# Solution Exercise 6

Current injected [I]: 2 mA

Apparent resistivity [ρ]: 250 ohm-m

Noise level: 15 mV

$\displaystyle K_{g}=\frac{2\pi }{\frac{1}{\overline{AM}}-\frac{1}{\overline{AN}}-\frac{1}{\overline{BM}}+\frac{1}{\overline{BN}}}$

$\displaystyle K_{g,Wenner}=\frac{2\pi }{\frac{1}{5\mathrm{m}}-\frac{1}{10\mathrm{m}}-\frac{1}{10\mathrm{m}}+\frac{1}{5\mathrm{m}}}$

= 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:

$\displaystyle K_{g,dipole-dipole}=\frac{2\pi }{\frac{1}{\overline{AM}}-\frac{1}{\overline{AN}}-\frac{1}{\overline{BM}}+\frac{1}{\overline{BN}}}$ = $\displaystyle \frac{2\pi }{\frac{1}{9\mathrm{m}}-\frac{1}{6\mathrm{m}}-\frac{1}{6\mathrm{m}}+\frac{1}{3\mathrm{m}}}$ = 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 = $\frac{31.4\ \textrm{m}\ V\ \textrm{mV}}{2\ \textrm{mA}}$

solving for V yields V = 15.9 mV

ρa_dipoledipole = 250 ohm-m = $\frac{56.5\ \textrm{m}\ V\ \textrm{mV}}{2\ \textrm{mA}}$

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