer resistivities.

Common arrays and their geometric factors
ArrayArrangementK
WennerA M N B equally spaced by a2π a
SchlumbergerAB = 2L, MN = 2l, with MN not above AB/5, centredπ (L² − l²)/(2l), about π L²/(2l)
Dipole–dipoleAB = MN = a, nearest electrodes B and M separated by n aπ n (n + 1)(n + 2) a

Derivation check, Wenner: AM = a, BM = 2a, AN = 2a, BN = a, so the bracket is 1/a − 1/2a − 1/2a + 1/a = 1/a and K = 2π a. Worked example: a = 20 m, ΔV = 50 mV, I = 0.5 A give K = 125.7 m and ρa = 125.7 × 0.05/0.5 = 12.57 Ω·m.

Depth of investigation has no sharp value, because current flows at all depths; it grows with electrode spacing. The usual measure is the median depth above which half the signal originates, about 0.5a for the Wenner array (0.17 of its total length 3a). For Schlumberger it rises with AB/2, and for dipole–dipole with n. To see a layer at depth z, the spacing must be a few times z, and a thin layer at great depth needs a large contrast.

3. Sounding, profiling, electrode arrangements and the theory of reciprocity

  • Sounding (VES): the array centre stays fixed while the spacing is increased, and ρa is plotted against spacing on log–log paper. It images vertical change and assumes horizontal layers (1D). In the Schlumberger array only A and B move for most readings, and M and N are widened now and then when ΔV becomes too small; a small step between overlapping segments flags a lateral inhomogeneity near MN. In the Wenner array all four electrodes move at every reading, which is slower and picks up near-surface lateral noise.
  • Profiling (constant-separation traversing): a fixed array is moved along a line, so ρa maps lateral change at a roughly constant depth: dikes, faults, cavities, buried channels. Wenner, Schlumberger and dipole–dipole are all used.
  • Imaging (ERT): many fixed electrodes, read in many combinations of spacing and position, are inverted in 2D or 3D, joining the two ideas.

Reciprocity. Interchanging the current pair and the potential pair leaves the measured resistance ΔV/I unchanged: R(AB; MN) = R(MN; AB). The geometric factor is symmetric too, so ρa is unchanged. Repeating a reading with the pairs swapped is therefore a check on data quality (the reciprocal error), and reciprocity lets the survey designer choose whichever arrangement is easier to lay out and underlies the sensitivity calculations used in imaging.

⚠️ Sounding is not profiling
A sounding changes spacing at a fixed centre to read depth; a profile changes position at a fixed spacing to read lateral change. A sounding interpreted as 1D layers is wrong near a vertical contact or fault, where lateral change is mistaken for depth change.

4. Apparent chargeability and induced polarisation

When the current is switched off, the voltage between M and N does not fall to zero at once but decays over seconds: the ground has stored charge, and this is induced polarisation (IP). Two causes dominate. Electrode polarisation occurs where ions in the pore path pile up against a metallic grain (disseminated sulphides, graphite, some oxides). Membrane polarisation occurs at clay particles and narrow pore throats that block ion movement. IP is therefore sensitive to the amount of polarisable mineral, while resistivity is sensitive to pore water.

In the time domain the decay voltage Vs(t) is compared with the steady voltage Vp: chargeability is M = (1/Vp) ∫ Vs(t) dt over a set window (in ms), or simply Vs/Vp read just after switch-off, in mV/V. Over a non-uniform earth the measured value is the apparent chargeability ma, the chargeability of the uniform earth that would give the same reading with the same array. Example: Vp = 800 mV and Vs = 24 mV give ma = 24/800 = 0.03 = 30 mV/V. In the frequency domain ρa is measured at a low and a higher frequency, and the percent frequency effect is PFE = 100 (ρlow − ρhigh)/ρhigh, which is positive because ρa falls as frequency rises.

