Question

A typical cell membrane is 8.15 nm thick and has an electrical resistivity of 1.35E+7 Ω·m. If the potential difference between the inner and outer surfaces of a cell membrane is 78.2 mV, how much current flows through a square area of membrane 1.07 μm on a side? Suppose the thickness of the membrane is quadrupled, but the resistivity and potential difference remain the same. How much current flows through the same area of membrane now?

Answer 1

here by using the ohm's law

I = V / R

and R = rho * L / A

I = V * A / rho * L

then by putting the values

I = 78.2 * 10^-3 * (1.07 * 10^-3)^2 / ( 1.35 * 10^7 * 8.15 * 10^-9)

I = 8.13 * 10^-7 amp

quadrupled the thickness the current will be 1/4 of the current