Consider a portion of a cell membrane that has a thickness of 7.50 nm and 1.1 m

Question

Consider a portion of a cell membrane that has a thickness of 7.50 nm and 1.1 m x 1.1 m in area. A measurement of the potential difference across the inner and outer surfaces of the membrane gives a reading of 97.6 mV. The resistivity of the membrane material is 1.30 x 107 m (a) Determine the amount of current that flows through this portion of the membrane. 1.43e-12X What is the resistance of the membrane? Which side of the membrane contributes to its resistance? Which sides of the membrane contribute to the the membrane? A cross sectional area of (b) By what factor does the current change if the side dimensions of the membrane portion is halved? The other values do not change. decrease by a factor of 4 decrease by a factor of 2 increase by a factor of 2 increase by a factor of 4 decrease by a factor of 8

Explanation / Answer

ressitance of the membrane R = rho*t/A

t = thickness

A = area of membrane = 1.1*10^-6*1.1*10^-6 m^2 = 1.21*10^-12 m^2

rho = resistivity = 1.3*10^7

from ohms law

current I = V/R = V*A/(rho*t)

current I = (97.6*10^-3*1.21*10^-12)/(1.3*10^7*7.5*10^-9) = 1.211*10^-12 A

============================================

If sides are halved

new area A1 = A/4

Inew = V*A1/(rho*t) = I/4

decreases by a factor of 4

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.