5. Layered earth: Dar Zarrouk parameters and sounding-curve types

For layers with thickness hᵢ and resistivity ρᵢ, the longitudinal conductance is S = Σ hᵢ/ρᵢ (siemens) and the transverse resistance is T = Σ hᵢ ρᵢ (Ω·m²). Current flowing along the layers meets them in parallel, giving the longitudinal resistivity ρL = H/S; current crossing them meets them in series, giving the transverse resistivity ρT = T/H, where H = Σ hᵢ. Always ρT ≥ ρL, so a stack is electrically anisotropic, with coefficient of anisotropy λ = √(ρT/ρL) ≥ 1. Example: 10 m of 20 Ω·m over 30 m of 5 Ω·m give S = 0.5 + 6 = 6.5 S, T = 200 + 150 = 350 Ω·m², ρL = 40/6.5 = 6.15 Ω·m, ρT = 8.75 Ω·m and λ = 1.19.

Three-layer sounding-curve types
TypeResistivitiesCurve shape
Hρ1 > ρ2 < ρ3Falls to a minimum, then rises
Kρ1 < ρ2 > ρ3Rises to a maximum, then falls
Aρ1 < ρ2 < ρ3Rises throughout
Qρ1 > ρ2 > ρ3Falls throughout

At small spacing ρa tends to ρ1, the first layer, and at very large spacing to the resistivity of the last layer. Over a very resistive base the late branch rises at 45° on log–log paper: for a layer of longitudinal conductance S1 above an insulator, ρa = (AB/2)/S1 for the Schlumberger array, which gives S1 directly. Plotted curves are smoothed, since each reading averages over a volume, so a boundary appears as a gradual change, bending at a spacing larger than the depth of the boundary.

6. Linear filter theory, interpretation of field data, equivalence and suppression

For a layered earth the resistivity transform T(λ) is built from the bottom up: Tₙ = ρₙ, and Tᵢ = (Tᵢ₊₁ + ρᵢ tanh(λ hᵢ))/(1 + Tᵢ₊₁ tanh(λ hᵢ)/ρᵢ). The Schlumberger curve is then ρa(s) = s² ∫ T(λ) J1(λ s) λ dλ, with s = AB/2. Writing s = e^x and λ = e^(−y) turns this into a convolution in log variables, ρa(x) = ∫ T(y) f(x − y) dy, so it can be evaluated as a short weighted sum of T at a few points per decade with fixed filter coefficients. This is linear filter theory: a forward model costs milliseconds instead of a slow Bessel integral, and the same filter run backwards converts an observed curve into T(λ), from which layers are peeled.

Interpretation of field data follows a routine: plot ρa against spacing on log–log paper, identify the curve type and the number of layers, build a starting model by partial curve matching against master curves and auxiliary charts, then refine it by least-squares inversion that adjusts ρᵢ and hᵢ until the filter-computed curve fits the data. Check the result against boreholes, lithology and water levels, and remember that the 1D assumption fails near lateral contacts.

Equivalence and suppression
PrincipleWhen it arisesWhat stays determined
EquivalenceA middle layer more conductive than both neighbours (H type)Only S2 = h2/ρ2; thickness and resistivity trade off
EquivalenceA middle layer more resistive than both neighbours (K type)Only T2 = ρ2 h2
SuppressionA thin layer whose resistivity lies between those above and below (A and Q types)Nothing: it leaves no separate bend and is missed

Why. Current runs along a thin conductive layer, so only its conductance S matters; current has to cross a thin resistive layer, so only its resistance T matters. Example: a middle layer of 10 Ω·m and 5 m (S = 0.5 S) beneath 10 m of 100 Ω·m and above a 500 Ω·m base gives nearly the same curve, within a few percent, as one of 20 Ω·m and 10 m (also S = 0.5 S), but not as one of 10 Ω·m and 10 m (S = 1 S). Equivalence is ambiguity in a layer that is seen; suppression is failure to see a layer at all, which makes the depths and resistivities interpreted for the layers that are seen wrong, because the hidden layer is merged with a neighbour.

