GATE-2018 (24) :Wyllie – Time average equation

 

64) In a time domain(T-D) induced polarization experiment with a steady voltage of 10 mV during the current flow interval, the voltage drop at after the current cut-off is given by v(t)=4.0 e^-0.3t. The Chargeability after current cut-off between t₁ =1s and t₂ = 4s is ____mS

 (Thanks to Chandrasekhar, ANU)

Solution:

The chargeability for the time domain IP measurement is defined as

$M=\frac{1}{V_{c}}\int_{t_{1}}^{t_{2}}V(t) dt $

 Where V(t) and V_c have the same units and the chargeability M is in milliseconds.

 

 Given that Vc is steady voltage= 10mV

            V(t) is voltage decay = 4.0 e^-0.3t

            Then the Chargeability $M=\frac{1}{V_{c}}\int_{t_{1}}^{t_{2}}V(t) dt $

 

$M=\frac{1}{10}\int_{1}^{4}4 e^{-0.3t} dt $

 

$M=\frac{4}{10}\int_{1}^{4} e^{-0.3t} dt $

 

$M=\frac{2}{5}\left\{\frac{1}{-0.3}\left[e^{-0.3t}\right]_1^4 \right\}$

 

$M=\frac{-2}{5\times 0.3}\left[e^{-0.3\times4}-e^{-0.3\times1}\right]$

 

$M=\frac{-2}{5\times 0.3}\left[0.3-0.74\right] dt $

 

$M=\frac{-2\times 0.44}{5\times 0.3} $

 

$M=\frac{0.88}{1.5}$

 

$M= 0.586 seconds$

$M= 586 milli seconds$