Calculating Current in a Mixed Tissue Conductor: A Weighted Average Approach

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SUMMARY

The discussion focuses on calculating the current in a mixed tissue conductor, specifically a 35-cm-long, 16-cm-diameter cylinder representing a person's upper leg composed of 82% muscle and 18% fat. The resistivities are 13 Ωm for muscle and 25 Ωm for fat. The correct approach involves calculating the weighted average resistivity based on the volume fractions of the tissues, rather than treating them as separate resistors in parallel. The final current calculation, given a 1.9V potential difference, requires careful consideration of the resistivity contributions from both tissues.

PREREQUISITES
  • Understanding of electrical resistance and Ohm's Law
  • Familiarity with the concept of resistivity in materials
  • Knowledge of geometric calculations for cylinders
  • Ability to perform weighted average calculations
NEXT STEPS
  • Study the derivation of the formula R = ρL/A for cylindrical conductors
  • Learn about the implications of mixing materials with different resistivities
  • Explore the concept of effective resistivity in composite materials
  • Investigate the application of weighted averages in electrical engineering contexts
USEFUL FOR

This discussion is beneficial for electrical engineers, biomedical engineers, and students studying bioelectrical impedance and tissue conductivity. It provides insights into the complexities of modeling mixed materials in electrical applications.

ricebowl07
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The conductive tissues of the upper leg can be modeled as a 35-cm-long, 16-cm-diameter cylinder of muscle and fat. The resistivities of muscle and fat are 13 Ωm and 25 Ωm, respectively. One person's upper leg is 82% muscle, 18% fat. The resistivity of the whole leg will be a weighted average of the resistivities of muscle and fat. This is not the same as treating the system as two resistors (one made of muscle and the other of fat) in parallel; the muscle and fat are mixed together, not separate side by side.

Part A

What current is measured if a 1.9V potential difference is applied between the person's hip and knee?
Express your answer to two significant figures and include the appropriate units.

Relevant Equations

R = ρL/A
I = ΔV/R
A = ∏r^(2)

I kept getting 7.67 * 10^(-3) A for my answer. My attempt to solve this problem attached to this post.
 

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Hello rice and welcome to PF.

I take it your question is: why is this wrong ?
Teachers instructions clearly state this is not the same as two resistors in parallel.

Teacher tells you to weigh the resistivities with percentage. Not the same as weighing 1/resisitivity by percentage !
 
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Just in case BvU's response is not completely clear to you...
You have added up the currents. Currents add up if the conductors are in parallel.
Similarly, you have taken the 18% and 82% as percentages of cross-sectional area. You should be taking them as percentages of length.
 
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