Key takeaways

  • Rocks conduct mainly through pore water (electrolytic), so resistivity falls with porosity, saturation and salinity. Archie: ρ₀ = a ρw φ^(−m) and ρt = ρ₀ Sw^(−n), with a ≈ 1, m ≈ n ≈ 2. Metals lose conductivity on heating; electrolytes gain it.
  • ρ = K ΔV/I, with K = 2π/(1/AM − 1/BM − 1/AN + 1/BN): Wenner 2π a; Schlumberger π(L² − l²)/(2l); dipole–dipole π n(n + 1)(n + 2) a. Apparent resistivity is a weighted average, and depth of investigation grows with spacing (about 0.5a for Wenner).
  • A sounding varies spacing at a fixed centre (vertical change, 1D); profiling moves a fixed array (lateral change). Reciprocity, R(AB; MN) = R(MN; AB), gives a data-quality check and design freedom.
  • IP is the voltage decay after current switch-off, from electrode polarisation (metallic grains) and membrane polarisation (clays). Apparent chargeability is Vs/Vp (mV/V); in the frequency domain PFE = 100 (ρlow − ρhigh)/ρhigh.
  • S = Σ h/ρ and T = Σ hρ; ρL = H/S ≤ ρT = T/H. Three-layer curve types: H (ρ1 > ρ2 < ρ3), K, A, Q. The curve starts at ρ1 and ends at the last layer’s resistivity.
  • The filter method turns the Bessel integral into a convolution in log variables, so forward models are fast. Equivalence: conductive middle layer fixes S, resistive middle layer fixes T. Suppression: a thin layer of intermediate resistivity is missed.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A metal wire and a water-saturated sandstone are both heated by 30 °C. How do their electrical conductivities change?

    1. Both fall, because greater thermal agitation scatters the charge carriers in each of them
    2. Both rise, because heating releases more charge carriers in each of them and widens the conduction paths
    3. The metal’s falls while the sandstone’s rises, because pore-water ions move more easily when warm
    4. The metal’s rises while the sandstone’s falls, because the pore water expands and dilutes its ions
    Show answer

    Answer: C — The metal’s falls while the sandstone’s rises, because pore-water ions move more easily when warm

    In a metal, heating increases scattering of the free electrons, so conductivity falls. A water-saturated rock conducts by ions in its pore water, whose mobility rises with temperature (about 2 percent per °C), so its conductivity rises. Treating the rock like the metal is the tempting error.

  2. Which ordering of typical resistivities, from lowest to highest, is correct?

    1. Sea water, wet clay, water-saturated sand, dry granite
    2. Wet clay, sea water, dry granite, water-saturated sand
    3. Sea water, water-saturated sand, wet clay, dry granite
    4. Water-saturated sand, sea water, dry granite, wet clay
    Show answer

    Answer: A — Sea water, wet clay, water-saturated sand, dry granite

    Sea water is the most conductive (about 0.2 Ω·m), wet clay is a few to a few tens of Ω·m because of its water and clay-surface conduction, saturated sand is tens to hundreds, and dry crystalline rock is 10³ Ω·m or more. Placing sand below clay ignores the surface conduction of clay minerals.

  3. In a DC resistivity survey the current is passed through M and N and the voltage is measured across A and B, which is the reverse of the original arrangement over the same ground. By the reciprocity theorem:

    1. ΔV/I changes sign but not magnitude, because the current direction relative to the line is reversed
    2. ΔV/I changes in proportion to the ratio of the lengths AB and MN, because the sampled volume differs
    3. ΔV/I changes unless the ground is homogeneous, because the current paths in the two cases differ
    4. ΔV/I is unchanged, so ρa is unchanged and the repeat reading serves as a check on data quality
    Show answer

    Answer: D — ΔV/I is unchanged, so ρa is unchanged and the repeat reading serves as a check on data quality

    Reciprocity says the transfer resistance is the same when source and receiver pairs are swapped, for any earth, uniform or not. The geometric factor is symmetric as well, so ρa is unchanged. The reciprocal error is therefore a standard data-quality check. The current paths do differ, but the resistance still matches.

