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

current = 0.0027 A

Step-by-step explanation:

the resistivity of upper leg

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

Resistance of upper leg

`R = (\rho L)/(A)`

`= (\rho L)/(\pi R^2)`

`= (15.16 * 0.40)/(\pi [(0.12)/(2)]^2)`

= 551.27 ohm

current
`i = (V)/(R)`

`current = (1.5)/(551.27)`

current = 0.0027 A