#(20a7) Your help is kindly appreciated Consider a portion of a cell membrane th

Question

#(20a7)

Your help is kindly appreciated

Consider a portion of a cell membrane that has a thickness of 7.50 nm and 1.5

#(20a7) Your help is kindly appreciated Consider a portion of a cell membrane that has a thickness of 7.50 nm and 1.5 µm ? 1.5 µm in area. A measurement of the potential difference across the inner and outer surfaces of the membrane gives a reading of 86.5 mV. The resistivity of the membrane material is 1.30 x10^7 ? A.m (a) Determine the amount of current that flows through this portion of the membrane. A (b) By what factor does the current change if the side dimensions of the membrane portion is doubled? The other values do not change. decrease by a factor of 4increase by a factor of 4 increase by a factor of 8increase by a factor of 2decrease by a factor of 2

Explanation / Answer

A =(1.5*10^-6)^2 =2.25*10^-12 m^2

R = pL/A = (1.3*10^7)(7.5*10^-9)/(2.25*10^-12) =4.33*10^10 ohms

I=V/R =86.5*10^-3/4.33*10^10

I=1.996*10^-12 A

= 2.0x10-12 A    (rounded)

b)

S = So/2 =1.5/2 =0.75 um

A=(0.75*10^-6)^2 = 5.625*10^-13 m^2

R = pL/A = 1.3*10^7*7.5*10^-9/(5.625*10^-13) =1.73*10^11 ohms

I = V/R = 86.5*10^-3/1.73*10^11 = 4.99*10^-13 A = 5*10^-13 A

I1/I2 = [2.0x10-12 A ]/ [5*10^-13 A] = 4

I2 =(1/4)I1

so decreases by a factor of 4

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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