  4. A buried vertical dike is to be located by its lateral resistivity contrast along a traverse. Which field procedure is the most suitable?

    1. A vertical electrical sounding with the spacing expanded about one fixed centre
    2. Constant-separation traversing, with a fixed array moved step by step along the line
    3. One long-spacing reading at the middle of the traverse, interpreted as horizontal layers
    4. A sounding at each end of the traverse only, interpreted as horizontal layers
    Show answer

    Answer: B — Constant-separation traversing, with a fixed array moved step by step along the line

    Lateral change at roughly constant depth is mapped by profiling: a fixed array is moved along the line and ρa is plotted against position. A sounding changes spacing at one centre and assumes horizontal layers, so it cannot locate a vertical contact, and it can mistake lateral change for depth change.

  5. A three-layer earth has ρ1 = 100 Ω·m, ρ2 = 10 Ω·m and ρ3 = 500 Ω·m, with layer 2 neither extremely thin nor extremely thick. Which description of its Schlumberger sounding curve is correct?

    1. A type, with ρa rising steadily and smoothly from about 100 Ω·m to about 500 Ω·m with no minimum
    2. K type, with ρa rising to a maximum near 500 Ω·m and then falling away to about 10 Ω·m at large spacing
    3. H type, with ρa starting near 100 Ω·m, passing a minimum above 10 Ω·m and rising towards 500 Ω·m
    4. Q type, with ρa falling from about 100 Ω·m to a minimum near 10 Ω·m and staying there at all larger spacings
    Show answer

    Answer: C — H type, with ρa starting near 100 Ω·m, passing a minimum above 10 Ω·m and rising towards 500 Ω·m

    ρ1 > ρ2 < ρ3 is the H type. The curve starts at ρ1 and ends at ρ3, and because a finite layer is averaged with its neighbours the minimum stays above ρ2. The Q type needs ρ1 > ρ2 > ρ3, and the K type needs the middle layer to be the most resistive.

  6. A sounding over 50 Ω·m ground shows a resistive middle layer (K type) of 500 Ω·m and 5 m thickness, beneath a 10 m top layer of 50 Ω·m and above a base that is again 50 Ω·m. Which alternative middle layer would reproduce the sounding curve most closely?

    1. 500 Ω·m and 10 m, which keeps the resistivity and makes the layer better resolved
    2. 250 Ω·m and 10 m, which keeps the transverse resistance ρh at 2500 Ω·m²
    3. 1000 Ω·m and 10 m, which keeps the longitudinal conductance h/ρ at 0.01 S
    4. 250 Ω·m and 5 m, which halves the resistivity but keeps the thickness
    Show answer

    Answer: B — 250 Ω·m and 10 m, which keeps the transverse resistance ρh at 2500 Ω·m²

    For a resistive middle layer, current has to cross it, so the curve depends mainly on its transverse resistance T = ρh. Here T = 500 × 5 = 2500 Ω·m², matched by 250 Ω·m and 10 m, whose curve differs by only a few percent. The model of 1000 Ω·m and 10 m preserves h/ρ = 0.01 S, the invariant of a conductive layer, so it is the tempting distractor, but its T is 10,000 Ω·m² and its curve differs greatly.

  7. In a resistivity sounding, which layer is most likely to be suppressed, that is, to leave no recognisable bend on the curve?

    1. A thick layer far more resistive than the layers above and below it, giving a clear maximum
    2. A thick layer far more conductive than both of its neighbours, giving a clear minimum
    3. The topmost layer, because ρa tends to ρ1 at the smallest spacing and so hides it
    4. A thin layer whose resistivity lies between those of the layers above and below it
    Show answer

    Answer: D — A thin layer whose resistivity lies between those of the layers above and below it

    A thin layer of intermediate resistivity (A and Q types) produces only a smooth change that is indistinguishable from a two-layer curve, so it is suppressed. A thick contrasting layer, resistive or conductive, makes a clear maximum or minimum, and the top layer is always seen at small spacing.

