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 {\rm cm} 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

Answer:

The resistivity is 1.31 Ωm

Explanation:

Diameter = 1.5 mm

Potential difference = 8.9 V

Current = 240 μA

length = 5.0 cm

We need to calculate the resistivity

Using ohm's law

$V= I\times R$

We know that,

Formula of resistance

$R = \dfrac{\rho l}{A}$

Put the value of R in ohm's law

$V=I\times\dfrac{\rho l}{A}$

Where, $\rho$ = resistivity

A = area

I = length

Put the value into the formula

$8.9=240\times10^{-6}\times\dfrac{\rho\times5.0\times10^{-2}}{\pi\times(0.75\times10^{-3})^2}$

$\rho=\dfrac{8.9\times\pi\times(0.75\times10^{-3})^2}{240\times10^{-6}\times5.0\times10^{-2}}$

$\rho=1.31\ \Omega m$

Hence, The resistivity is 1.31 Ωm

Answer 2

The resistivity is 1.31 Ωm.

What is Resistivity?

This is a property which helps to measure how a material strongly resists electric current.

Parameters

Diameter = 1.5 mm

Potential difference = 8.9 V

Current = 240 μA

Length = 5.0 cm

V = IR

Resistance(R) = ρl/A

V = I × ρl/A

where Where, ρ = resistivity, A = area and I = length

Substitute the values into the equation

8.9 = 240× 10⁻⁶ ₓ ρₓ 5 ₓ 10⁻² / (π ₓ ((0.75 ₓ 10⁻³))²

Resistivity is 1.31 Ωm

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