Calculation of true thickness of the bed

Gate_2019

82) In an electrical resistivity imaging survey, Axial Dipole-dipole array is placed over an in-homogeneous structure. The centers of current and potential dipoles are separated by a distance of 100 m. The length of each dipole is 10 m. If 5 Amp current flows through the subsurface and 50 mV potential difference is measured across the potential dipole then apparent resistivity will be ______ohm-meters. (Round off to the nearest integer) (Use pi =3.14)

(Thanks to Chandra Sekhar, ANU)

Solution:

In Axial Dipole- dipole array the apparent resistivity is 

$\rho_{a}=\pi \frac{V}{I} \frac{R(R^{2}-a^{2})}{a^2}---(1)$

To simplify above formula, 

 

$\rho_{a}=(\frac{\triangle V}{I})(\frac{R^{3}}{a^{2}}-R)\pi$

Given that R = 100m

a = 10m

I = 5 Amp

V = 50 mV= 0.05 V

By substituting the above formulas in the equation 1 we get 

 

$\rho_{a}=(\frac{\triangle V}{I})(\frac{R^{3}}{a^{2}}-R)\pi$ 

 

$\rho_{a}=(\frac{0.05}{5})(\frac{100^{3}}{10^{2}}-100)(3.14)$

$\rho_{a}=(0.01)(\frac{1000000}{100}-100)(3.14)$

$\rho_{a}=(0.01)(10000-100)(3.14)$

$\rho_{a}=(0.01)(9900)(3.14)$

$\rho_{a}=310.86$ ohm-meter.

Therefore the apparent resistivity $\rho_{a}=310.86$ ohm-meter.

Reference: http://fsc.scuegypt.edu.eg/attach/2/ELECTRICAL%20GEOPHYSICS.pdf