Question

Variations in the resistivity of blood can give valuable clues to changes in the blood's viscosity and other properties. The resistivity is measured by applying a small potential difference and measuring the current. Suppose a medical device attaches electrodes into a 1.5-mm-diameter vein at two points 5.0 apart.

What is the blood resistivity if a 8.9 V potential difference causes a 240 mu A current through the blood in the vein (in omega*m)?

Answer 1

Concepts and reason

Use the concept of resistance and ohms law to solve this problem.

First calculate the potential difference that is applied to the blood by using the concept of ohms law.

Later find the resistance of the blood vessel by using the relation between the resistivity and the resistance.

Finally find the resistivity of the blood by using the concept of resistance.

Fundamentals

The current passing through the conductor is directly proportional to the voltage between two points. This is called ohms law.

The expression for the ohms law is as follows:

$V = IR$

Here, V is the potential difference, I is the current, and R is the resistance.

The expression for the resistance is as follows:

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

Here,$\rho$is the resistivity of the blood, L is the separation of electrodes at two points, and A is the area of the vein.

The expression for the potential difference that is applied to the blood vessel is,

$V = IR$

Substitute 8.9 V for V, $240\;\mu {\rm{A}}$for I.

$8.9\;{\rm{V}} = \left( {240\;\mu {\rm{A}}\left( {\frac{{1\;{\rm{A}}}}{{{{10}^6}\;\mu {\rm{A}}}}} \right)} \right)R$ …… (1)

The expression for the resistance is,

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

Substitute 5.00 cm for L, and $\left( {\pi {r^2}} \right)$for A

$R = \frac{{\rho \left( {5.00\;{\rm{cm}}\left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)} \right)}}{{\left( {\pi {r^2}} \right)}}$

Here, r is the radius of the vein.

Substitute$\left( {\frac{d}{2}} \right)$for r

$R = \frac{{\rho \left( {5.00\;{\rm{cm}}\left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)} \right)}}{{\left( {\pi {{\left( {\frac{d}{2}} \right)}^2}} \right)}}$

Here, d is the diameter of the vein.

Substitute 1.5 mm for d

$\begin{array}{c}\\R = \frac{{\rho \left( {5.00\;{\rm{cm}}\left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)} \right)}}{{\left( {\pi {{\left( {\left( {\frac{{1.5\;{\rm{mm}}}}{2}} \right)\left( {\frac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}}} \right)} \right)}^2}} \right)}}\\\\ = \frac{{\rho \left( {5.00 \times {{10}^{ - 2}}\;{\rm{m}}} \right)}}{{1.7663 \times {{10}^{ - 6}}\;{\rm{m}}}}\\\\ = \left( {2.831 \times {{10}^4}} \right)\rho \\\end{array}$ …… (2)

The resistivity of the blood is,

$8.9\;{\rm{V}} = \left( {240\;\mu {\rm{A}}\left( {\frac{{1\;{\rm{A}}}}{{{{10}^6}\;\mu {\rm{A}}}}} \right)} \right)R$

Substitute (2) in (1) for R as follows:

$8.9\;{\rm{V}} = \left( {240\;\mu {\rm{A}}\left( {\frac{{1\;{\rm{A}}}}{{{{10}^6}\;\mu {\rm{A}}}}} \right)} \right)\left( {2.831 \times {{10}^4}} \right)\rho$

Rearrange the above equation for$\rho$.

$\begin{array}{c}\\\rho = \frac{{8.9\;{\rm{V}}}}{{\left( {240\;\mu {\rm{A}}\left( {\frac{{1\;{\rm{A}}}}{{{{10}^6}\;\mu {\rm{A}}}}} \right)} \right)\left( {2.831 \times {{10}^4}} \right)}}\\\\ = \frac{{8.9\;{\rm{V}}}}{{\left( {240 \times {{10}^{ - \;6}}\;{\rm{A}}} \right)\left( {2.831 \times {{10}^4}} \right)}}\\\\ = 1.31\;\Omega \cdot {\rm{m}}\\\end{array}$

Ans:

The resistivity of the blood is$1.31\;\Omega \cdot {\rm{m}}$.