Solution Exercise 6
Current injected [I]: 2 mA
Apparent resistivity [ρ]: 250 ohm-m
Noise level: 15 mV
[latex]\displaystyle K_{g}=\frac{2\pi }{\frac{1}{\overline{AM}}-\frac{1}{\overline{AN}}-\frac{1}{\overline{BM}}+\frac{1}{\overline{BN}}}[/latex]
[latex]\displaystyle K_{g,Wenner}=\frac{2\pi }{\frac{1}{5\mathrm{m}}-\frac{1}{10\mathrm{m}}-\frac{1}{10\mathrm{m}}+\frac{1}{5\mathrm{m}}}[/latex]
= 31.4 m
For a positive geometric factor with a dipole-dipole array, you need to have the electrodes configured as A B N M (this also gives a positive resistance). If you configure the electrodes as A B M N (as often written) the geometric factor is negative (so is the resistance). Using A B N M the geometric factor is:
[latex]\displaystyle K_{g,dipole-dipole}=\frac{2\pi }{\frac{1}{\overline{AM}}-\frac{1}{\overline{AN}}-\frac{1}{\overline{BM}}+\frac{1}{\overline{BN}}}[/latex] = [latex]\displaystyle \frac{2\pi }{\frac{1}{9\mathrm{m}}-\frac{1}{6\mathrm{m}}-\frac{1}{6\mathrm{m}}+\frac{1}{3\mathrm{m}}}[/latex] = 56.5 m
If you do the calculation as A B M N, you get a geometric factor of -56.5.
We then need to calculate the apparent resistivity:
ρa = KgΔV/I
ρa_Wenner = 250 ohm-m = [latex]\frac{31.4\ \textrm{m}\ V\ \textrm{mV}}{2\ \textrm{mA}}[/latex]
solving for V yields V = 15.9 mV
ρa_dipole–dipole = 250 ohm-m = [latex]\frac{56.5\ \textrm{m}\ V\ \textrm{mV}}{2\ \textrm{mA}}[/latex]
solving for V yields V = 8.8 mV
A negative voltage would be correct for A B M N configuration, as the geometric factor is negative.
In summary:
| Array | a(m) | n | Kg | V(mV) | SNR |
| Wenner | 5 | – | 31.4 | 15.9 | 1.06 |
| Dipole-dipole | 3 | 3 | 56.5 | 8.8 | 0.59 |