Question

The conductivity of pure water in a conductivity cell with electrodes of cross-sectional area $4c{m}^2$ and $2cm$ apart is $8\times {10}^{-1}Sc{m}^{-1}$.

i) What is the resistance of the conductivity cell?

ii) What current would flow through the cell under an applied difference of $1volt$?

Answer 1

$K\left(\right)=8\times {10}^{-1}5c{m}^{-1}$

$K=\frac{1}{R}\times \frac{l}{A}$

$l=2cm$

$A=4c{m}^2$

$8\times {10}^{-1}=\frac{1}{R}\times \frac{2}{4}$

$8\times {10}^{-1}=\frac{1}{R}\times \frac{1}{2}$

$R=\frac{1}{2\times 8\times {10}^{-1}}=\frac{1}{16}\times 10$

$R=0.0625\times 10$

$R=0.625\Omega$

Acc. to ohm's law

$V=IR$

$I=\frac{V}{R}=\frac{1}{0.625}$

$I=1.6A$