Question

The conductive tissues of the upper leg can be modeled as a 40- cm-long, 12-cm-diameter cylinder of muscle and fat. The resistivities of muscle and fat are 13 Ω m and 25 Ω m, respectively. One person’s upper leg is 82% muscle, 18% fat.
What current is measured if a 1.5 V potential difference is applied between the person’s hip and knee?

Answer 1

Answer:

current = 0.0027 A

Explanation:

the resistivity of upper leg

$\rho = 0.82 (13) + 0.18(25) = 15.16 ohm . m$

Resistance of upper leg

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

   $= \frac{\rho L}{\pi R^2}$

  $= \frac{15.16 \times 0.40}{\pi [\frac{0.12}{2}]^2}$

  = 551.27 ohm

current$i = \frac{V}{R}$

$current = \frac{1.5}{551.27}$

current = 0.0027 A