  8. In the linear filter method for a Schlumberger sounding, the apparent resistivity is obtained from the resistivity transform T(λ) by:

    1. a convolution in logarithmic variables, computed as a short weighted sum with fixed filter coefficients
    2. a Fourier transform in time of the voltage decay recorded after the current is switched off
    3. a least-squares straight-line fit to segments of the logarithmic sounding curve
    4. a graphical overlay of the observed curve on printed master curves of fixed thickness and resistivity ratios
    Show answer

    Answer: A — a convolution in logarithmic variables, computed as a short weighted sum with fixed filter coefficients

    Writing s = e^x and λ = e^(−y) turns ρa(s) = s² ∫ T(λ) J1(λ s) λ dλ into a convolution, which is computed as a short sum of T at a few points per decade with fixed coefficients. The time-domain decay belongs to IP, and curve matching and line fitting are different, non-filter procedures.

  9. Which statement about induced polarisation (IP) is correct?

    1. The decay voltage after switch-off is caused by eddy currents that continue to flow in the ground for some time
    2. Electrode polarisation arises at metallic grains, so IP can detect sulphides that barely change the resistivity
    3. Apparent chargeability is measured in Ω·m, like apparent resistivity, because both are computed from ΔV/I and K
    4. Membrane polarisation occurs only in clean rocks with wide pores, where ions move freely and cannot be blocked
    Show answer

    Answer: B — Electrode polarisation arises at metallic grains, so IP can detect sulphides that barely change the resistivity

    Electrode polarisation at metallic grains is the main IP mechanism for disseminated ore, and it responds to the amount of polarisable mineral rather than to pore-water resistivity. IP decay is a stored-charge effect, not eddy currents, which belong to EM methods. Chargeability is in mV/V or ms, and membrane polarisation needs clays or narrow pore throats.

  10. Which of the following statements about electrode arrays and apparent resistivity are correct?

    1. For the Wenner array with spacing a, ρa = 2π a ΔV/I
    2. In a Schlumberger sounding all four electrodes are moved at every reading
    3. The depth of investigation of an array increases as its electrode spacing increases
    4. Apparent resistivity is the arithmetic mean of the true resistivities of the layers present
    Show answer

    Answer: A — For the Wenner array with spacing a, ρa = 2π a ΔV/I; C — The depth of investigation of an array increases as its electrode spacing increases

    The Wenner geometric factor is 2π a, and a larger spacing samples deeper. In a Schlumberger sounding mainly A and B are expanded, with M and N widened only occasionally. Apparent resistivity is a weighted average over the sensed volume that depends on the array, not a simple arithmetic mean of layer values.

  11. Which of the following statements about the Dar Zarrouk parameters of a stack of horizontal layers are correct?

    1. The longitudinal conductance S is the sum of the products h × ρ over the layers
    2. The longitudinal resistivity of the stack never exceeds its transverse resistivity
    3. A stack of 10 m at 20 Ω·m over 30 m at 5 Ω·m has a total longitudinal conductance of 6.5 S
    4. A stack of layers with different resistivities is electrically isotropic on the scale of the whole stack
    Show answer

    Answer: B — The longitudinal resistivity of the stack never exceeds its transverse resistivity; C — A stack of 10 m at 20 Ω·m over 30 m at 5 Ω·m has a total longitudinal conductance of 6.5 S

    S is the sum of h/ρ (not hρ, which gives the transverse resistance T). Current along the layers sees them in parallel and across them in series, so ρL = H/S never exceeds ρT = T/H and the stack is anisotropic. Numerically S = 10/20 + 30/5 = 0.5 + 6 = 6.5 S.

