Variations in the resistivity of blood can give valuable clues about changes in various properties of the blood. Suppose a medical device attaches two electrodes into a $1.5mm$diameter vein at positions role="math" localid="1649141775053" $5.0cm$apart. What is the blood resistivity if a $9.0V$potential difference causes a$230\mu A$ current through the blood in the vein?

Two electrodes diameter$d=1.5\mathrm{mm}$

Distance$=5.0\mathrm{cm}$

Potential differencelocalid="1649142178730" $=9.0V$

Current $I=230\mu A$

We can use following equation to find the length:

$I=\frac{\Delta V}{R}$

Where $R$is resistance. This yields,

$R=\frac{\Delta V}{I}$

Here $\Delta V$is potential difference and $I$is the current

Calculate the resistivity,

$R=\frac{\rho L}{A}$

$A$, is the area of the cross section of the vein,

$L$is the length of vein and $\rho$is resistivity of vein.

The expression for the resistivity is,

$\frac{\Delta V}{I}=\frac{\rho L}{A}$

$\rho =\frac{\Delta VA}{IL}$

The diameter of blood vein is $1.5\mathrm{mm}$.

Convert units of diameter from $\mathrm{mm}$to $\mathrm{m}$

$d=(1.5\mathrm{mm})\left(\frac{1\mathrm{m}}{1000\mathrm{mm}}\right)$

$=1.5\times {10}^{-3}\mathrm{m}$

Radius of the vein is

$r=\frac{d}{2}$

$=\frac{1.5\times {10}^{-3}\mathrm{m}}{2}$

$=0.00075\mathrm{m}$

Area of blood vein is

$A=\pi r^2$

$=\pi (0.00075\mathrm{m}{)}^2$

$=1.767\times {10}^{-6}{\mathrm{m}}^2$

The two electrodes are a length, L, of $5.0\mathrm{cm}$

Convert units of length from $\mathrm{cm}$to $\mathrm{m}$

$L=(5.0\mathrm{cm})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$

$=5.0\times {10}^{-2}\mathrm{m}$

$\rho =\frac{(9.0\mathrm{V})\left(1.767\times {10}^{-6}{\mathrm{m}}^2\right)}{\left(230\times {10}^{-6}\mathrm{A}\right)\left(5.0\times {10}^{-2}\mathrm{m}\right)}$

$\rho =1.4\Omega \mathrm{m}$

Therefore, the blood resistivity if a $9.0\mathrm{V}$potential difference causes a $230\mathrm{mA}$current through the blood in the vein is $1.4\Omega \mathrm{m}$.