Potential difference and current

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SUMMARY

The discussion focuses on calculating the current flowing through a cell membrane given its thickness, resistivity, and potential difference. The cell membrane is 8.0 nm thick with a resistivity of 1.3 x 107 Ω·m, and the potential difference is 75 mV. To find the current, the resistance is calculated using the formula R = ρ(L/A), where L is the thickness of the membrane and A is the area of the membrane (1.3 µm on a side). The correct application of these formulas leads to the determination of the current flowing through the membrane.

PREREQUISITES
  • Understanding of electrical resistivity and its units (Ω·m)
  • Familiarity with Ohm's Law (V = IR)
  • Knowledge of geometric calculations for area (A = L x W)
  • Basic concepts of cell membrane structure and properties
NEXT STEPS
  • Learn about the implications of electrical properties in biological membranes
  • Study the effects of varying resistivity on current flow in different materials
  • Explore advanced applications of Ohm's Law in biological systems
  • Investigate the role of ion channels in cellular electrical activity
USEFUL FOR

Students in biology or physics, electrical engineers, and researchers interested in biophysics or cellular physiology will benefit from this discussion.

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Homework Statement



A typical cell membrane is 8.0 nm thick and has an electrical resistivity of 1.3 10^7 m. If the potential difference between the inner and outer surfaces of a cell membrane is 75 mV, how much current flows through a square area of membrane 1.3 µm on a side?

Homework Equations



I was thinking to use R=p(L/A) and then V=IR to solve for the current, but I wasn't sure what to use for numbers. I know that V=75 mV and p=1.3 x 10^7, but would L=1.3 micrometers and A=L x 8.0 nm ?
 
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First find the resistance, R, by using the formula you wrote down: R=p(L/A). You know the length, L, which is like the thickness. The area, A, is also given and you know the resistivity of the specimen. Whats the problem?
 

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