  12. Which of the following statements about Archie’s law for a clean, clay-free rock are correct?

    1. The formation factor F = ρ₀/ρw depends on porosity and cementation, not on the water resistivity
    2. The law applies unchanged to clay-rich rock, because clay minerals add no conduction of their own
    3. Warming the pore water raises its resistivity, so the resistivity of the whole rock also rises
    4. Reducing the water saturation Sw raises the resistivity above its saturated value by the factor Sw^(−n)
    Show answer

    Answer: A — The formation factor F = ρ₀/ρw depends on porosity and cementation, not on the water resistivity; D — Reducing the water saturation Sw raises the resistivity above its saturated value by the factor Sw^(−n)

    F = a φ^(−m) involves only the pore system, and ρt = ρ₀ Sw^(−n) shows that draining raises resistivity. Clay adds surface conduction in parallel, so the clay-free law fails there, and warming pore water lowers its resistivity.

  13. A Wenner array with spacing a = 20 m measures ΔV = 50 mV for a current I = 0.5 A. The apparent resistivity, in Ω·m to two decimal places, is ____ .

    Numerical answer — type the value.

    Show answer

    Answer: 12.57

    K = 2π a = 2π × 20 = 125.66 m. ρa = K ΔV/I = 125.66 × 0.050/0.5 = 12.566, which is 12.57 Ω·m. The usual slip is to forget to convert mV to V.

  14. A Schlumberger array has AB = 200 m and MN = 10 m, with the geometric factor K = π (L² − l²)/(2l), where L = AB/2 and l = MN/2. For a current of 0.8 A the measured voltage is 20 mV. The apparent resistivity, in Ω·m to one decimal place, is ____ .

    Numerical answer — type the value.

    Show answer

    Answer: 78.3

    L = 100 m and l = 5 m, so K = π (10000 − 25)/(2 × 5) = π × 997.5 = 3133.7 m. ρa = K ΔV/I = 3133.7 × 0.020/0.8 = 78.34 Ω·m, which is 78.3. Using L = 200 (the whole AB) is the common slip.

  15. A dipole–dipole array has AB = MN = a = 10 m, and the nearest electrodes B and M are n a apart with n = 2. Using the geometric factor K = π n (n + 1)(n + 2) a, K in metres, to the nearest integer, is ____ .

    Numerical answer — type the value.

    Show answer

    Answer: 754

    K = π × 2 × 3 × 4 × 10 = 240π = 753.98 m, so 754. The factor n(n + 1)(n + 2) grows fast with n, which is why a large n needs a large current to keep ΔV measurable.

  16. For a clean water-saturated rock, Archie’s law is ρ₀ = a ρw φ^(−m), and at partial saturation ρt = ρ₀ Sw^(−n). Take a = 1, m = 2, n = 2, ρw = 0.5 Ω·m, porosity φ = 0.2 and Sw = 0.5. The rock resistivity ρt, in Ω·m, is ____ .

    Numerical answer — type the value.

    Show answer

    Answer: 50

    ρ₀ = 1 × 0.5 × 0.2⁻² = 0.5/0.04 = 12.5 Ω·m. Then ρt = 12.5 × 0.5⁻² = 12.5 × 4 = 50 Ω·m. Draining half the water quadruples the resistivity for n = 2.

  17. A stack has a 10 m layer of 20 Ω·m above a 30 m layer of 5 Ω·m. The longitudinal resistivity of the stack, ρL = H/S with H the total thickness and S the total longitudinal conductance, in Ω·m to two decimal places, is ____ .

    Numerical answer — type the value.

    Show answer

    Answer: 6.15

    H = 40 m and S = 10/20 + 30/5 = 0.5 + 6 = 6.5 S, so ρL = 40/6.5 = 6.154, which is 6.15 Ω·m. The arithmetic mean (12.5) and the thickness-weighted mean of ρ (8.75, which is ρT) are the wrong averages for current flowing along the layers.

  18. In a time-domain IP measurement the steady primary voltage is Vp = 1250 mV and the secondary voltage read just after switch-off is Vs = 50 mV. The apparent chargeability Vs/Vp, in mV/V, is ____ .

    Numerical answer — type the value.

    Show answer

    Answer: 40

    Vs/Vp = 50/1250 = 0.04, which is 40 mV per V (equivalently 4 percent). The ratio is dimensionless, so mV over mV scaled to mV/V gives 40, not 0.04 mV/V.