Question

Wenner DC resistivity survey. The operator forgot to cancel self-potential (SP) between potential electrodes before injecting current. With equal current magnitude: measured \(\Delta V_1=+158\ \text{mV}\) for current \(C1\rightarrow C2\) and \(\Delta V_2=-214\ \text{mV}\) for reversed current \(C2\rightarrow C1\). Determine the SP (in mV, integer) that existed between potential electrodes before current injection.

Hint:

To recover SP quickly: take two readings with reversed current; the SP equals \(\dfrac{\Delta V_1+\Delta V_2}{2}\), the ohmic part equals \(\dfrac{\Delta V_1-\Delta V_2}{2}\).

Answer 1

Step 1: Build a superposition model of measured voltage.
Let \(V_{SP}\) be the static electrode SP (does not change with current reversal) and \(V_I\) be the ohmic voltage due to injected current (changes sign when current reverses). Then \[ \Delta V_{\text{meas}} = V_{SP} \pm V_I. \] With given signs: \[ \begin{aligned} V_{SP}+V_I &= +158\ \text{mV}\quad\text{(forward)},
V_{SP}-V_I &= -214\ \text{mV}\quad\text{(reverse)}. \end{aligned} \]

Step 2: Solve the 2\(\times\)2 linear system.
Add equations to eliminate \(V_I\): \[ 2V_{SP} = (+158) + (-214) = -56 \Rightarrow V_{SP} = -28\ \text{mV}. \] (Then \(V_I=158-(-28)=186\ \text{mV}\) — optional check.)

Step 3: Consistency verification.
Insert into the second equation: \[ V_{SP}-V_I = -28-186 = -214\ \text{mV}\quad\checkmark \] Thus the SP offset was \(-28\ \text{mV}\) (negative electrode at the second potential stake).

Final Answer:\ \(\boxed{-28\ \text{mV}}\)